how to write slope intercept equation line passing through poin t (4,7) parallel to another equation

The point (2,5) satisfies only one of the two inequalities in the system, it is not included in the solution area for the system.

To see if the point (2,5) is in the system's solution area, we need to see if it satisfies both inequalities in the system:

4x + 2y ≤ 16

4(2) + 2(5) ≤ 16

8 + 10 ≤ 16

18 ≤ 16

Because this inequality is false, point (2,5) fails to satisfy the first inequality.

x + y ≥ 4

2 + 5 ≥ 4

7 ≥ 4

Because this inequality holds, the point (2,5) satisfies the second inequality.

Because point (2,5) satisfies only one of the system's two inequalities, it is not included in the system's solution area. Landon cannot, therefore, spend $16 on two sandwiches and five bags of chips to feed his four teammates.

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The mass and volume of two metals is 93.98g, 165.56g and 39.37cm3,81.89cm3.

Suppose that a finite amount of substance is there having its properties as:

mass of substance = m kg

density of substance = d kg/m³

volume of that substance = v m³

Then, they are related as:

[tex]d = \dfrac{m}{v}[/tex]

Given that;

Mass of 100cm3 of metal = 254g

Mass of b volume with 37cm3 and 64cm3 is 100g and 208g.

Now,

m=pv

254=p*100

p=2.54g/cm3

m=2.54v

Part A;

V1 =37cm3

V2=64cm3

m1= 2.54 x 37= 93.98g

m2= 2.54 x 64= 165.56g

Part B:

V1=m1/2.54 = 100/2.54 = 39.37cm3

V2=208/2.54 = 81.89cm3

Therefore, the volumes will be 39.37cm3 and 81.89cm3

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Answer:

Step-by-step explanation:

translation:  (x+4, y-6)

original    image

A(5,4)     A'(9,-2)

B(-9,6)    B'(-5,0)

C(-6,2)   C'(-2,-4)

a. The two lines intersect at the point (-1/2, 8, 3/2).

b. The equation of the plane that passes through the two lines is -2y - 4z + 20 = 0.

a) To find the point where the two lines intersect, we need to solve the system of equations:

x = -2 + t

y = 3 + 2t

z = 4 - t

x = 3 - t

y = 4 - 2t

z = t

Equating x from the two equations, we get:

-2 + t = 3 - t

2t = 5

t = 5/2

Substituting t in either equation, we get:

x = -2 + 5/2 = -1/2

y = 3 + 2(5/2) = 8

z = 4 - 5/2 = 3/2

Therefore, the two lines intersect at the point (-1/2, 8, 3/2).

b) To obtain the equation of the plane that passes through the two lines, we first find the direction vectors of the lines. These are given by the coefficients of t in the equations:

Line 1: (-2, 3, 4) + t(1, 2, -1)

Line 2: (3, 4, 0) + t(-1, -2, 1)

The direction vectors are (1, 2, -1) and (-1, -2, 1), respectively.

The normal vector of the plane can be found by taking the cross-product of these two direction vectors:

(1, 2, -1) × (-1, -2, 1) = (0, -2, -4)

Now we use the point (-1/2, 8, 3/2) that we found in part (a) and the normal vector (0, -2, -4) to write the equation of the plane in the point-normal form:

0(x + 1/2) - 2(y - 8) - 4(z - 3/2) = 0

Simplifying, we get:

-2y - 4z + 20 = 0

Therefore, the equation of the plane that passes through the two lines is -2y - 4z + 20 = 0.

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The question is -

a) determine the point where the two lines

x=−2+t, y=3+2t, z=4−t and x=3−t, y=4−2t, z=t intersect.

b) use your results from part (a) to obtain the equation of the plane that passes through the two lines.

1) Multiply both sides of the equation by 2:

2(A)=2[1/2h(b+B)

2A=2/2h(b+B)

2A=1h(b+B)

2A=h(b+B)

2) Divide both sides of the equation by (b+B):

(2A)/(b+B)=[h(b+B)]/(b+B)

2A/(b+B)=h

h=2A/(b+B)

Answer: [tex]h=2\frac{A}{(b+B)}[/tex]

hope this helped!!

The minimum value of the function P over the feasible region is -3, which occurs at the point (1,2).

