Edgar decided to add a second gate. He removes 2 yards t foot of fencing from his section of 13 yards. How much fencing is left?

Answer:

X=4 easy blah blah blah it’s 4. jdjjdjfnfvbdnnfn

The transformation for circle 1 exists (x+9), (y+5) and the scale factor of circle 1 exists 2.

Scale exists described as the ratio of the length of any object on a model to the actual length of the exact object in the entire world.

The function transformation rules: f(x)+b changes the function b units upward. f(x)−b shifts the function b units downward. f(x + b) shifts the function b units to the left.

Circle 1

center: (−4, −2)

Radius: 3 centimeters

Circle 2

center: (5, 3)

Radius: 6 centimeters

Transformations can be used to circle 1 to verify that the circles exist similar.

As circle 2 and circle 1 do not contain the exact coordinates

So, circle 1 has to utilize transformation rules.

The difference between the coordinates of circle 2 and circle 1 exists

[tex]$x_2 - x_1 = 5 - (-4 ) = 9[/tex]

[tex]y_2 - y_1 = 3 - (-2) = 5[/tex]

Therefore, transformation for circle 1: (x+9), (y + 5)

The scale factor between circle 2 and circle 1 exists

The radius of Circle 2 = 2 [tex]*[/tex] radius of circle 1

Therefore, the scale factor of circle 1 = 2

Therefore, the transformation for circle 1 exists (x+9), (y+5), and the scale factor of circle 1 exists 2.

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Answer:

[tex]y^6[/tex]

Step-by-step explanation:

So there is an exponent identity that states: [tex](x^b)^a = x^{a*b}[/tex] which means this question becomes: [tex](y^2)^3 = y^{2*3} = y^6[/tex].

Just so you completely understand why this works, it might help to express y^2, as what it truly represents: [tex]y^2=y*y[/tex]. So using this definition we can substitute it into the equation so it becomes: [tex](y*y)^3[/tex]. Now let's use the definition of exponents like we just did with the y, to redefine this in terms of multiplication: [tex](y*y)^3 = (y * y) * (y * y) * (y * y)[/tex]. We can just multiply all these out, and we get: [tex]y * y * y * y * y * y =y^6[/tex].

To make it a bit more general when we have the exponent: [tex]x^b[/tex] it can be expressed as: [tex](x*x*x...\text{ b amount of times})[/tex] now when we raise it to the power of a. we get: [tex](x * x * x...\text{ b amount of times})^a[/tex] which can further be rewritten using the definition of an exponent to become the equation: [tex](x*x*x\text... \text{ b amount of times}) * (x * x * x...\text{ b amount of times})...\text{ a amount of times}[/tex] which just simplifies to: [tex]x*x*x*x...\text{ a times b amount of times}[/tex]. Hopefully this makes the identity a bit more understandable

1a) 256/9

2a) 0.0048

3a) 0.64

4a) 2/5

5a) 4.6875

1b) 125/343

2b) 32/3125

3b) 0.003125

4b) 48.875

5b) 1.39

Whew! Hope this helps :)

Answer: (x + 11) cm²

Step-by-step explanation:

The formula for the area of a trapezoid is [tex]\frac{1}{2}(a+b)h[/tex], where

For this figure, [tex]a = 5[/tex] and [tex]b= x+6[/tex] (the top and bottom sides), while [tex]h=2[/tex] (the height).

If we suppose [tex]A[/tex] is the area, plugging these values in, we get

[tex]A = \frac{1}{2}(5 + x + 6)*2[/tex]

[tex]A = \frac{1}{2} * 2 * (x + 11)[/tex]

[tex]A = x + 11[/tex]

All the lengths are in cm, so the area will be (x + 11) cm².

Answer:

4ab2(7a2b2 5a - 4b)

Step-by-step explanation:

28a3b4 20a2b2 − 16ab3

The GCF of 28, 20 and 16 is 4.

GCF of the variables is ab2

So the answer is

4ab2(7a2b2 5a - 4b)

Answer:

4ab2(7a2b2 5a - 4b)

Step-by-step explanation:

Answer:

27

Step-by-step explanation:

(3^−9)(3^6)(3^6)

=1/19683(3^6)(3^6)

=1/19683(3^6)(3^6)

=(1/19683(729))(3^6)

=1/27(3^6)

=1/27(729)

=27

Answer:

27^3

Step-by-step explanation:

Answer: 17.3 cm

Step-by-step explanation:

[tex]\tan 68^{\circ}=\frac{AC}{7}\\\\AC=7\tan 68^{\circ}\\\\AC \approx 17.3[/tex]

Answer: 6x

Step-by-step explanation:

The  expression represents the radius of the cone's base is √111/π x = r

a three dimensional shape can be defined as a solid figure or an object or shape that has three dimensions—length, width, and height.

