A. X = -6 and X = -5.5
B. x = -0.5 and x = -5
C. X = 0 and x = 5
D. X = 5 and x = 6

Step-by-step explanation:

1. if the number of pages in the 1st day is 'x', then the 2d day - 'x+10', the 3d day - 'x+20', the 4th day - 'x+30', the 5th day - 'x+40' and the last day - 'x+50' pages;

2. if the sum of all the pages is 300, then it is possible to make up the equation:

x+x+10+x+20+x+30+x+40+x+50=300;

3. x=25, it means:

1st day - 25;

2d day - 35;

3d day - 45;

4th day - 55;

5th day - 65;

6th day - 75 pages.

Answer:

20 x 20 x 20 = 8000

Step-by-step explanation:

The formula for the angle between vectors

[tex] \alpha = \cos {}^{ - 1} ( \frac{uv}{ |u| |v| } ) [/tex]

To multiply vectors, multiply the first component and multiply the second component and add them.

(-4*-1) + (-3*5)= 4-15=-11.

To find magnitude of vectors, use the Pythagorean theorem

[tex]u = \sqrt{ { - 4}^{2} + { - 3}^{2} } = 5[/tex]

[tex]v = \sqrt{ - 1 {}^{2} + 5 {}^{2} } = \sqrt{26} [/tex]

so

[tex] |u| |v| = 5 \sqrt{26} [/tex]

Know we have,

[tex] \alpha = \cos {}^{ - 1} ( \frac{7}{5 \sqrt{26} } ) [/tex]

[tex] \alpha = 105.94[/tex]

in degrees,

[tex] \alpha = 1.849[/tex]

in radians

Answer:

115.6° (1 d.p.)

Step-by-step explanation:

To find the angle between two vectors:

**Please see attached for the triangle diagram**

Given vectors:

[tex]\textbf{u}=-4\textbf{i}-3\textbf{j}[/tex]

[tex]\textbf{v}=-\textbf{i}+5\textbf{j}[/tex]

Use Pythagoras Theorem to find the magnitude of each vector:

[tex]\implies |\textbf{u}|=\sqrt{(-4)^2+(-3)^2}=5[/tex]

[tex]\implies |\textbf{v}|=\sqrt{(-1)^2+5^2}=\sqrt{26}[/tex]

[tex]\overrightarrow{\text{UV}}=\textbf{v}-\textbf{u}=(-\textbf{i}+5\textbf{j})-(-4\textbf{i}-3\textbf{j})=3\textbf{i}+8\textbf{j}[/tex]

[tex]|\overrightarrow{\text{UV}}|=\sqrt{3^2+8^2}=\sqrt{73}[/tex]

Cosine Rule (for finding angles)

[tex]\sf \cos(C)=\dfrac{a^2+b^2-c^2}{2ab}[/tex]

where:

Find angle θ using the cosine rule:

[tex]\implies \cos(\theta)=\dfrac{|\textbf{u}|^2+|\textbf{v}|^2-|\overrightarrow{\text{UV}}|^2}{2|\textbf{u}||\textbf{v}|}[/tex]

[tex]\implies \cos(\theta)=\dfrac{5^2+\left(\sqrt{26}\right)^2-\left(\sqrt{73}\right)^2}{2(5)\left(\sqrt{26}\right)}[/tex]

[tex]\implies \cos(\theta)=\dfrac{-22}{10\sqrt{26}}[/tex]

[tex]\implies \theta=\cos^{-1}\left(\dfrac{-22}{10\sqrt{26}}\right)[/tex]

[tex]\implies \theta=115.5599652...^{\circ}[/tex]

Therefore, the angle between the vectors is 115.6° (1 d.p.).

The first table with the following values of x and y, shows a linear function.

x = -4, y = 8

x = -1, y = 2

x = 1, y = 2

x = 2, y = 4

x = 3, y = 6

Definition of a Linear Function:

A polynomial function of degree zero or one that has a straight line as its graph is referred to as a linear function in calculus and related fields.

For the above chosen option, the value of y corresponding to a value of x is twice that of its x value. This linear function can be represented as follows,

y = | 2x | .......... (1)

In this linear function, mod represents, that y is always maintained positive. Besides, the value of y is always twice that of the x. Hence, the linear function can also be written as, y ± 2x = 0.

Since, equation (1) is of the form, y = mx + c, it produces a straight line. Here, m = 2 and c = 0. Thus, it is a linear function.

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So to run 10km, you need to run 6.25 times the trail. (Or 7 if you only accept whole numbers as answers, we need to round up).

If you run the trail x times, then you will run:

y = x*1mi

Now remember that:

1 mi = 1.6 km

Replacing that, we get the linear equation:

y = x*1.6km

Now we want to find the value of x such that:

x*1.6km = 10km

Dividing both sides by 1.6km

x = 10km/1.6km = 6.25

So to run 10km, you need to run 6.25 times the trail. (Or 7 if you only accept whole numbers as answers, we need to round up).