The expression that established the relationship between the dependent variable and independent variable is referred to as a function. In the function as the value of the independent variable varies the value of the dependent variable also varies.

To find the minimum value of the function P = x - 2y over the feasible region determined by the vertices (1,2), (7,3), and (5,1), we need to evaluate the function at each of these points and find the lowest value.

P(1,2) = 1 - 2(2) = -3

P(7,3) = 7 - 2(3) = 1

P(5,1) = 5 - 2(1) = 3

Therefore, the minimum value of the function P over the feasible region is -3, which occurs at the point (1,2).

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Answer:

Quadrant IV

Step-by-step explanation:

cos(∅) > 0 means cos (∅) is positive

tan (∅) < 0 means tan (∅) is negative

thus,

∅ lies in Quadrant IV

After the final audit of sales results, the West Region needs to have their sales updated for Q3 to $15,740 thus is obtained by clicking D3. Option A is correct answer.

A row and a column are joined to create a cell. It is, in other words, where a row and a column meet. In general, rows are labelled with numbers like 1, 2, and 3, whereas columns are labelled with letters like A, B, and C. Each cell has a unique name or cell address based on its column and row.

A cell range in Microsoft Excel is a collection of cells. It may appear in a formula.

After the final audit of sales statistics, the West Region's sales for Q3 must be revised to $15,740, so the cell that would be clicked is D3.

Hence, after the final audit of sales results, the West Region needs to have their sales updated for Q3 to $15,740 thus is obtained by clicking D3.

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Answer:

[tex]A)81x^{-12}y^{20}\\\\B)5\\\\C) x^\frac{2}{3}[/tex]

Step-by-step explanation:

[tex]A) (3x^{-3}y^5)^4=(3)^4 \,\cdot\, (x^{-3})^4\,\cdot\,(y^5)^4=81x^{-12}y^{20}\\\\B) 5x^0=5(1)=5\\\\C) \frac{x^{\frac{5}{6} }}{x^{\frac{1}{6} }} =x^{\frac{5}{6} -\frac{1}{6} }=x^{\frac{5-1}{6} }=x^{\frac{4}{6} }=x^\frac{2}{3}[/tex]

At y there is a minimum for single-slit diffraction and a maximum for double-slit interference, as noted in the question.

For single-slit diffraction, the first minimum occurs when

sin θ = λ/a.

and double-slit interference maxima occur when

(m) λ / d = sin θ --------- (2)

The first diffraction minimum for single-slit diffraction and the fourth double-slit interference maximum (m = 4) occur at the same position y, as seen in the figure below.

On the left of the screen, the dashed curve is due to single-slit interference. On the right of the screen, the dashed curve is due to double-slit interference. The positive direction is reflected on each of the two sides of the screen and the screen position is zero amplitude.

Since the single-slit diffraction minimum masks the fourth double-slit interference maxima, one must estimate where the fourth double-slit maxima occur using the spacing between the double-slit interference pattern appearing on the right-hand side of the display screen, as seen in the figure above.

Using Eq. 1 and 2, we have

sin θ / λ = 1/ a = (m) 1/ d

d/ a = (4)

= 4.

Therefore, At y there is a minimum for single-slit diffraction and a maximum for double-slit interference, as noted in the question.

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The correct question is:

Consider the setup of double-slit experiment in the schematic drawing below. One of the double-slit interference minima is located at the first single-slit diffraction minimum. Determine the ratio d/a ; i.e., the slit separation d compared to the slit width a. Use a small angle approximation; e.g., sin θ ≈ tan θ ≈ θ and cos θ ≈ 1.

1. d/a = 13/2

2. d/a = 6

3. d/a = 15/2

4. d/a = 5

5. d/a = 7

6. d/a = 9/2

7. d/a = 11/2

8. d/a = 17/2

9. d/a = 4

10. d/a = 8

A box plot is a special type of diagram that shows the quartiles in a box and the line extending from the lowest to the highest value.

What is Box plot?

Box plots, also known as box-and-whisker plots or box-whisker plots, provide a clear pictorial representation of the distribution of the data. They also demonstrate how remote the extreme numbers are from the majority of the data. Five values are used to create a box plot: the minimum value, the first quartile, the median, the third quartile, and the maximum value. These numbers are used to gauge how closely other data points adhere to them. Use a horizontal or vertical number line along with a rectangular box to create a box plot. The axis' ends are identified by the smallest and greatest data values. The first quartile designates one end of the box, while the third quartile designates the opposite end.

