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vichka [17]
3 years ago
15

jared is painting a wall in his living room that is 20 feet high and 18 feet long there are 3 windows on the wall each window is

3 feet wide and 4 feet tall how many square feet will jered need to paint
Mathematics
1 answer:
SSSSS [86.1K]3 years ago
3 0

20 high x 18 long = 360 square feet

3 windows x (4 high x 3 long) = 3 windows x 12 square feet = 36 square feet

360 - 36 (because we obviously don't paint windows) = 324

324 square feet to paint on one wall

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Hello people ~
Luden [163]

Cone details:

  • height: h cm
  • radius: r cm

Sphere details:

  • radius: 10 cm

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

From the endpoints (EO, UO) of the circle to the center of the circle (O), the radius is will be always the same.

<u>Using Pythagoras Theorem</u>

(a)

TO² + TU² = OU²

(h-10)² + r² = 10²                                   [insert values]

r² = 10² - (h-10)²                                     [change sides]

r² = 100 - (h² -20h + 100)                       [expand]

r² = 100 - h² + 20h -100                        [simplify]

r² = 20h - h²                                          [shown]

r = √20h - h²                                       ["r" in terms of "h"]

(b)

volume of cone = 1/3 * π * r² * h

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

\longrightarrow \sf V = \dfrac{1}{3}  * \pi  * (\sqrt{20h - h^2})^2  \  ( h)

\longrightarrow \sf V = \dfrac{1}{3}  * \pi  * (20h - h^2)  (h)

\longrightarrow \sf V = \dfrac{1}{3}  * \pi  * (20 - h) (h) ( h)

\longrightarrow \sf V = \dfrac{1}{3} \pi h^2(20-h)

To find maximum/minimum, we have to find first derivative.

(c)

<u>First derivative</u>

\Longrightarrow \sf V' =\dfrac{d}{dx} ( \dfrac{1}{3} \pi h^2(20-h) )

<u>apply chain rule</u>

\sf \Longrightarrow V'=\dfrac{\pi \left(40h-3h^2\right)}{3}

<u>Equate the first derivative to zero, that is V'(x) = 0</u>

\Longrightarrow \sf \dfrac{\pi \left(40h-3h^2\right)}{3}=0

\Longrightarrow \sf 40h-3h^2=0

\Longrightarrow \sf h(40-3h)=0

\Longrightarrow \sf h=0, \ 40-3h=0

\Longrightarrow \sf  h=0,\:h=\dfrac{40}{3}<u />

<u>maximum volume:</u>                <u>when h = 40/3</u>

\sf \Longrightarrow max=  \dfrac{1}{3} \pi (\dfrac{40}{3} )^2(20-\dfrac{40}{3} )

\sf \Longrightarrow maximum= 1241.123 \ cm^3

<u>minimum volume:</u>                 <u>when h = 0</u>

\sf \Longrightarrow min=  \dfrac{1}{3} \pi (0)^2(20-0)

\sf \Longrightarrow minimum=0 \ cm^3

6 0
2 years ago
Read 2 more answers
A rectangular prism has a height of 3,a length of 4,and a width of 12,what is the surface area?
NNADVOKAT [17]
It would be 144 to the 2 power

4 0
3 years ago
Read 2 more answers
Please answer correctly !!!!!!! Will mark brainliest answer !!!!!!!!
Allisa [31]

Answer:

A is the answer i believe

Step-by-step explanation:

3 0
3 years ago
4times 1 1/5 as a product of a mixed number
natka813 [3]
The answer is 24/5 or 4 4/5 but if it is 11 on 5, the answer would be 44/5 or 8 4/5
7 0
3 years ago
Find the standard equation of the circle with center (5, -3
Mrac [35]

Answer:

○ C

Explanation:

Accourding to one of the circle equations, \displaystyle (x - h)^2 + (y - k)^2 = r^2,the centre of the circle is represented by \displaystyle (h, k).Moreover, all negative symbols give you the OPPOCITE TERMS OF WHAT THEY <em>REALLY</em> ARE, so you must pay cloce attention to which term gets which symbol. Another thing you need to know is that the radius will ALWAYS be squared, so no matter how your equation comes about, make sure that the radius is squared. Now, in case you did not know how to define the radius, you can choose between either method:

Pythagorean Theorem

\displaystyle a^2 + b^2 = c^2 \\ \\ 6^2 + 8^2 = r^2 \hookrightarrow 36 + 64 = r^2 \hookrightarrow \sqrt{100} = \sqrt{r^2} \\ \\ \boxed{10 = r}

Sinse we are dealing with <em>length</em>, we only desire the NON-NEGATIVE root.

Distanse Equation

\displaystyle \sqrt{[-x_1 + x_2]^2 + [-y_1 + y_2]^2} = d \\ \\ \sqrt{[-5 - 3]^2 + [3 + 3]^2} = r \hookrightarrow \sqrt{[-8]^2 + 6^2} = r \hookrightarrow \sqrt{64 + 36} = r; \sqrt{100} = r \\ \\ \boxed{10 = r}

Sinse we are dealing with <em>distanse</em>, we only desire the NON-NEGATIVE root.

I am joyous to assist you at any time.

5 0
2 years ago
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