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3 years ago

8

What is the length of the hypotenuse of the triangle below?

1 answer:

Lubov Fominskaja [6]3 years ago

4 0

Answer:

D

Step-by-step explanation:

Using Pythagoras' identity in the right triangle

The square oh the hypotenuse is equal to the sum of the squares on the other 2 sides, that is

h² = (9 $\sqrt{2}$ )² + (9 $\sqrt{2}$ )² = 162 + 162 = 324 ( take the square root of both sides )

h = $\sqrt{324}$ = 18 → D

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Evaluate M/q + np2 for m = 24, n = 6, p = 9, and q = 8.

irina1246 [14]

Answer:

489

Step-by-step explanation:

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4 years ago

In trapezoid ABCD, find m∠A if m∠B= 25°, m∠C = 155° and m∠ D= 117°.

telo118 [61]

Hello there! A trapezoid is a quadrilateral, and the angles of a quadrilateral have a sum of 360°. We don't know the angle measurement for A, so let's start off by adding the measurements of the other angles together. 117 + 155 + 25 is 297. Now, we subtract that number from 360 to find the angle measurement. 360 - 297 is 63. There. The measurement of angle A is 63°.

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3 years ago

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DiKsa [7]

Answer:

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3 years ago

Area of parallelogram

Aliun [14]

The formula for the area of a parallelogram is base times height or bh.

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3 years ago

Read 2 more answers

Two solutions to y'' – 2y' – 35y = 0 are yı = e, Y2 = e -5t a) Find the Wronskian. W = 0 Preview b) Find the solution satisfying

pashok25 [27]

Answer:

a. $w(t)=-12e^{2t}$

b. $y(t)=-\frac{9}{2}e^{7t}-\frac{5}{2}e^{-5t}$

Step-by-step explanation:

We have a differential equation

y''-2 y'-35 y=0

Auxillary equation

$(D^2-2D-35)=0$

By factorization method we are finding the solution

$D^2-7D+5D-35=0$

$(D-7)(D+5)=0$

Substitute each factor equal to zero

D-7=0 and D+5=0

D=7 and D=-5

Therefore ,

General solution is

$y(x)=C_1e^{7t}+C_2e^{-5t}$

Let $y_1=e^{7t} \;and \;y_2=e^{-5t}$

We have to find Wronskian

$w(t)=\begin{vmatrix}y_1&y_2\\y'_1&y'_2\end{vmatrix}$

Substitute values then we get

$w(t)=\begin{vmatrix}e^{7t}&e^{-5t}\\7e^{7t}&-5e^{-5t}\end{vmatrix}$

$w(t)=-5e^{7t}\cdot e^{-5t}-7e^{7t}\cdot e^{-5t}=-5e^{7t-5t}-7e^[7t-5t}$

$w(t)=-5e^{2t}-7e^{2t}=-12e^{2t}$

a. $w(t)=-12e^{2t}$

We are given that y(0)=-7 and y'(0)=23

Substitute the value in general solution the we get

$y(0)=C_1+C_2$

$C_1+C_2=-7$ ....(equation I)

$y'(t)=7C_1e^{7t}-5C_2e^{-5t}$

$y'(0)=7C_1-5C_2$

$7C_1-5C_2=23$ ......(equation II)

Equation I is multiply by 5 then we subtract equation II from equation I

Using elimination method we eliminate $C_1$

Then we get $C_2=-\frac{5}{2}$

Substitute the value of $C_2$ in I equation then we get

$C_1-\frac{5}{2}=-7$

$C_1=-7+\frac{5}{2}=\frac{-14+5}{2}=-\frac{9}{2}$

Hence, the general solution is

b. $y(t)=-\frac{9}{2}e^{7t}-\frac{5}{2}e^{-5t}$

7 0

3 years ago