The box plot distribution will demonstrate how skewed, tightly packed, and symmetrical the data are.

In the box and whisker plot:

The box's upper and lower quartiles serve as its ends, allowing it to cross the interquartile range. The vertical line inside the box denotes the median, and the two lines outside the box serve as its whiskers, extending to the highest and lowest observations.

Hence, A box plot is a special type of diagram that shows the quartiles in a box and the line extending from the lowest to the highest value.

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When interpreting pairwise confidence intervals to ANOVA test with either Fisher, Tukey or Bonferroni methods, we determine that there are differences between two treatments whenever  F. the interval does NOT include zero.

Using the given techniques, the pairwise confidence intervals and test can be easily generated. These can state whether the disparity among the two groups' means is statistically significant. To compare the means of two groups, pairwise confidence intervals using Fisher, Tukey, or Bonferroni procedures are created as a result of ANOVA testing.

The disparity is statistically significant at the selected level of confidence if the confidence interval for the difference among the means of the two groups does not include zero. This suggests that the two therapies being compared significantly differ from one another. If the confidence interval includes zero, the variation is not statistically significant, and we should not claim that there is a purposeful disparity among the two treatments.

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The value of angle x based on the information will be 131°.

Alternate interior angles are the angles formed when a transversal intersects two coplanar lines.

The two marked angles are on opposite sides of the transversal, so they are called "alternate" angles. They are both between the parallel lines, so they are called "interior" angles.

In this geometry, alternate interior angles are congruent. They have the same measure. x° = 131°.

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The algorithm to write the binary representation of the positive integer "64" is explained below and the binary representation of 64 is 1000000 .

Step(i) : 64 divided by 2 gives a quotient of 32 with a remainder of 0.

Step(ii) : 32 divided by 2 gives a quotient of 16 with a remainder of 0.

Step(iii) : 16 divided by 2 gives a quotient of 8 with a remainder of 0.

Step(iv) : 8 divided by 2 gives a quotient of 4 with a remainder of 0.

Step(v) : 4 divided by 2 gives a quotient of 2 with a remainder of 0.

Step(vi) : 2 divided by 2 gives a quotient of 1 with a remainder of 0.

Step (vii) : 1 divided by 2 gives a quotient of 0 with a remainder of 1.

So , we observe that  the remainders, read from bottom to top, are 1000000.

Therefore, the binary representation of "64" is 1000000 in binary.

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The given question is incomplete , the complete question is

Find the binary representation of the positive integer "64" by working through an algorithm by hand .

Answer:

6 Shirts

Step-by-step explanation:

if you buy 1 pair of jeans you can buy 6 pairs of shirts since 6 x 11 = 66

66 + 40 = 106

Nine cuboidal boxes and eight cuboidal boxes can each hold 153 and 136 toy cars, respectively.

How is the average calculated?

The ratio of the sum of all provided observations to the total number of observations is the arithmetic mean, which is another name for the average formula. So, any sample of data may have its arithmetic mean determined using the average formula.

What does arithmetic mean?

The mean or arithmetic average are terms that are frequently used to refer to the arithmetic mean. It is determined by adding up all the numbers in a given data collection, then dividing that total by the number of items in the data set. For uniformly distributed integers, the middle number is the arithmetic mean (AM).

Given: There are two storage options for 288 toy cars: 16 and 18 cuboidal boxes.

supposing there were 288 toy cars stored in 16 toy cars.

There are 18 toy cars in a box of 288 (288 / 16) toy cars.

if 288 toy cars were housed in 18 toy cars.

There are 16 toy cars in a box that holds 288 toy cars.

Average: 18 + 16 / 2 = 34 / 2 = 17 toy cars.

Nine boxes of toy cars include 153 toy cars (9 * 17).

eight boxes of toy cars include 136 toy cars (8 * 17)

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answer :

P(t) = 3000 / (1 + 29*e^(-0.637t))

steps

logistic function or logistic curve is a common S-shape curve (sigmoid curve) with equation:

P(t) = K / (1 + A*e^(-rt))

1. There is a type of animal or plant

species that started with 100

individuals.

