A scale drawing of the side view of a house is shown at the right. Find the total area (in square inches) of the side of the house in the drawing PLEASE HELP!!

The question is about the mathematical modeling of a vineyard's grape production. As the number of vines increases, the individual yield of each vine decreases by 1%. An equation, such as P = 5000 - 50(n-500), is a possible mathematical model to represent this situation.

This question appears to require a detailed understanding of mathematical modeling and percentage decrease concept. The problem presented describes the decrease in grape production per vine as the number of vines planted per acre increases. It's an example of an inverse relationship, when one variable increases the other variable decreases.

The initial production quantity is 10 lb of grapes per vine when there are 500 vines per acre. However, for every additional vine planted, there is a subsequent 1% drop per vine. This means that if 501 vines are planted, each vine then produces only 99% of 10 lbs, or 9.9 lbs, and so on.

To model this mathematically, an equation could possibly be P = 5000 - 50(n-500), where P is the production of grapes in pounds, and n is the number of vines. This formula might help to calculate the maximum yield that could be obtained according to the number of vines.

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Final answer:

The range of Set A is 5 and Set B is 16. The mean of Set A is 15.2 and Set B is 12. Set B has a larger range but a smaller mean compared to Set A.

Explanation:

The range of a data set is calculated by subtracting the smallest value from the largest value. For Set A, the range is 18 - 13 = 5. For Set B, the range is 20 - 4 = 16.

The mean of a data set is calculated by summing all the values and dividing by the number of values. For Set A, the mean is (13 + 15 + 16 + 18 + 14) / 5 = 76 / 5 = 15.2. For Set B, the mean is (4 + 10 + 8 + 18 + 20) / 5 = 60 / 5 = 12.

Comparing the two data sets, we can see that Set B has a larger range than Set A, indicating greater variability in the data. However, Set B has a smaller mean than Set A, indicating that the values in Set B are generally lower than those in Set A.

First, lets start with the formula for the volume of a cone

[tex] \frac{1}{3} (\pi \times {r}^{2})h [/tex]

Then lets find the diameter so we can get the radius...

Since we have circumference, all we need to do is use this formula-

[tex] c \div \pi[/tex]

C being circumference, the PI being the 3.14, we plug it in to the formula as the number we have to get the diameter...

[tex]{75.36} \div 3.14[/tex]

And it comes out to...

24.

Now we need to divide it by two to get the radius, and we come up with 12.

Now that we have our radius, we can finally plug in all of the numbers into our original formula...

[tex] \frac{1}{3} (3.14 \times \: {12}^{2} ) \times 18[/tex]

The answer of the entire problem turns out to be... Drum roll please...

2712.96 cubic centimeters.

Answer:

y = 1 for the line.

Step-by-step explanation:

All values under y = 1. Surprisingly 1^0 is still 1. So just fill the table in with 1s under y.

I've drawn the line in desmos for you. I'm not sure whether you can extend the question enough to graph a line segment containing these 5 points (which is what I have done) or if you should just submit a graph with 4 points on. If it does not cost you anything to submit it twice, I would try the line first and the points alone the second time.

[tex]\bf ~\hspace{7em}\textit{rational exponents} \\\\ a^{\frac{ n}{ m}} \implies \sqrt[ m]{a^ n} ~\hspace{10em} a^{-\frac{ n}{ m}} \implies \cfrac{1}{a^{\frac{ n}{ m}}} \implies \cfrac{1}{\sqrt[ m]{a^ n}} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ 9^{\frac{2}{3}}\implies (3^2)^{\frac{2}{3}}\implies 3^{2\cdot \frac{2}{3}}\implies 3^{\frac{4}{3}}\implies \sqrt[3]{3^4}\implies \sqrt[3]{3^3\cdot 3^1}\implies 3\sqrt[3]{\stackrel{\textit{this one}}{3}}[/tex]

Answer:

\bf ~\hspace{7em}\textit{rational exponents} \\\\ a^{\frac{ n}{ m}} \implies \sqrt[ m]{a^ n} ~\hspace{10em} a^{-\frac{ n}{ m}} \implies \cfrac{1}{a^{\frac{ n}{ m}}} \implies \cfrac{1}{\sqrt[ m]{a^ n}} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ 9^{\frac{2}{3}}\implies (3^2)^{\frac{2}{3}}\implies 3^{2\cdot \frac{2}{3}}\implies 3^{\frac{4}{3}}\implies \sqrt[3]{3^4}\implies \sqrt[3]{3^3\cdot 3^1}\implies 3\sqrt[3]{\stackrel{\textit{this one}}{3}}

