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Oduvanchick [21]

2 years ago

11

Law of Sines (please help!)

1 answer:

love history [14]2 years ago

5 0

Hope this helps! :))))

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Find an equation of the line passing through the points (2, 2) and (-1,-6)?

Margarita [4]

Brainiliest pleaseeeeee

8 0

3 years ago

What is negative 1/4 to the 3rd power

Y_Kistochka [10]

Answer:

-1/64

Step-by-step explanation:

Just do 4 to the 3rd and make that the denominator

4^3 = 64

-1 / 64

7 0

2 years ago

Read 2 more answers

Find the solution for the system of linear equations by elimination: 2x+y=9,3x-y=16

alexandr402 [8]

Answer:

(-5,-1)

Step-by-step explanation:

2x+y=9 -> y=-2x+9

3x-y=16

3x-(-2x+9)=16

3x+2x-9=16

5x-9=16

5x-9 (+9)=16 (+9)

5x=25

x=5

no plug in 5 to x in any equation above

2(5)+y=9

10+y (-10)=9 (-10)

y=-1

so the answer is (-5,-1)

hope it's correct! and hope i helped

3 0

2 years ago

Read 2 more answers

Verify by substitution whetherthe given functions are solutions of the given DE. Primes denote derivatives with respect to x.y!!

julia-pushkina [17]

Complete Question

The complete question is shown on the first uploaded

Answer:

$y_1$ is not a solution of the differential equation

$y_2$ is not a solution of the differential equation

$y_3$ is not a solution of the differential equation

Step-by-step explanation:

The differential equation given is $y'' + y' = cos2x$

Let consider the first equation to substitute

$y_1 = cosx +sinx$

$y_1' = -sinx +cosx$

$y_1'' = -cosx -sinx$

So

$y_1'' - y_1' = -cosx -sinx -sinx +cosx$

$y_1'' + y_1' = -2sinx$

So

$-2sinx \ne cos2x$

This means that $y_1$ is not a solution of the differential equation

Let consider the second equation to substitute

$y_2 = cos2x$

$y_2' = -2sin2x$

$y_2'' = -4cos2x$

So

$y_2'' + y_2' = -4cos2x-2sin2x$

So

$-4cos2x-2sin2x \ne cos2x$

This means that $y_2$ is not a solution of the differential equation

Let consider the third equation to substitute

$y_3 = sin 2x$

$y_3' = 2cos 2x$

$y_3'' = -4sin2x$

So

$y_3'' + y_3' = -4sin2x - 2cos2x$

So

$-4sin2x - 2cos2x \ne cos2x$

This means that $y_3$ is not a solution of the differential equation

6 0

3 years ago

Irina-Kira [14]

Answer:the answer is b

Step-by-step explanation:

7 0

3 years ago