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2 years ago

I need help FAST!!!!

Mathematics

1 answer:

Pavel [41]2 years ago

6 0

Answer:

Step-by-step explanation:

okay, so here are a few methods you can use.

___

1) +

x + 7 = 10 so because this is addition we use the reverse, subtraction. So

x - 7 = 10 - 7 and that equals to 3.

2) -

x - 2 = 10 so because this is subtraction we do the reverse, addition.

x + 2 = 10 + 2 and that equals to 12.

3) x

x times 3 = 21 because this is multiplying we divide.

x/3 = 21/3 and that would be 7.

4) /

x/2 = 10 same as above, reverse is multiplying.

x times 2 = 10 times 2 and thats 20.

GL!

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Which linear inequality represents the graph below

Contact [7]

Answer:

your worse nightmare MATH

Step-by-step explanation:

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2 years ago

Need help with 13 asap

Whitepunk [10]

I’m hoping you are talking about:
Factor -1/2 out of -1/2x + 6
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3 years ago

URGENT PLEASE HELP 45 POINTS! Complete this activity. Given: f(x) = x + 2 <br><br> [f(x)] 2 =

ivanzaharov [21]

We are told that f(x) = x + 2. We want to find $(f(x))^{2}$ .

Squaring f(x) on one side means we square x + 2 on the other.

So,

f(x) = x + 2

(f(x))² = (x + 2)²

= (x + 2)(x + 2)

We can use FOIL to square x + 2. The first terms multiply to x², the outside terms to 2x, the inside terms to 2x and the last terms to 4.

= x² + 2x + 2x + 4

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So [f(x)]² = x² + 4x + 4

3 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

$y_1 = cosx +sinx$

$y_1' = -sinx +cosx$

$y_1'' = -cosx -sinx$

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

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

$-2sinx \ne cos2x$

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

Let consider the second equation to substitute

$y_2 = cos2x$

$y_2' = -2sin2x$

$y_2'' = -4cos2x$

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

$-4cos2x-2sin2x \ne cos2x$

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

Let consider the third equation to substitute

$y_3 = sin 2x$

$y_3' = 2cos 2x$

$y_3'' = -4sin2x$

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

$-4sin2x - 2cos2x \ne cos2x$

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

6 0

2 years ago

Find the area of the figure below

horrorfan [7]

Answer:

99 $cm^{2}$

Step-by-step explanation:

12x4=48

12x3=36

12-4-5=3

3x5=15

48+36+15=99

5 0

2 years ago