Step-by-step explanation:

Answer:

d. $43.00

Step-by-step explanation:

Karli's total cost is ...

total cost = material cost + labor cost

= ($8.00/10 gal)·(5 gal) + ($13.00/h)·(3 h)

= $4.00 + $39.00

= $43.00

Answer:

 the answer is B (5 square root 2, 315°), (-5 square root 2, 135°)

Step-by-step explanation:

1) Let A be the point (x, y) = (5, - 5)  

=> x = 5 and y = - 5  

r = √(x² + y²) = √(25 + 25) = √50 = ± 5√2  

tan Θ = - 5/5 = - 1  

=> Θ = (i) 315º or - 45º ; (ii) 135º or - 225  

Hence, the Polar Coordinates of A are (i) (5√2, 315º) (ii) (- 5√2, 135º)

Two pairs of polar coordinates for the point is option b,

Here we assume that A be the point (x, y) = (5, - 5)  

So,

x = 5 and y = - 5  

Now

[tex]r = \sqrt(x^2 + y^2) = \sqrt(25 + 25) = \sqrt50 = \pm 5\sqrt2[/tex]

tan Θ = - 5/5 = - 1  

Now

(i) 315º or - 45º ; (ii) 135º or - 225  

So, the polar coordinates should be (5 square root 2, 315°), (-5 square root 2, 135°)

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Answer:

It would be A. 40 square units (:

Step-by-step explanation:

Answer:

  968π ≈ 3041 . . . cubic units

Step-by-step explanation:

The usual formula for the volume of a cylinder is ...

  V = πr²h

For your given dimensions, the volume is found by putting the values into the formula and doing the arithmetic.

  V = π(11²)(8) = 968π . . . . cubic units

 V ≈ 3041 cubic units

I think a but I’m not quite sure

Assume the car starts at the origin, so that its initial position is [tex]x(0)=0[/tex]. The car's displacement at any time [tex]t[/tex] over the 20 second interval is

[tex]\displaystyle x(0)+\int_0^t(44+2.2u)\,\mathrm du=0+\left(44u+1.1u^2\right)\bigg|_{u=0}^{u=t}=44t+1.1t^2[/tex]

so that after 20 seconds the car has moved 1320 ft.

###

Without using calculus, recall that under constant acceleration, the average velocity of the car over the 20 second interval satisfies

[tex]v_{\rm avg}=\dfrac{v_f+v_i}2[/tex]

and that, by definition, we have

[tex]v_{\rm avg}=\dfrac{\Delta x}{\Delta t}[/tex]

where [tex]v_f[/tex] and [tex]v_i[/tex] are the final/initial speeds of the car and [tex]\Delta x[/tex] is the displacement it undergoes. It starts with a speed of 44 ft/s and ends with a speed of 88 ft/s, so we have

[tex]\dfrac{88\frac{\rm ft}{\rm s}+44\frac{\rm ft}{\rm s}}2=\dfrac{\Delta x}{20\,\rm s}\implies\Delta x=1320\,\mathrm{ft}[/tex]

same as before.

To find the distance traveled by the car during the 20 seconds, we integrate the velocity function and solve for the distance using the given values. The car travels a distance of 1320 feet.

To find the distance traveled by the car during the 20 seconds, we need to calculate the area under the velocity-time graph. The velocity function given is v(t)=44+2.2t. To find the distance, we integrate the velocity function from 0 to 20 seconds:

d = ∫(44+2.2t) dt

Applying integration, we get: d = 44t + 1.1t^2

Substituting the values t=0 and t=20 into the equation, we can find the distance traveled by the car:

d = 44(20) + 1.1(20)^2

Solving this equation, we get d = 880 + 440

So, the car has traveled a distance of 1320 feet during the 20 seconds.