2. The environment can support up to

3000 individuals, which is called the

"carrying capacity".

3. After 4 years, the population has

grown to 600 individuals.

4. We need to find a "logistic function"

that can show how the population

changes over time.

A species with an initial population of 100 is growing in an environment where the carrying capacity is 3000.

After 4 years, the population is up to 600.

We want to find the logistic function that models this population as a function of time.

The logistic function is a model of population growth that takes into account the carrying capacity of the environment. It is given by the formula:

P(t) = K / (1 + A e^(-r(t-t0)))

Where:

P(t) is the population size at time t

K is the carrying capacity of the environment

r is the growth rate of the population

t0 is the time at which the population starts to grow

A is a constant that determines the initial population size

To find the logistic function that models the population described in the problem, we need to determine the values of K, r, t0, and A.

We know that the initial population size is 100, so A = 100.

After 4 years, the population is up to 600, so P(4) = 600. We can use this information to solve for r:

600 = K / (1 + A e^(-r(4-t0)))

600 = K / (1 + 100 e^(-4r))

600(1 + 100 e^(-4r)) = K

K = 600 + 60000 e^(-4r)

We also know that the carrying capacity is 3000, so K = 3000.

3000 = 600 + 60000 e^(-4r)

2400 = 60000 e^(-4r)

0.04 = e^(-4r)

ln(0.04) = -4r

r = ln(0.04) / -4

r ≈ 0.693

Now we can use the value of r to solve for t0. We know that the initial population size is 100, so we can use this to find the value of A at t0:

100 = K / (1 + A)

100 = 3000 / (1 + A)

1 + A = 30

A = 29

We can now use this value of A to solve for t0:

100 = 3000 / (1 + 29 e^(-r(t0)))

1 + 29 e^(-r(t0)) = 30

e^(-r(t0)) = 1/29

ln(1/29) = -r(t0)

t0 = ln(1/29) / -r

t0 ≈ 2.48

Now we have all the values we need to write the logistic function:

P(t) = 3000 / (1 + 29 e^(-0.693(t-2.48)))

This is the logistic function that models the population as a function of time. It predicts that the population will grow exponentially at first, but then level off as it approaches the carrying capacity of the environment.

We can model the population growth of the species using the logistic equation:

dP/dt = rP(1 - P/K)

where P is the population size, t is time, r is the growth rate, and K is the carrying capacity.

To find the logistic function that models this population as a function of time, we need to determine the values of r and K. We can use the information given in the problem to solve for these values:

The initial population size is 100, so P(0) = 100.

The carrying capacity is 3000.

After 4 years, the population size is 600, so P(4) = 600.

Using these values, we can solve for r and K:

P(t) = K / (1 + A*e^(-rt))

where A is a constant determined by the initial population size, and e is the base of the natural logarithm.

From the initial population size, we know that:

A = (K - P(0)) / P(0) = (3000 - 100) / 100 = 29

We can use the population size after 4 years to solve for r and K:

600 = K / (1 + 29*e^(-4r))

Multiplying both sides by the denominator:

600 + 29600e^(-4r) = K

Substituting K = 3000:

3000 = 600 + 29600e^(-4r)

Dividing both sides by 600:

5 = 29*e^(-4r)

Taking the natural logarithm of both sides:

ln(5) = ln(29) - 4r

Solving for r:

r = (ln(29) - ln(5)) / 4

r ≈ 0.637

Now that we have r and K, we can plug them into the logistic equation to get the logistic function:

P(t) = 3000 / (1 + 29*e^(-0.637t))

This function models the population of the species as a function of time, where P(t) is the population size at time t.

ChatGPT

From the given data, the company can produce 500 units per month at a minimum cost of $140,000.

To minimize the total monthly cost of production while producing 500 units per month, we can use the method of Lagrange multipliers. Let L(x, y, λ) be the Lagrangian function:

L(x, y, λ) = 2x² + xy + 4y² + 1600 + λ(500 - x - y)

We need to find the values of x and y that minimize L(x, y, λ), subject to the constraint that 500 units are produced per month. Taking partial derivatives of L with respect to x, y, and λ and setting them to zero, we get:

∂L/∂x = 4x + y - λ = 0

∂L/∂y = x + 8y - λ = 0

∂L/∂λ = 500 - x - y = 0

Solving these equations simultaneously, we get:

x = 200

y = 150

λ = 26/3

Therefore, to minimize the total monthly cost of production while producing 500 units per month, the company should produce 200 units at Factory X and 150 units at Factory Y.

