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2 years ago

7

Help, I have a time limit for this

1 answer:

salantis [7]2 years ago

5 0

Answer:

I believe that it is the first one.

Step-by-step explanation:

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20 points!~ And thanks if you help me.

katovenus [111]

Answer:

the answer is b

-18 -18

x6 x-6

____ _____

-108 108

3 0

2 years ago

aleksklad [387]

618 hope this helps you have an amazing week

4 0

2 years ago

I will mark you the brainiest for the correct answer, please be correct, Have a good day and take care, thanks. ( VIEW THE IMAGE

makkiz [27]

Mistake in Line 2 : Was in the determining the values for a ,b,c

Mistake in Line 4 : Not taking the square roots of both sides.

Step-by-step explanation:

${a}^{2} + {b}^{2} = {c}^{2}$

First Mistake : Line 2

$a = x \\ b = 6 \\ c = 10$

Second Mistake : Line 4

$136 = {x}^{2} \\ not \\ 136 = x$

Correct Solution :

${x}^{2} + {6}^{2} = {10}^{2} \\ {x}^{2} = {10}^{2} - {6}^{2} \\ {x}^{2} = 100 - 36 \\$

${x}^{2} = 64 \\ \sqrt{ {x}^{2} } = \sqrt{64} \\ x = 6$

3 0

3 years ago

When lines are at the same distance apart over their lengths they are ?

svlad2 [7]

Answer:

parallel lines

Step-by-step explanation:

parallel lines are lines that do not meet each other

7 0

3 years ago

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