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Answer:

The answer is (b)

Step-by-step explanation:

* Lets check how to find the inverse of the matrix,

 its dimensions is 2 × 2

* To know if the inverse of the matrix exist find the determinant

- If its not equal 0, then it exist

* How to find the determinant

- It is the difference between the multiplication of

 the diagonals of the matrix

Ex: If the matrix is [tex]\left[\begin{array}{ccc}a&b\\c&d\end{array}\right][/tex]

     its determinant = ad - bc

- After that lets swap the positions of a and d, put negatives

 in front of b and c, and divide everything by the determinant

- The inverse will be [tex]\left[\begin{array}{ccc}\frac{d}{ad-bc} &\frac{-b}{ad-bc}\\\frac{-c}{ad-bc} &\frac{a}{ad-bc}\end{array}\right][/tex]

* Lets do that with our problem

∵ The determinant = (9 × 9) - (-2 × -10) = 81 - 20 = 61

- The determinant ≠ 0, then the inverse is exist

∴ The inverse is [tex]\frac{1}{61}\left[\begin{array}{ccc}9&2\\10&9\end{array}\right][/tex]=

   [tex]\left[\begin{array}{ccc}\frac{9}{61}&\frac{2}{61}\\\frac{10}{61} &\frac{9}{61}\end{array}\right][/tex]

* The answer is (b)

Answer:

The residual value is -0.75

Step-by-step explanation:

we know that  

The residual value is the observed value minus the predicted value.

RESIDUAL VALUE=[OBSERVED VALUE-PREDICTED VALUE]

where

Predicted value.--> the predicted value given the current regression equation

Observed value. --> The observed value for the dependent variable.

in this problem

we have the point (1,4)

so

The observed value is 4

Find the predicted value  for x=1

[tex]y =1.5(1)+3.25=4.75[/tex]

predicted value is 4.75

so

RESIDUAL VALUE=(4-4.75)=-0.75

Answer:

-0.75

Step-by-step explanation:

since the numerator is x² + 5x - 3, and therefore has a degree of 2, whilst the denominator, 4x¹ - 1, has a degree of 1, therefore, there's no horizontal asymptote.

recall, we only get a horizontal asymptote if the denominator's expression degree is equals or greater than that of the numerator's.

The function [tex]f(x) = (x^2+5x-3)/(4x-1)[/tex] does not have a horizontal asymptote because the degree of the numerator is higher than the degree of the denominator.

To identify the horizontal asymptote of the function

[tex]f(x) = \frac{{x^2+5x-3}}{{4x-1}}[/tex], you can examine the degrees of the polynomial in the numerator and the polynomial in the denominator. Since the degree of the numerator (which is 2) is higher than the degree of the denominator (which is 1), this function does not have a horizontal asymptote. However, for functions like

[tex]f(x) = \frac{{x^2+3}}{{x^2+4}}[/tex], where the degrees of the numerator and denominator are the same, the horizontal asymptote is determined by the leading coefficients of the numerator and denominator. Specifically, the horizontal asymptote is

[tex]y = \frac{{1}}{{1}}[/tex] = 1,

since the coefficients of the x^2 terms are both 1.

2/3 is the answer because 6 in is the model and 4 ft is the actual

Answer:

The  ratio of the height of the actual table to the height of the model table is [tex]\frac{8}{1}[/tex] .

Step-by-step explanation:

As given

A table is 4 ft high. A model of the table is 6 in. high.

As

1 foot = 12 inch

Now convert 4 ft into inches .

4 ft = 4 × 12

     = 48 inches

Height of the actual table = 48 inches

Now the ratio of the height of the actual table to the height of the model

table .

[tex]Ratio\ of\ the\ height\ of\ the\ actual\ table\ to\ the\ height\ of\ the\ model\ table =\frac{48}{6}[/tex]

[tex]Ratio\ of\ the\ height\ of\ the\ actual\ table\ to\ the\ height\ of\ the\ model\ table =\frac{8}{1}[/tex]

Therefore the  ratio of the height of the actual table to the height of the model table is [tex]\frac{8}{1}[/tex] .