To find the minimum cost, we substitute the values of x and y into the joint cost function:

C(x, y) = 2x² + xy + 4y² + 1600

C(200, 150) = 2(200)² + (200)(150) + 4(150)² + 1600

C(200, 150) = 140,000

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Answer:

? = 77°

Step-by-step explanation:

the 3 angles in a triangle sum to 180° , that is

? + 35° + 68° = 180°

? + 103° = 180° ( subtract 103° from both sides )

? = 77°

Answer:77

Step-by-step explanation:

180-103=77

Answer:

Step-by-step explanation:

The expression for the team time is:

[tex]11\frac{3}{5} =3\frac{3}{10} +2\frac{4}{5} +2\frac{1}{10} +X\\[/tex]

Where X= Cindy's Time

Isolating X from the equation:

[tex]X=11\frac{3}{5}-(3\frac{3}{10} +2\frac{4}{5}+2\frac{1}{10})\\\\ X=11\frac{3}{5}-(8\frac{1}{5})\\ \\ X=3\frac{2}{5}[/tex]

Cindy's time was [tex]3\frac{2}{5}[/tex] minutes.

The probability density function is of E(Y1) and E(Y2) is 1/6 and variance is 1/18.

To find the expected value E(Y1+Y2), we can use the linearity of expectation and the fact that the expected value of a constant is that constant:

E(Y1+Y2) = E(Y1) + E(Y2)

To find E(Y1) and E(Y2), we need to integrate y1 and y2 respectively over their joint probability density function:

E(Y1) = ∫∫ y1 f(y1,y2) dy1 dy2

= ∫0^1 ∫0^(1-y1) 2y1 dy2 dy1

= ∫0^1 2y1(1-y1)/2 dy1

= ∫0^1 y1-y1^2 dy1

= [y1^2/2 - y1^3/3] from 0 to 1

= 1/6

Similarly,

E(Y2) = ∫∫ y2 f(y1,y2) dy1 dy2

= ∫0^1 ∫0^(1-y2) 2y2 dy1 dy2

= ∫0^1 2y2(1-y2)/2 dy2

= ∫0^1 y2-y2^2 dy2

= [y2^2/2 - y2^3/3] from 0 to 1

= 1/6

Therefore, E(Y1+Y2) = E(Y1) + E(Y2) = 1/6 + 1/6 = 1/3.

To find the variance V(Y1+Y2), we can use the formula:

V(Y1+Y2) = E((Y1+Y2)^2) - [E(Y1+Y2)]^2

To find E((Y1+Y2)^2), we need to integrate (y1+y2)^2 over their joint probability density function:

E((Y1+Y2)^2) = ∫∫ (y1+y2)^2 f(y1,y2) dy1 dy2

= ∫0^1 ∫0^(1-y1) (y1+y2)^2 2 dy2 dy1

= ∫0^1 [(2/3)y1^3 + y1^2 + (1/3)y1] dy1

= 5/18

Therefore, V(Y1+Y2) = E((Y1+Y2)^2) - [E(Y1+Y2)]^2 = 5/18 - (1/3)^2 = 5/18 - 1/9 = 1/18

The variance is 1/18.

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According to the information, he gets 6 pieces in total after cutting his boards.

To find the number of pieces Nathan gets, we must divide the total length of the boards by the length of the pieces. Below is the procedure:

According to the above, from one table he gets 4 2-feet pieces, from the other he gets 2 4-feet pieces. Additionally, in total he would have 6 pieces.

Note: This question is incomplete because there is some information missing. Here is the complete information:

Nathan has two 8-foot boards. He cuts one board into 2-foot pieces, and he cuts the other board into 4-foot pieces.

How many pieces does he get?

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The area of the figure is calculated as: 536 km².

The area of the figure given can be found by decomposing the figure into three rectangles.

