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1 year ago

14

What is the slope of the line?

1 answer:

12345 [234]1 year ago

8 0

You can find the slope using the following formula:

$\begin{gathered} m=\frac{y2-y1}{x2-x1} \\ where\colon \\ (x1,y1)=(0,2) \\ (x2,y2)=(1,0) \\ m=\frac{0-2}{1-0}=\frac{-2}{1}=-2 \end{gathered}$

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Emma has 7 pens in a box. 5 of the pens are blue. 2 of the pens are red. emma takes at random a pen from the box and writes down

alexandr402 [8]

Answer:

There is no question stated. If one assume the question is "Does the probability of picking the same color pen change after the first try?," the answer is "No."

Step-by-step explanation:

When the pen is replaced, the probability reverts back to the original value:

There are 7 pens:

5 Blue

2 Red

The probability of picking a red or blue pen on the first try is:

Red = 2/7

Blue = 5/7

When the pen is replaced, the probabilities return to the original values.

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1 year ago

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sasho [114]

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2 years ago

Is 6/12 equivalent to 1/2?

PIT_PIT [208]

Answer:

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2 years ago

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2 years ago

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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

2 years ago