Answer:

[tex]\large\boxed{N-T=-a^2+a-11}[/tex]

Step-by-step explanation:

[tex]T=-2a^2+a+6\\N=-3a^2+2a-5\\\\N-T=?\\\\\text{Substitute:}\\\\N-T=(-3a^2+2a-5)-(-2a^2+a+6)\\\\N-T=-3a^2+2a-5-(-2a^2)-a-6\\\\N-T=-3a^2+2a-5+2a^2-a-6\qquad\text{combine like terms}\\\\N-T=(-3a^2+2a^2)+(2a-a)+(-5-6)\\\\N-T=-a^2+a-11[/tex]

The difference between the functions is [tex]-a^2+a-11[/tex]

Given the following expression:

[tex]T=-2a^2+a+6\\N=-3a^2+2a-5[/tex]

We are to take the difference between N and T and this is as shown:

[tex]N - T= -3a^2+2a-5-(-2a^2+a+6)\\Expand\\N - T= -3a^2+2a-5+2a^2-a-6\\\\Collect \ the \ like \ terms\\N-T=-3a^2+2a^2+2a-a-5-6\\N-T=-a^2+a-11[/tex]

Hence the difference between the functions is [tex]-a^2+a-11[/tex]

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False. That does not satify the equation

Answer:

False

Step-by-step explanation:

The given inequality  is [tex]-3 \:<\:x\:<\:14[/tex].

Since both boundaries of the inequalities are not inclusive , we use the parenthesis for open interval."()".

We write the given inequality in interval notation as;

[tex](-3,14)[/tex].

The correct choice is false

➷ Pythagoras' theorem is only suitable for right triangles, which this isn't.

The Sine rule would not be applicable as there isn't any side and paired angle given

The best option for this would be the Law of Cosines as it is suitable for when you are given two sides and an angle between the.

➶ Hope This Helps You!

➶ Good Luck (:

➶ Have A Great Day ^-^

↬ ʜᴀɴɴᴀʜ ♡

Answer:  Third option is correct.

Step-by-step explanation:

Since we have given that

ABC is triangle with its dimensions:

AB = 11

AC = 13

∠A = 108°

BC = a

We need to find the length of 'a'.

So, we can use "Law of cosines" as we have given two sides and one angle.

So, it becomes,

[tex]\cos A=\dfrac{b^2+c^2-a^2}{2bc}\\\\\cos 108^\circ=\dfrac{11^2+13^2-a^2}{2\times 13\times 11}\\\\-0.3=\dfrac{121+169-a^2}{286}\\\\-0.3\times 286=290-a^2\\\\-85.8=290-a^2\\\\-85.8-290=-a^2\\\\375.8=a^2\\\\a=\sqrt{375.8}\\\\a=19.38[/tex]

Hence, Third option is correct.

Answer:

10 servings

Step-by-step explanation:

Divide the total juice available, 2 quarts, by the serving size, (1/5) quart per serving:

 2 quarts                     2        5

--------------------------- = ----  ·  ------ servings = 10 servings

(1/5) quart/serving       1         1

Answer:

 x² - 3x - 10 = 0

Step-by-step explanation:

Given there are 2 solutions then the equation is a quadratic.

Since the solutions are x = - 2 and x = 5 then

the factors are (x + 2) and (x - 5) and

f(x) = (x + 2)(x - 5) ← expand factors

     = x² - 3x - 10, hence the equation is

x² - 3x - 10 = 0

The equation that yields the solutions x = -2 and x = 5 is: x^2 + 0.00088x - 0.000484 = 0. We can solve this equation using the quadratic formula.

The equation that yields the solutions x = -2 and x = 5 is:

x^2 + 0.00088x - 0.000484 = 0

To solve this equation, we can use the quadratic formula:

x = (-b +/- sqrt(b^2 - 4ac))/(2a)

Plugging in the values from the equation, we get:

x = (-0.00088 +/- sqrt((0.00088)^2 - 4(1)(-0.000484)))/(2(1))

Simplifying further, we have:

x = (-0.00088 +/- sqrt(0.0000007744 + 0.001936))/0.002

Continuing to simplify, we get:

x = (-0.00088 +/- sqrt(0.0027104))/0.002

Finally, we have the two possible solutions:

x = (-0.00088 + sqrt(0.0027104))/0.002 and x = (-0.00088 - sqrt(0.0027104))/0.002

the answer is A, what they changed is the (x) with (a+h), so the right side equation should be changed the same way just like A.