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Using the binomial distribution, we have that:

It is the probability of exactly x successes on n repeated trials, with p probability of a success on each trial.

The expected value of the binomial distribution is:

[tex]\mu = np[/tex]

The standard deviation of the binomial distribution is:

[tex]\sqrt{V(X)} = \sqrt{np(1-p)}[/tex]

For this problem, the parameters are:

n = 20, p = 0.3.

Hence:

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Answer:3522

Step-by-step explanation:222

2

Answer:

Step-by-step explanation:

$274 for 40 hours of work as a unit rate

in practice you look for the hourly wage, you find it by dividing the wages divided by the working hours

274 : 40 = 6.85 units per hour

The figure is represented by the inequality y ≥ 5 · x² - 40 · x - 45. (Correct choice: C)

In accordance with the figure, we have an inequation of the form y ≥ f(x). Now we proceed to find the quadratic equation of the parabola:

f(x) = a · (x + 1) · (x - 9)

- 125 = a · (4 + 1) · (4 - 9)

- 125 = a · 5 · (- 5)

- 125 = - 25 · a

a = - 5

f(x) = 5 · (x + 1) · (x - 9)

f(x) = 5 · (x² - 8 · x - 9)

f(x) = 5 · x² - 40 · x - 45

The figure is represented by the inequality y ≥ 5 · x² - 40 · x - 45. (Correct choice: C)

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The amount of the radioactive material in the vault after 140 years is 210 pounds

We have that the function is given as a model;

f(x) = 300(0.5)x/100

Where

Let's substitute the value of 'x' in the model

f(x) = 300(0.5)x/100

[tex]f(x) = \frac{300(0.5) * 140}{100}[/tex]

[tex]f(x) =\frac{21000}{100}[/tex]

f(140) = 210 pounds

This mean that the function of 149 years would give an amount of 210 pounds rounded up to the nearest whole number.

Thus, the amount of the radioactive material in the vault after 140 years is 210 pounds

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The distribution is skewed because it's on the right side of the polygon and this illustrates that's it's positively skewed.

It should be noted that the polygon can be gotten by plotting the data.

After the analysis of the distribution, the distribution is skewed to the right.

This illustrates that it is skewed.

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Answer:

11664

Step by Step:

Evaluate for x=6,y=1

3(6^2)(6^2)(6−(3)(1))

3(6^2)(6^2)(6−(3)(1))

=11664

A quadratic equation that models something from one's daily life is when calculating areas, and determining a product's profit.

It should be noted that a quadratic equation means an algebraic equation of the second degree in x.

The quadratic equation in its standard form is ax² + bx + c = 0. Here, a and b are the coefficients, x is the variable, while c is the constant term.

Quadratic equations are used in everyday life, as when calculating areas, and the determination of a product's profit and formulating the speed of an object.

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Total amount of money in the savings account after 11 weeks is $1310.

Total money in the account =

 Initial Money + Money added per week x Number of weeks

Given:

Initial Money = $650

Money added per week = $60

Total money in the account = S

Number of weeks = W

Substituting it in the above equation we get,

S = $650 + $60xW                                          (General Equation)

Total amount of money in the savings account after 11 weeks

S = $650 + $60x11

S = $650 + $660

S = $1310

Thus total amount of money in the savings account after 11 weeks is $1310.

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Answer:

No

Step-by-step explanation:

This table does not represent a function. The domain(x) can have only one range(y) in order for a set of data to represent a function. However, the domain value 2 has the range of both 1 and 4. Therefore, the table does not represent a function.

The score Macro needs in his next game to have a mean of 130 is 150.

Mean is the average of a set of numbers. It is determined by adding the numbers together and dividing it by the total number

Mean = sum of the numbers / total number

130 = (128 + 127 + 123 + 122 + x) / 5

Where x represents the fifth score

130 = (500 + x) / 5

130 x 5 = 500 + x

650 = 500 + x

650 - 500 = x

x = 150

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a) The exponential function that models the number of cells is: [tex]A(t) = 200(1.75)^{\frac{t}{3}}[/tex]

b) There will be 1292 cells after 10 hours.

An increasing exponential function is modeled by:

[tex]A(t) = A(0)(1 + r)^{\frac{t}{n}}[/tex]

In which:

For this problem, the parameters are:

A(0) = 200, r = 0.75, n = 3.