The formula for the volume of a cone is V = (1/3)πr²h, where V is the volume,

r is the radius of the base,

h is the height, and π is a constant approximately equal to 3.14.

We are given that the volume of the cone is 37x³ cubic units and the height is x units.

Using the formula for the volume of a cone, we can write:

37x³ = (1/3)πr²x

37x² = πr²/3

37×3 x²/π = r²

111x²/π  = r²

Take square root on both sides

√111/π x = r

Hence, the  expression represents the radius of the cone's base is √111/π x = r

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No this does not indicate that the population proportion of women athletes who graduate from the university is now less than 67%.

Given long term graduation rate of 67%, sample size of 36  and the women athletes graduated is 23.

We have to find whether  the given information shows  that the population proportion is less than 67%.

First we have to create hypothesis for this :

[tex]H_{0}[/tex]:P=0.67

[tex]H_{1}[/tex]:P<0.67

Under null hypothesis the test statistic is

z=p bar-p/[tex]\sqrt{p(1-p)/n}[/tex]

where p bar=23/36

=0.638

z=(0.638-0.67)/[tex]\sqrt{0.67(1-0.67)/36[/tex]

=-0.032/[tex]\sqrt{0.67*0.33/36}[/tex]

=-0.032/[tex]\sqrt{0.0064}[/tex]

=-0.032/0.078

=-0.41

Now we have to find the left tailed critical at 0.01 significance level using z table.

z=-2.33

Since the z value does not fall in the critical region,therefore we fail to reject the null hypothesis. So we can conclude that there is not sufficient evidence to say that the population proportion of women athletes who graduate from the university is now less than 67%.

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Question is incomplete as it should specify the signficance level of 0.01 to be used.

Two possibilities  --- see below

Step-by-step explanation:

So it could be GREATER than 46

  .85 x = 46    x = 54.11 cm^3

Or it could be  smaller than 46

  46= (1.15)  x      x = 40 cm^3

Using the t-distribution, it is found that there is not enough evidence that the diet is helping people lose weight.

At the null hypothesis, it is tested if the mean hasn't decreased, that is:

[tex]H_0: \mu_B \leq \mu_A[/tex]

[tex]H_0: \mu_B - \mu_A \leq 0[qtex]

At the alternative hypothesis, it is tested if the mean has decreased, that is:

[tex]H_1: \mu_B - \mu_A > 0[qtex]

For each sample, they are given as follows:

Hence, for the distribution of differences, they are:

The test statistic is:

[tex]t = \frac{\overline{x} - \mu}{s}[/tex]

In which [tex]\mu = 0[/tex] is the value tested at the null hypothesis.

Hence:

[tex]t = \frac{\overline{x} - \mu}{s}[/tex]

[tex]t = \frac{9.88 - 0}{11.66}[/tex]

t = 0.85.

Considering a right-tailed test with 9 + 9 - 2 = 16 df, with a standard significance level of 0.05, the critical value is t = 1.7459. Since t = 0.85 < 0.85, there is not enough evidence that the diet is helping people lose weight.

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Answer:

7.7 km

Explanation:

Use cosine rule as here given two sides and one angle.

Cosine rule states:

a² = b² + c² - 2bc cos(A)

While solving, treat a = 7.5 km as to that opposite angle is given of 68°

then b = missing side, c = 5.2 km, A = 68°

Applying rule:

7.5² = b² + 5.2² - 2(b)(5.2) cos(68)

56.25 = b² + 27.04 - 3.8959b

56.25 - 27.04 = b² - 3.8959b

b² - 3.8959b = 29.21

b² - 3.8959b - 29.21 = 0

apply quadratic equation, Here [a = 1, b = - 3.8959, c = -29.21]

[tex]\sf b = \dfrac{ -b \pm \sqrt{b^2 - 4ac}}{2a} \quad\:when \:\ ax^2 + bx + c = 0[/tex]

[tex]\sf b = \dfrac{ -(-3.8959) \pm \sqrt{(-3.8959)^2 - 4(1)(-29.21)}}{2(1)}[/tex]

[tex]\sf b = 7.69 291 \quad or \quad b = -3.797[/tex]

[tex]\sf b = 7.7 \quad (rounded \ to \ nearest \ tenth)[/tex]

As length cannot be negative. Hence the value of b is only 7.7 km

The answer is 7.7 km.