Answer:

[tex]\large\boxed{1.\ f(1)=135}\\\boxed{2.\ f(6)=115}\\\boxed{3.\ f(26)=25}\\\boxed{4.\ f(n)=139-4n}[/tex]

Step-by-step explanation:

[tex]f(1)=135\\f(2)=135-4=131\\f(3)=131-4=127\\f(4)=127-4=123\\f(5)=123-4=119\\\vdots\\\\\text{It's an arithmetic sequence with firs term = 135 and the common}\\\text{difference d = -4.}\\\text{The formula of arithmetic sequence: }\\\\f(n)=f(1)+(n-1)d\\\\\text{We have}\ f(1)=135\ \text{and}\ =-4.\ \text{Substitute:}\\\\f(n)=135+(n-1)(-4)=135+(n)(-4)+(-1)(-4)\\=135-4n+4=139-4n\\\\\boxed{f(n)=139-4n}[/tex]

[tex]\text{Put n = 6, n=26 to the formula:}\\\\f(6)=139-4(6)=139-24=115\\\\f(26)=139-4(26)=139-104=25[/tex]

Answer:

108π cm^3/min

Step-by-step explanation:

At a time of t min, let the height be h cm

The volume of a cylinder;

V = π r^2 h

   = 36π h

differentiating both sides with respect to t;

dV/dt = 36π dh/dt

but dh/dt = 3 cm/min

dV/dt = 36π(3) = 108π cm^3/min

Answer:

The rate of change of the volume of the cylinder when the height is 20 cm is [tex]\frac{dV}{dt}=108\pi \:{\frac{cm^3}{min} }[/tex]

Step-by-step explanation:

This is a related rates problem. In this problem, you need to find a relationship between the quantity whose rate of change you want to find, the volume in this case, and the quantity whose rates of change you know, the height of the cylinder.

We know that the volume of the cylinder is

[tex]V=\pi r^2h[/tex]

We also know that the radius is a constant, 6 cm and thus

[tex]V=\pi (6)^2h=36\pi h[/tex]

V and h both vary with time so you can differentiate both sides with respect to time, t, to get

[tex]\frac{dV}{dt}=36\pi \frac{dh}{dt}[/tex]

Now use the fact that [tex]\frac{dh}{dt}=3 \:{\frac{cm}{min}[/tex] to find [tex]\frac{dV}{dt}[/tex].

[tex]\frac{dV}{dt}=36\pi (3)=108\pi[/tex]

Answer:  [tex]266,407.057\ km[/tex]

Step-by-step explanation:

The formula used to calculate the circumference of a circle is:

[tex]C=2\pi r[/tex]

The radius of the circle is r.

In the diagram you can observe that the radius of the satellite's orbit (r2) is the sum of the radius of the Earth (r1) and the distance from the Earth's surface to the satellite's orbit:

[tex]r2=r1+36,000\ km\\r2=6,400\ km+36,000\ km\\r2=42,400\ km[/tex]

Then, the circumference of the satellite's orbit is:

[tex]C=2\pi (42,400\ km)\\C=266,407.057\ km[/tex]

The circumference of the satellite's orbit is calculated by adding Earth's radius to the satellite's distance from Earth's surface to determine the orbit radius. The circumference is then found by using the formula for the circumference of a circle, 2πr, giving approximately 266,433 kilometers.

To find the circumference of the satellite's orbit, we need to first calculate the total distance from the center of Earth to the satellite. This is the sum of the Earth's radius (6400 kilometers) and the satellite's distance from the Earth's surface (36000 kilometers), which totals 42400 kilometers.

Once we have the radius of the orbit, we can calculate the circumference using the formula for the circumference of a circle, which is 2πr (two times Pi times the radius). Using this formula, the circumference of the satellite's orbit is approximately 266,433 kilometers.

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Answer:

Step-by-step explanation:

[tex]3\sqrt2\cos\theta+2=-1\\\\\text{Method 1}\\\\\text{Put}\ \theta=\dfrac{\pi}{4}\ \text{to the equation and check the equality:}\\\\\cos\dfrac{\pi}{4}=\dfrac{\sqrt2}{2}\\\\L_s=3\sqrt2\cos\dfrac{\pi}{4}+2=3\sqrt2\left(\dfrac{\sqrt2}{2}\right)+2=\dfrac{(3\sqrt2)(\sqrt2)}{2}+2\\\\=\dfrac{(3)(2)}{2}+2=3+2=5\\\\R_s=-1\\\\L_s\neq R_s\\\\\boxed{FALSE}[/tex]