Area of rectangle 1:

Length = 15 km

Width = 11 km

Area = length * width = 15 * 11 = 165 km²

Area of rectangle 2:

Length = 6 + 11 = 17 km

Width = 11 km

Area = 17 * 11 = 187 km²

Area of rectangle 3:

Length = 23 km

Width = 8 km

Area = 23 * 8 = 184 km²

The area of the figure = 165 + 187 + 184 = 536 km²

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The probability of drawing the ace of spades is 1/52 for the first card and 1/51 for the second card. P(B|As) ≈ 0.1176.

The likelihood of drawing the trump card on the primary draw is for sure 1/52. Be that as it may, when the principal card is drawn, there are presently 51 cards remaining, and only one of them is the trump card. In this way, the likelihood of drawing the trump card on the subsequent draw, considering that the principal card isn't the trump card, is 1/51.

With regards to the issue, the occasion As is characterized as the trump card being picked, whether or not it is the first or second card. Since the issue determines that the two cards are drawn without substitution, there are two cases to consider: either the trump card is drawn first, with likelihood 1/52, or it is drawn second, considering that the principal card isn't the trump card, with likelihood (51/52)*(1/51) = 1/52. Accordingly, the likelihood of occasion Similar to the amount of these two probabilities, which is 1/52 + 1/52 = 1/26.

Utilizing Bayes' recipe, we have:

P(B|As) = P(As|B) * P(B)/P(As)

We previously determined P(As) to be 1/26. To ascertain P(As|B), note that in the event that the two cards are aces, the trump card should be one of them, so P(As|B) = 1. At last, to compute P(B), note that there are (4 nCr 2) = 6 methods for picking two aces out of the four in the deck, and there are (52 nCr 2) = 1326 methods for picking two cards out of the deck without substitution, so P(B) = 6/1326 = 1/221.

Subbing these qualities into Bayes' recipe, we get:

P(B|As) = 1 * (1/221)/(1/26) = 26/221 ≈ 0.1176

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The area of the triangle ΔABC is given by A = √275 units²

A triangle is a plane figure or polygon with three sides and three angles.

A Triangle has three vertices and the sum of the interior angles add up to 180°

Let the Triangle be ΔABC , such that

∠A + ∠B + ∠C = 180°

The area of the triangle = ( 1/2 ) x Length x Base

For a right angle triangle

From the Pythagoras Theorem , The hypotenuse² = base² + height²

if a² + b² = c² , it is a right triangle

if a² + b² < c² , it is an obtuse triangle

if a² + b² > c² , it is an acute triangle

Given data ,

Let the area of the triangle ΔABC be A

Now , the measure of side BD = 5

The measure of side ED = 11

So , the measure of side EB = 11 - 5 = 6

The measure of side AD = 10

And , the measure of side CE = 4

For a right angle triangle

From the Pythagoras Theorem , The hypotenuse² = base² + height²

From the triangle ΔABD ,

The measure of side AB = √ ( AD )² - ( BD )²

The measure of side AB = √ ( 100 - 25 )

The measure of side AB = √75 units

And , from the triangle ΔBCE

The measure of side CB = √ ( EB )² - ( CE )²

The measure of side CB = √ ( 36 - 16 )

The measure of side CB = √20 units

And , from the triangle ΔABC

The measure of side AC = √ ( AB )² - ( CB )²

The measure of side AC = √ ( 75 - 20 )

The measure of side AC = √55 units

And , the area of triangle ΔABC = ( 1/2 ) x AC x BC

The area of triangle ΔABC = ( 1/2 ) x √55 x √20

The area of triangle ΔABC = ( 1/2 ) x √1100

The area of triangle ΔABC , A = ( 1/2 ) x 2√275

The area of triangle ΔABC , A = √275 units²

Hence , the area of triangle ΔABC is √275 units²

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The number of marshmallows and graham crackers will Janna use to make 13 s'mores is 26 and 36, the correct option is 36.

Algebra is the study of abstract symbols, while logic is the manipulation of all those ideas.

The acronym PEMDAS stands for Parenthesis, Exponent, Multiplication, Division, Addition, and Subtraction. This approach is used to answer the problem correctly and completely.

We are given that;

S'mores Marshmallows Graham Crackers 4 8 12 13

Now,

8/4 = 2 Marshmallows for each S'mores,

12/4 = 3 Graham Crackers for each S'mores.

