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Svetradugi [14.3K]

3 years ago

8

Which statement can be proved if you are given that SK≌LR?

1 answer:

MaRussiya [10]3 years ago

6 0

I believe it would be D) ST ≈ TR

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H + 14.8 = −14.8 solve for h

Slav-nsk [51]

Answer:

h= -29.6

Step-by-step explanation:

trust me!!

7 0

2 years ago

Read 2 more answers

evablogger [386]

A. 36/a
B. 55-b
C.18+c
D.49d
E.e/62

7 0

3 years ago

Polygon MNOPQ is dilated by a scale factor of 0.8 with the origin as the center of dilation, resulting in the image M′N′O′P′Q′.

Goryan [66]

M` = ( 0.8 * 2, 0.8 *4 ) = 0.8 * ( 2, 4 );
N` = ( 0.8 * 3, 0.8 * 5 ) = 0.8 * ( 3 , 5 ).
The slope of line M`N` ( after dilation ):
m = Rise / Run = ( y 2 - y 1 ) / ( x 2 - x 1 ) =
= 0.8 * ( 5 - 4 ) / ( 0.8 ( 3 - 2 ) = 0.8 / 0.8 = 1
m = 1
Answer:
The slope of line M`N` is : B ) 1

4 0

3 years ago

Solve y′′=sin(x) if y(0)=0 and y′(0)=5.<br><br> y(x)=?

Ksenya-84 [330]

Answer:

$y(x)=6x-sin(x)$

Step-by-step explanation:

Rewrite the differential equation as:

$\frac{d^{2} y }{dx^{2} } =sin(x)$

Integrate both sides with respect to x:

$\int\ \frac{d^{2} y }{dx^{2} } dx = \int\ sin(x) dx$

$\frac{dy}{dx} =-cos(x)+C_1$

Integrate one more time both sides with respect to x:

$\int\ \frac{dy}{dx} = \int\ -cos(x)+C_1 dx$

$y(x)=-sin(x)+C_1x+C_2$

Now that we find the solution, let's find its derivate:

$y'(x)=C_1-cos(x)$

Evaluating the initial conditions:

$y(0)=C_1(0)+C_2-sin(0)=0\\C_2=0$

$y'(0)=C_1-cos(0)=5\\C_1=5+1=6$

Replacing the value of the constants that we found in the differential equation solution:

$y(x)=6x-sin(x)$

6 0

3 years ago

Find the value of X.

cestrela7 [59]

Answer: x does not have a value

Step-by-step explanation: its a "variable" or sometimes an "unknown".

5 0

2 years ago