[tex]\text{Method 2}\\\\\text{Solve the equation:}\\\\3\sqrt2\cos\theta+2=-1\qquad\text{subtract 2 from both sides}\\\\3\sqrt2\cos\theta=-3\qquad\text{divide both sides by}\ 3\sqrt2\\\\\cos\theta=-\dfrac{3}{3\sqrt2}\\\\\cos\theta=-\dfrac{1}{\sqrt2}\cdot\dfrac{\sqrt2}{\sqrt2}\\\\\cos\theta=-\dfrac{\sqrt2}{2}\to\theta=\dfrac{3\pi}{4}+2k\pi\ \vee\ \theta=-\dfrac{3\pi}{4}+2k\pi\ \text{for}\ k\in\mathbb{Z}\\\\\text{It's not equal to}\ \dfrac{\pi}{4}\ \text{for any value of }\ k.[/tex]

By writing the quotient in lowest terms, 3 and 1/8 divided by 1 and 2/3 equals 5 and 5/24.

To divide 3 and 1/8 by 1 and 2/3, we can follow these steps:

Step 1: Convert the mixed numbers to improper fractions.

3 and 1/8 = (3 * 8 + 1) / 8 = 25 / 8

1 and 2/3 = (1 * 3 + 2) / 3 = 5 / 3

Step 2: Invert the divisor (the second fraction) and multiply.

25/8 ÷ 3/5 = 25/8 * 5/3

Step 3: Simplify the fractions if possible.

The numerator of 25/8 and the denominator of 5/3 have a common factor of 5.

25/8 * 5/3 = (5 * 25) / (8 * 3) = 125/24

Step 4: Express the improper fraction as a mixed number (if necessary).

125/24 can be expressed as 5 and 5/24.

Therefore, 3 and 1/8 divided by 1 and 2/3 equals 5 and 5/24.

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The solution to [tex]\(3\dfrac{1}{8} \div 1\dfrac23\) is \(\frac{15}{8}\),[/tex] expressed as a fraction in its simplest form after converting the mixed numbers to improper fractions and performing division.

Let's solve [tex]\(3\dfrac{1}{8} \div 1\dfrac23\)[/tex]

convert the mixed numbers into improper fractions:

[tex]\(3\dfrac{1}{8} = \frac{3 \times 8 + 1}{8} = \frac{24 + 1}{8} = \frac{25}{8}\)[/tex]

[tex]\(1\dfrac23 = \frac{1 \times 3 + 2}{3} = \frac{3 + 2}{3} = \frac{5}{3}\)[/tex]

Now, we have:

[tex]\(\frac{25}{8} \div \frac{5}{3}\)[/tex]

To divide by a fraction, we multiply by its reciprocal:

[tex]\(\frac{25}{8} \times \frac{3}{5}\)[/tex]

Multiply the numerators and denominators:

Numerator:[tex]\(25 \times 3 = 75\)[/tex]

Denominator: [tex]\(8 \times 5 = 40\)[/tex]

Therefore, [tex]\(3\dfrac{1}{8} \div 1\dfrac23 = \frac{75}{40}\)[/tex]

Reduce the fraction to its simplest form by dividing both the numerator and denominator by their greatest common divisor, which is 5:

[tex]\(\frac{75}{40} = \frac{75 \div 5}{40 \div 5} = \frac{15}{8}\)[/tex]

Hence,[tex]\(3\dfrac{1}{8} \div 1\dfrac23 = \frac{15}{8}\).[/tex]

To answer the student's question, we list each possible pair of marble colors drawn without replacement from a bag with a white, red, and blue marble: White-Red, White-Blue, Red-White, Red-Blue, Blue-White, and Blue-Red.

The question asks for the display of all possible outcomes when two marbles are drawn from a bag containing a white, a red, and a blue marble, without replacement. To show all possible outcomes, we can list them in an organized manner, considering each color once it is drawn, is not put back into the bag. The first marble drawn can be any one of the three colors. Once a marble is drawn, there are only two colors left for the second draw.

angle GAL would be the same as MLA

Answer:

the answer is 5

Step-by-step explanation:

25/5=5 & 5*5=25

Answer:

The Answer Is 5 because every square number has to equal the number by multiplying by 2 to get your Answer 5 x 5 = 25 which 5 is multiplied 2 times 5 and 5 which gives you your answer 25.

Step-by-step explanation:

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