Let's apply the Cosine Law in this situation.

a² = b² + c² - 2bc cos(A)

Now, substitute the values based on the given diagram.

Here, using the Quadratic Equation, we can solve :

Point-slope form

[tex]y-2=4(x+1)[/tex]

Slope-intercept form

[tex]y-2=4x+4 \\ \\ y=4x+6[/tex]

Standard form

[tex]4x-y+6=0[/tex]

Answer:

[tex]\textsf{Point-slope form}: \quad \sf y-2=4(x+1)[/tex]

[tex]\textsf{Slope-intercept form}: \quad \sf y=4x+6[/tex]

[tex]\textsf{Standard form}: \quad \sf 4x-y=-6[/tex]

Step-by-step explanation:

Given information:

Point-slope form of linear equation:  

[tex]\sf y-y_1=m(x-x_1)[/tex]

(where m is the slope and (x₁, y₁) is a point on the line)

Substitute the given slope and point into the formula:

[tex]\implies \sf y-2=4(x-(-1))[/tex]

[tex]\implies \sf y-2=4(x+1)[/tex]

Slope-intercept form of a linear equation:

[tex]\sf y=mx+b[/tex]

(where m is the slope and b is the y-intercept)

Substitute the given slope and point into the formula and solve for b:

[tex]\implies \sf 2=4(-1)+b[/tex]

[tex]\implies \sf b=6[/tex]

Therefore:

[tex]\sf y=4x+6[/tex]

Standard form of a linear equation:

[tex]\sf Ax+By=C[/tex]

(where A, B and C are constants and A must be positive)

Rearrange the found slope-intercept form of the equation into standard form:

[tex]\implies \sf y=4x+6[/tex]

[tex]\implies \sf 4x-y+6=0[/tex]

[tex]\implies \sf 4x-y=-6[/tex]

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Using the laws of indices , 4⁵ ÷ 4² can rewritten as 4³.

Given the expression; 4⁵ ÷ 4²

To perform the division operation, we use one of the laws of indices;

[tex]n^a/n^b = n^{a-b}[/tex]

Given that;

Now, we apply the laws of indices ;

4⁵ ÷ 4² = 4⁵⁻² = 4³

Therefore, using the laws of indices, 4⁵ ÷ 4² can rewritten as 4³.

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Rolle's Theorem does not apply to the function because there are points on the interval (a,b) where f is not differentiable.

Given the function is [tex]f(x)=\sqrt{(2-x^{\frac{2}{3}})^{3}}[/tex]  and the Rolle's Theorem does not apply to the function.

Rolle's theorem is used to determine if a function is continuous and also differentiable.

The condition for Rolle's theorem to be true as:

To apply the Rolle’s Theorem we need to have function that is differentiable on the given open interval.

If we look closely at the given function we can see that the first derivative of the given function is:

[tex]\begin{aligned}f(x)&=\sqrt{(2-x^{\frac{2}{3}})^3}\\ f(x)&=(2-x^{\frac{2}{3}})^{\frac{3}{2}}\\ f'(x)&=\frac{3}{2}(2-x^{\frac{2}{3}})^{\frac{1}{2}}\cdot \frac{2}{3}\cdot (-x)^{\frac{1}{3}}\\ f'(x)&=\frac{-\sqrt{2-x^{\frac{2}{3}}}}{\sqrt[3]{x}}\end[/tex]

From this point of view we can see that the given function is not defined for x=0.

Hence, all the assumptions are not satisfied we can reach a conclusion that we cannot apply the Rolle's Theorem.

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Given that (p - 1/p) = 4, the value of p² + 1/p² is 18. Detail below

(p - 1/p) = 4

Square both sides

(p - 1/p)² = (4)²

(p - 1/p)² = 16 ....(1)

Recall

(a - b)² = a² + b² - 2ab

Thus,

(p - 1/p)² = p² + 1/p² - (2 × p × 1/p)

(p - 1/p)² = p² + 1/p² - 2

From equation (1) above,

(p - 1/p)² = 16

Therefore,

p² + 1/p² - 2 = 16

Rearrange

p² + 1/p² = 16 + 2

p² + 1/p² = 18

Thus, the value of p² + 1/p² is 18

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The expression which is equivalent to [tex]5 log_{10}x+log_{10}20-log_{10}10[/tex] is

[tex]$5 log_{10}x+log_{10}20-log_{10}10 = log_{10}(20x^5)-1[/tex].