For, 13 S'mores,

2*13=26

3*13=39

Therefore, by algebra the answer will be 26 and 39.

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The ratio of students to classrooms  is 26:1

A ratio is defined as the comparison of two or more numbers indicating their sizes in relation to each other.

It  shows how many times one number contains another.

For example, if there are 20 oranges and 15 lemons in a bowl of fruit, then the ratio of oranges to lemons is 20 to 15.

From the information given, we have that;

234 students in 9 different classrooms

The ratio of students to classrooms would be;

= 234: 9

Find the least common factor

= 26: 1

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Brayden's total earning by walking lap around a track is $35, if he walks 2+1/2 = 5/2 laps per day for 7 days.

Earnings are the net benefits of a corporation's operation. Earnings is also the amount on which corporate tax is due. For an analysis of specific aspects of corporate operations several more specific terms are used as EBIT and EBITDA. Many alternative terms for earnings are in common use, such as income and profit.

here, we have,

Per day walking of brayden = 5/2 laps

7 days walking of brayden

= 5/2 × 7

= 35/2 laps

As per the question statement he earns $1 per lap if he walk less than 16 laps and $2 per lap if hi walks 16 or more lap.

Since the total walking of brayden is more than 16 laps, so he will earn at the rate of $2 per lap.

So total earning will be,

= 2× 35/2

= $35

Hence, Brayden's total earning by walking lap around a track is $35, if he walks 2+1/2 = 5/2 laps per day for 7 days.

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Answer:

14.5 laps



There is an even number of data points. So the median is the mean of the two middle numbers.

The recurrence relation for the number of ternary strings of length n that contain two consecutive zeros looks correct.

You can see that the first few terms match with your calculations:

a_1 = 0 (no possible strings)

a_2 = 1 (only one possible string: "00")

a_3 = 5 (possible strings: "000", "001", "002", "100", "200")

To explain your reasoning:

You can add "1" or "2" to the end of all the a_{n-1} strings, so there are 2a_{n-1} possible strings with n-1 length that end with "1" or "2".

You can add "01" or "02" to the end of all the a_{n-2} strings, so there are 2a_{n-2} possible strings with n-2 length that end with "01" or "02".

To count the number of strings of length n that contain two consecutive zeros, you need to add the strings that end with "00". There are 3^{n-2} possible strings of length n-2 that you can append "00" to.

Therefore, the recurrence relation is:

a_n = 2a_{n-1} + 2a_{n-2} + 3^{n-2}

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The diver can go 44.11 feet deep without having to resort to use of the artificial light .

The intensity function L(x) satisfy the differential equation : dL/dx = -kL  ;

On solving the differential equation by Integrating both sides, we get:

⇒ln |L| = -kx + C ; where C is the constant of integration.

Taking exponential of both sides,

we have ;

⇒ |L| = [tex]e^{-kx+C}[/tex] = [tex]e^{C}\times e^{-kx}[/tex] ;

⇒ L = [tex]Ce^{-kx}[/tex] ;  where C is a constant of integration.

We know that when x = 19, the intensity (L) is halved. That means :

⇒ L(19) = (1/2)L(0)

Substituting L = [tex]Ce^{-kx}[/tex]  into this equation, we get:

⇒ [tex]Ce^{-k\times19} = (\frac{1}{2} )Ce^{0}[/tex]

Simplifying further , we get ;

⇒ [tex]e^{-19k} = \frac{1}{2}[/tex] ;

Taking natural log(ln) on both sides,

⇒ -19k = ln(1/2)

⇒ k = ln(2)/(19)    ....equation(1)

We also know that the diver must stop diving when L = (1/5)L(0).

Which means :

⇒ L(x) = (1/5)L(0)

Substituting L = [tex]Ce^{-kx}[/tex] into this equation, we get:

⇒ [tex]Cx^{-kx} = (\frac{1}{5}) C[/tex]  ;

⇒ [tex]e^{-kx}[/tex] = 1/5 ;

Taking natural log(ln) on both sides,

⇒ -kx = ln(1/5)

⇒ x = -ln(1/5)/k

Substituting k = ln(2)/(19)  from equation(1) , we get:

⇒ x = [19×ln(5)]/ln(2) .

⇒ x = 44.11 feet  .

Therefore, the diver can go approximately 44.11 feet deep .

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