Hence the equation is:

[tex]A(t) = A(0)(1 + r)^{\frac{t}{n}}[/tex]

[tex]A(t) = 200(1 + 0.75)^{\frac{t}{3}}[/tex]

[tex]A(t) = 200(1.75)^{\frac{t}{3}}[/tex]

The number of cells after 10 hours is:

[tex]A(t) = 200(1.75)^{\frac{t}{3}}[/tex]

[tex]A(10) = 200(1.75)^{\frac{10}{3}}[/tex]

A(10) = 1292.

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Answer: 0

Step-by-step explanation:

[tex]125x^3 =0\\\\x^3 =0\\\\x=0[/tex]

There are no real solutions to solve Z. If there may be a typo or something, there are no solutions.

The equation of the exponential function is [tex]y = 3^{-x -5} - 3[/tex]

An exponential equation is represented as:

y = b^x

Where b represents the base.

The base is 3.

So, we have:

[tex]y = 3^x[/tex]

When reflected over the y-axis, we have:

[tex]y = 3^{-x[/tex]

When shifted right by 5 units, we have:

[tex]y = 3^{-x -5[/tex]

Lastly, the function has an asymptote of y=-3

So, we have:

[tex]y = 3^{-x -5} - 3[/tex]

Hence, the equation of the exponential function is [tex]y = 3^{-x -5} - 3[/tex]

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Answer:Principal P = 2000

Amount= A

years=n 10.00

compounded 12 times a year t

Rate = 2.00 0.02

Amount = P*((n+r)/n)^n*t

Amount = = 2000 *( 1 + 0.02 /t)^ 10 * 12

Amount = 2000 *( 1 + 0 )^ 120

2000 *( 1 )^ 120

Amount = 2442.4

Step-by-step explanation:

we conclude that 900 cubes enter in the given box.

We need to find the volumes of both figures.

The blocks are cubes of 2cm by 2cm by 2cm, then the volume of each block is:

[tex]V = (2cm)^3 = 8cm^3[/tex]

The box has dimensions of 10cm by 12cm by 60cm, then the volume is:

[tex]V = 10cm*12cm*60cm = 7,200 cm^3[/tex]

The number of cubes that enter in the box is N such that:

[tex]N*8cm^3 = 7,200cm^3[/tex]

Solving for N, we get:

[tex]N = 7200cm^3/8cm^3 = 900[/tex]

In this way, we conclude that 900 cubes enter in the given box.

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Answer:

16/25

Step-by-step explanation:

(at least 1 prime selected)=1−(no primes selected)

There are 3 numbers in the set that are not prime (6,8,9)

(no primes selected)=(35)2=925

(at least 1 prime selected)=1−925

=1625

The following expressions set up as an equation to solve for the unknown numbers in the cases given below:

let

5x + 6 = 46

5x = 46 - 6

5x = 40

x = 40/5

x = 8

2/3x - 5 = 11

2/3x = 11 + 5

2/3x = 16

x = 16 ÷ 2/3

x = 16 × 3/2

x = 48/2

x = 24

1/3x + 4 = 45

1/3x = 45 - 4

1/3x = 41

x = 41 ÷ 1/3

x = 41 × 3/1

x = 123

3x - 12 = 18

3x = 18 + 12

3x = 30

x = 30/3

x = 10

40 - 2x = 16

-2x = 16 - 40

-2x = -24

x = -24/-2

x = 12

(x + 18) / 6 = 7

(x + 18) = 7 × 6

x + 18 = 42

x = 42 - 18

x = 24

2/3x - 8 = 6

2/3x = 6 + 8

2/3x = 14

x = 14 ÷ 2/3

x = 14 × 3/2

= 42/2

x = 21

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The value of the expression C(6,3) is 20

The expression is given as:

C(6, 3)

This means 6 combination 3.

And it can be written as:

[tex]^6C_3[/tex]

Apply the combination formula

[tex]^6C_3 = \frac{6!}{3!3!}[/tex]

Evaluate the expression

[tex]^6C_3 = 20[/tex]

Hence, the value of the expression C(6,3) is 20

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The given polynomial is a Quartic trinomial

From the question, we are to determine the type of the given polynomial

The given polynomial is

3x⁴−2x³−2

The highest degree of the polynomial is 4. Thus, it is a Quartic Polynomial.

Also, the expression contains exactly 3 terms. Thus, it is a trinomial.

Hence, the given polynomial is a Quartic trinomial

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Answer:

c

Step-by-step explanation:

Aight, let's hop to it:

So we got

[tex] \frac{5x}{45} = \frac{20}{36} = = > \\ \frac{x}{9} = \frac{5}{9} = = > \\ x = 5[/tex]

and boom

Answer:

a.

2 / 3 + 5 / 6 - 1 / 2a.

2 / 3 + 5 / 6 - 1 / 2

= 0,66... + 0,833... - 0,5

= 1,5 - 0,5

= 1

Step-by-step explanation:

sorry I only know the letter A hope I helped anyway