A logarithmic equation exists as an equation that applies the logarithm of an expression having a variable.

Product Rule Law: [tex]log_{a} (MN) = log_{a} M + log_{a} N[/tex]

Quotient Rule Law: [tex]log_{a} (M/N) = log_{a} M - log_{a} N[/tex]

Power Rule Law: [tex]log_{a}M^{n} = n log_{a} M[/tex]

Given: [tex]5 log_{10}x+log_{10}20-log_{10}10[/tex]

[tex]5 log_{10}x+log_{10}20-log_{10}10 = log_{10}(x^5) +log_{10}20 -log_{10}10[/tex]

apply the law of logarithm

[tex]5 log_{10}x+log_{10}20-log_{10}10 = log_{10}(x^5*20/10)[/tex]

[tex]5 log_{10}x+log_{10}20-log_{10}10 = log_{10}(x^5*2)[/tex]

[tex]5 log_{10}x+log_{10}20-log_{10}10 = log_{10}(2x^5)[/tex]

Another possible equivalent expression is:

[tex]$5 log_{10}x+log_{10}20-log_{10}10 = log_{10}(x^5*20)-log_{10}10[/tex]

[tex]$5 log_{10}x+log_{10}20-log_{10}10 = log_{10}(20x^5)-log_{10}10[/tex]

Substitute the value of [tex]log_{10}10 = 1[/tex]

[tex]$5 log_{10}x+log_{10}20-log_{10}10 = log_{10}(20x^5)-1[/tex]

Therefore, the correct answer

[tex]$5 log_{10}x+log_{10}20-log_{10}10 = log_{10}(20x^5)-1[/tex].

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The equation will be like this,

6.89 + t = 8.00

t = 8 - 6.89

t = 1.11

tips = $1.11

6.89 + t = 8.00

To sum up the total amount of cash and coins you have, first, sort each invoice and coin by denomination. Create a separate stack for each denomination and count how many denominations or coins you have.

For each denomination of banknotes and coins, multiply the number you have by the denomination. For example, if you have 4 out of $ 10, multiply by 4x10 to get $ 40. If you have three $ 5 bills, multiply by 3x5 to get $ 15.

Sum the totals to calculate the total amount.

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Answer:

Translation of 3 units to the left.

Vertical stretch by a factor of 2.

Translation of 5 units down.

Step-by-step explanation:

Transformations

For a > 0

[tex]f(x+a) \implies f(x) \: \textsf{translated}\:a\:\textsf{units left}[/tex]

[tex]y=a\:f(x) \implies f(x) \: \textsf{stretched parallel to the y-axis (vertically) by a factor of}\:a[/tex]

[tex]f(x)-a \implies f(x) \: \textsf{translated}\:a\:\textsf{units down}[/tex]

Parent function:

[tex]y=x^2[/tex]

Translate 3 units left

Add 3 to the variable of the function

[tex]\implies y=(x+3)^2[/tex]

Stretch vertically by a factor of 2

Multiply the whole function by 2:

[tex]\implies y=2(x+3)^2[/tex]

Translate 5 units down

Subtract 5 from the whole function:

[tex]\implies y=2(x+3)^2-5[/tex]

Please see the attached graphs for the final transformed function (as well as the graphed steps).

Answer:

  a) vertical expansion by a factor of 2; translation 3 units left and 5 units down

  b) see attached

Step-by-step explanation:

Describing transformations is all about matching patterns. The elements of the transformed function are matched with the elements of a transformation.

A function is scaled vertically by multiplying each function value by some scale factor. In generic terms, the function f(x) is scaled vertically by the factor 'c' in this way:

If we want the function f(x) = x² scaled vertically by a factor of 2, then we have

  f(x) = x² . . . . . . . original function

  2·f(x) = 2x² . . . . scaled vertically by a factor of 2

On a graph, each point is vertically twice as far vertically from some reference point (the vertex, for example) as it is in the original function graph.

A function is translated to the right by 'h' units when x is replaced by (x -h).

If we want the function f(x) = x² translated right by 3 units, we will have ...

  f(x) = x² . . . . . . . . . . . original function

  f(x -3) = (x -3)² . . . . . .translated right 3 units

Note that translation left by 3 units would give ...

  f(x -(-3)) = f(x +3) = (x +3)² . . .  . translated left 3 units

On a graph, each point of the left-translated function is 3 units left of where it was on the original function graph.

A function is translated upward by 'k' units when k is added to the function value.

The value of k will be negative for a translation downward.

