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3 years ago

Find the missing side of triangle

Mathematics

1 answer:

Maksim231197 [3]3 years ago

7 0

Final Answer:

x = 20

Step-by-step explanation:

we'll be using the pythagorean theorem method, in this triangle the missing letter is a.

formula: $a=\sqrt{c^2-b^2}$

a = x

b = 21

c = 29

a = $\sqrt{29^2-21^2}$

a = $\sqrt{841-441}$ (note: 29² = 29 × 29 = 841 and 21² = 21 × 21 = 441)

a = $\sqrt{400}$

a = 20

x = 20

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What is the area of the piece of red paper after the hole for

MakcuM [25]

Answer:

$\large\boxed{\large\boxed{47unit^2}}$

Explanation:

The complete question is:

A hole the size of a photograph is cut from a red piece of paper to use in a picture frame.

On a coordinate plane, 2 squares are shown. The photograph has points (-3, -2), (- 2, 2), (2, 1), and (1, -3). The red paper has points (- 4, 4), (4, 4), (4, -4), and (-4, -4).

What is the area of the piece of red paper after the hole for the photograph has been cut?

<h2>Solution</h2>

The area of the piece of redpaper after the hole has been cut is equal to the area of the big square less the area of the small rectangle.

1. Area of the big rectangle

Vertices:

(- 4, 4)
(4, 4)
(4, -4)
(-4, -4)

The side length is the distance between two consecutive vertices:

The two points (-4,4) and (4,4) has the same y-coordinate, thus the length is just the absolute value of the difference of the x-coordinates:

$4-(-4)=4+4=8$

The area of this square is $(8unit)^2=64unit^2$

2. Area of the small square

Vertices:

(-3, -2)
(- 2, 2)
(2, 1)
(1, -3)

To find the side length use the distance formula for two consecutive points, for instance (2,1) and (-2,2):

$d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\\ \\d=\sqrt{(2-(-2))^2+(2-1)^2}=\sqrt{4^2+1^2}=\sqrt{17}$

The area is:

$(\sqrt{17}unit)^2}=17unit^2$

3. Area of the piece of red paper after the holw for the photograph has been cut:

Find the difference:

$64unit^2-17unit^2=47unit^2$

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Explain how to find the missing side of a right triangle when given an acute angle and one side

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Answer:

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Step-by-step explanation:

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Drupady [299]

Answer:

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Step-by-step explanation:

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3 years ago

99 POINT QUESTION, PLUS BRAINLIEST!!!

sattari [20]

We know, that the area of the surface generated by revolving the curve y about the x-axis is given by:

$\boxed{A=2\pi\cdot\int\limits_a^by\sqrt{1+\left(y'\right)^2}\, dx}$

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So there will be:

$A=2\pi\cdot\int\limits_0^{15}\dfrac{x^3}{15}\cdot\sqrt{1+\left(\dfrac{x^2}{5}\right)^2}\, dx=\dfrac{2\pi}{15}\cdot\int\limits_0^{15}x^3\cdot\sqrt{1+\dfrac{x^4}{25}}\,\, dx=\left(\star\right)\\\\-------------------------------\\\\
\int x^3\cdot\sqrt{1+\dfrac{x^4}{25}}\,\,dx=\int\sqrt{1+\dfrac{x^4}{25}}\cdot x^3\,dx=\left|\begin{array}{c}t=1+\dfrac{x^4}{25}\\\\dt=\dfrac{4x^3}{25}\,dx\\\\\dfrac{25}{4}\,dt=x^3\,dx\end{array}\right|=\\\\\\$

$=\int\sqrt{t}\cdot\dfrac{25}{4}\,dt=\dfrac{25}{4}\int\sqrt{t}\,dt=\dfrac{25}{4}\int t^\frac{1}{2}\,dt=\dfrac{25}{4}\cdot\dfrac{t^{\frac{1}{2}+1}}{\frac{1}{2}+1}= \dfrac{25}{4}\cdot\dfrac{t^{\frac{3}{2}}}{\frac{3}{2}}=\\\\\\=\dfrac{25\cdot2}{4\cdot3}\,t^\frac{3}{2}=\boxed{\dfrac{25}{6}\,\left(1+\dfrac{x^4}{25}\right)^\frac{3}{2}}\\\\-------------------------------\\\\$

$\left(\star\right)=\dfrac{2\pi}{15}\cdot\int\limits_0^{15}x^3\cdot\sqrt{1+\dfrac{x^4}{25}}\,\, dx=\dfrac{2\pi}{15}\cdot\dfrac{25}{6}\cdot\left[\left(1+\dfrac{x^4}{25}\right)^\frac{3}{2}\right]_0^{15}=\\\\\\=
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\boxed{\dfrac{5\Big(2026^\frac{3}{2}-1\Big)}{9}\pi}$

Answer C.

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3 years ago