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3 years ago

Please help with this! Will give brainliest if the answer is right!

Mathematics

2 answers:

Lesechka [4]3 years ago

8 0

Answer:

336 square feet.

Step-by-step explanation:

Remark

The question is what to do about the roof and the floor. Does the roof get painted or not? Does he need to include paint for the floor which we cannot see?

I will assume no to the roof and no for the floor. That's arbitary. Lateral Area is the perimeter of the base * height.

Solution

Front and Back

Area of the Front and Back is 8*14. Since there are 2 it is 2 * 8 * 14 = 224

Area of the left and right side = 2 * 7 *8 = 112

Total area = 112 + 224 = 336

Total 336 square feet

musickatia [10]3 years ago

4 0

Answer:

784

Step-by-step explanation:

caculator. of course.

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The answer to this is .960

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Which real world image represents a negative slope?

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Endpoint 1: (-1,2) Midpoint: (9,-6) Endpoint 2=

Vlad [161]

Answer:

(19 , -14)

Step-by-step explanation:

Find the distance in between each x & y for a coordinate.

Let: (x₁ , y₁) = (-1 , 2)

Let: (x₂ , y₂) = (9 , -6)

From x₁ ⇒ x₂: 9 - (-1) = 10

From y₁ ⇒ y₂: -6 - 2 = -8 = 8*

*Remember that distance cannot be negative, but for the sake of this question, we will leave it as -8.

The distance between the x points are in intervals of 10. The distance between the y points are in intervals of 8. Add 10 & subtract 8 to their respective numbers to get endpoint 2:

(9 (+ 10) , -6 (- 8)) = (19 , -14)

Endpoint 2 = (19 , -14)

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3 years ago

A wire b units long is cut into two pieces. One piece is bent into an equilateral triangle and the other is bent into a circle.

mezya [45]

1. Divide wire b in parts x and b-x.

2. Bend the b-x piece to form a triangle with side (b-x)/3

There are many ways to find the area of the equilateral triangle. One is by the formula A= $\frac{1}{2}sin60^{o}side*side= \frac{1}{2} \frac{ \sqrt{3} }{2} (\frac{b-x}{3}) ^{2}= \frac{ \sqrt{3} }{36}(b-x)^{2}$
A= $\frac{ \sqrt{3} }{36}(b-x)^{2}=\frac{ \sqrt{3} }{36}( b^{2}-2bx+ x^{2} )=\frac{ \sqrt{3} }{36}b^{2}-\frac{ \sqrt{3} }{18}bx+ \frac{ \sqrt{3} }{36}x^{2}$

Another way is apply the formula A=1/2*base*altitude,
where the altitude can be found by applying the pythagorean theorem on the triangle with hypothenuse (b-x)/3 and side (b-x)/6

3. Let x be the circumference of the circle.

$2 \pi r=x$

so $r= \frac{x}{2 \pi }$

Area of circle = $\pi r^{2}= \pi ( \frac{x}{2 \pi } )^{2} = \frac{ \pi }{ 4 \pi ^{2} }* x^{2} = \frac{1}{4 \pi } x^{2}$

4. Let f(x)= $\frac{ \sqrt{3} }{36}b^{2}-\frac{ \sqrt{3} }{18}bx+ \frac{ \sqrt{3} }{36}x^{2}+\frac{1}{4 \pi } x^{2}$

be the function of the sum of the areas of the triangle and circle.

5. f(x) is a minimum means f'(x)=0

f'(x)= $\frac{ -\sqrt{3} }{18}b+ \frac{ \sqrt{3} }{18}x+\frac{1}{2 \pi } x$ =0

$\frac{ -\sqrt{3} }{18}b+ \frac{ \sqrt{3} }{18}x+\frac{1}{2 \pi } x=0$

$(\frac{ \sqrt{3} }{18}+\frac{1}{2 \pi }) x=\frac{ \sqrt{3} }{18}b$

$x= \frac{\frac{ \sqrt{3} }{18}b}{(\frac{ \sqrt{3} }{18}+\frac{1}{2 \pi }) }$

6. So one part is $\frac{\frac{ \sqrt{3} }{18}b}{(\frac{ \sqrt{3} }{18}+\frac{1}{2 \pi }) }$ and the other part is b- $\frac{\frac{ \sqrt{3} }{18}b}{(\frac{ \sqrt{3} }{18}+\frac{1}{2 \pi }) }$

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3 years ago

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HELPSolve for x X = A: 1 B: 3 C: 7

OverLord2011 [107]

X equals 7

Hope that helped

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3 years ago