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2 years ago

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Selina is planning to paint a large picture on a wall. She draws a smaller version first. The drawing is 8 inches by 6 inches. I

f the scale of the drawing is 2 in : 1 ft, what is the area of the actual picture on the wall?
If it's possible, can you list step-by-step the procedures? Thanks!

1 answer:

stich3 [128]2 years ago

6 0

Do you have a graph or of any sort with this question? If so that would help a lot!

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The square pyramid pictured below has a surface area of

Evgen [1.6K]

Answer:

Step-by-step explanation:

Triangles

Area of a triangle = 1/2 b * slanted height

b = 6 m

h = 9 m

Area of 1 triangle = 1/2 * 6 * 9

Area of 1 triangle = 27

Area of all 4 triangles = 4*27 = 108 m^2

==================

Base area

The base is a square. All 4 sides are equal.

The formula for a square is s^2

Area = (6 m)^2

Area = 36 m^2

Total area

Total Area = 108 m^2 + 36 m ^2

Total Area = 144 m^2

8 0

2 years ago

Read 2 more answers

David’s bowling score is 5 less than 3 times Aaron’s score. The sum of their score is 215. Find the score of each student. Write

xenn [34]

Answer:

David's score= 160 ; Aaron's score = 55

Step-by-step explanation:

let Aaron's score be 'x'.

so, David's score will be 3x - 5

(3x - 5) + x = 215

3x + x - 5 = 215

4x = 215 + 5

4x = 220

x = 220/4

x = 55

therefore, Aaron's score = 55

David's score is (3x - 5) = (3*55) - 5 = 165 - 5 = 160

8 0

3 years ago

strojnjashka [21]

Answer:

The candle has a radius of 8 centimeters and 16 centimeters and uses an amount of approximately 1206.372 square centimeters.

Step-by-step explanation:

The volume ( $V$ ), in cubic centimeters, and surface area ( $A_{s}$ ), in square centimeters, formulas for the candle are described below:

$V = \pi\cdot r^{2}\cdot h$ (1)

$A_{s} = 2\pi\cdot r^{2} + 2\pi\cdot r \cdot h$ (2)

Where:

$r$ - Radius, in centimeters.

$h$ - Height, in centimeters.

By (1) we have an expression of the height in terms of the volume and the radius of the candle:

$h = \frac{V}{\pi\cdot r^{2}}$

By substitution in (2) we get the following formula:

$A_{s} = 2\pi \cdot r^{2} + 2\pi\cdot r\cdot \left(\frac{V}{\pi\cdot r^{2}} \right)$

$A_{s} = 2\pi \cdot r^{2} +\frac{2\cdot V}{r}$

Then, we derive the formulas for the First and Second Derivative Tests:

First Derivative Test

$4\pi\cdot r -\frac{2\cdot V}{r^{2}} = 0$

$4\pi\cdot r^{3} - 2\cdot V = 0$

$2\pi\cdot r^{3} = V$

$r = \sqrt[3]{\frac{V}{2\pi} }$

There is just one result, since volume is a positive variable.

Second Derivative Test

$A_{s}'' = 4\pi + \frac{4\cdot V}{r^{3}}$

If $\left(r = \sqrt[3]{\frac{V}{2\pi}}\right)$ :

$A_{s} = 4\pi + \frac{4\cdot V}{\frac{V}{2\pi} }$

$A_{s} = 12\pi$ (which means that the critical value leads to a minimum)

If we know that $V = 3217\,cm^{3}$ , then the dimensions for the minimum amount of plastic are:

$r = \sqrt[3]{\frac{V}{2\pi} }$

$r = \sqrt[3]{\frac{3217\,cm^{3}}{2\pi}}$

$r = 8\,cm$

$h = \frac{V}{\pi\cdot r^{2}}$

$h = \frac{3217\,cm^{3}}{\pi\cdot (8\,cm)^{2}}$

$h = 16\,cm$

And the amount of plastic needed to cover the outside of the candle for packaging is:

$A_{s} = 2\pi\cdot r^{2} + 2\pi\cdot r \cdot h$

$A_{s} = 2\pi\cdot (8\,cm)^{2} + 2\pi\cdot (8\,cm)\cdot (16\,cm)$

$A_{s} \approx 1206.372\,cm^{2}$

The candle has a radius of 8 centimeters and 16 centimeters and uses an amount of approximately 1206.372 square centimeters.

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3 years ago

A like has two-fold Symmetry

Vikki [24]

Answer:

yes

Step-by-step explanation:

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Find the DISTANCE of line segment AB with point A

Lelu [443]

Answer:

34

Step-by-step explanation:

x-y

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