If we want the function f(x) = x² translated down by 5 units, we will have ...

  f(x) = x² . . . . . . . . . . . original function

  f(x) = x² -5 . . . . . . . . .translated down 5 units

Using all of these transformations at once, we have ...

  f(x) = x² . . . . . . . . . . . . . . . . . original function

  c·f(x -h) +k = c·(x -h)² +k . . . scaled by 'c', translated h right and k up

Compare this to the given function:

 y = 2(x +3)² -5

and we can see that ...

This is the pattern matching that is described at the beginning.

__

When graphing a transformed function, it is often useful to start with a distinctive feature and work from there. The vertex of a parabola is one such distinctive feature.

The transformations move the vertex 3 units left and 5 units down from its original position at (0, 0). The location of the vertex on the transformed function graph will be at (x, y) = (-3, -5).

The graph of the parent function parabola (y= x²) goes up from the vertex by the square of the number of units right or left. That is, 1 unit right or left of the vertex, the graph is 1 unit above the vertex. 2 units right or left, the graph is 2² = 4 units above the vertex.

The scaled graph will have these vertical distances multiplied by 2:

The graph of the transformed function is shown in blue in the attachment.

__

Additional comment

The vertical scale factor 'c' may have any non-zero value, positive or negative, greater than 1 or less than 1. When the magnitude is less than 1, the scaling is a compression, rather than an expansion. When the sign is negative, the graph is also reflected across the x-axis, before everything else.

We can learn a lot about a population if we select a subset of it.

One kind of set is a sample space. It is a clear listing of every event that could occur in a statistical experiment. A statistical experiment's events are a subset of the sample space.

A subset is a smaller group of results that are part of the bigger group.

Subsets are events, and events are subsets. A subset is an event of a sample space, and an event is a potential result of an experiment. A random experiment's sample space is a set (S) that contains all of the experiment's potential outcomes.

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The weighted mean grade for the semester is 8.22

It is calculated by dividing the total number of values in a set of data, such as measurements or numbers, by the total number of values.

No. of credits for Spanish, History and English = 6

No. of credits for Statistics = 8

Grade points for A = 4

Grade points for B = 3

Grade points for C = 2

Weighted mean = (6 * 2 + 6 * 3 + 8 * 4 + 6 * 2) / 9

Weighted mean = (12  + 18 + 32 + 12) / 9

Weighted mean = 74 / 9 = 8.22

Hence, the weighted mean grade for the semester is 8.22.

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Answer:

c 1:4

Step-by-step explanation:

The length and width of the rectangle with an area of 120 units² are 12 units and 10 units respectively.

An equation is an expression that shows the relationship between two or more numbers and variables.

Let x represent the width, hence:

length = x + 2

The area of a rectangle is the product of its length and its width.

Area = length * width

Area = x(x + 2)

120 = x² + 2x

x² + 2x - 120 = 0

x = 10, length = 10 + 2 = 12

The length and width of the rectangle with an area of 120 units² are 12 units and 10 units respectively.

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Gradient = rise/ run

rise is d

run is the difference in our x coordinates of c & e

(or c - e)

So, it's d/C - e

(3rd option)

Hope this helps!

Answer:

3rd option

Step-by-step explanation:

calculate the slope m using the slope formula

m = [tex]\frac{y_{2}-y_{1} }{x_{2}-x_{1} }[/tex]

with (x₁, y₁ ) = E (e, 0 ) and (x₂, y₂ ) = (c, d )

m = [tex]\frac{d-0}{c-e}[/tex] = [tex]\frac{d}{c-e}[/tex]

Answer:

Greetings !

check the attachment above☝️ but i haven't done the second question wait a moment. thx

The following system of equation can be solved by first substituting the dependent variable in the other equation . The solution yields x = -1.5 and y = 6.5 .

The two equations are given as -

4x-2y=7

3x-3y=15

First substituting the value of x from the first equation and then putting that value in the second equation for the following system of equation.

From first equation,

⇒ 4x = 7 + 2y

∴  x = (7 + 2y)/4

Putting this value of x in second equation,

⇒ 3*(7 + 2y)/4 - 3y = 15

⇒ 3*(7 + 2y) - 12y = 60

⇒ 21 + 6y - 12y = 60

⇒ -6y = 39

∴  y = -6.5

∴ x = (7 + 2y)/4 = -1.5

Thus x = -1.5 and y = -6.5

Therefore, the following system of equation can be solved by first substituting the dependent variable in the other equation . The solution yields x = -1.5 and y = 6.5 .

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