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Sunny_sXe [5.5K]
3 years ago
6

-9x=-36 what is the answer

Mathematics
1 answer:
12345 [234]3 years ago
8 0

-9x=-36

divide both sides by -9

to get x by itself

to get +x

-9x/-9= -36/-9

cross out -9 and -9, then becomes -1*-1*x= x

x= -36/-9

answer:

x= 4

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Mr. Fox purchased 76 bags of candy to make treat bags each bag contains 48 pieces of candy how many pieces of candy did Mr. Fox
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Determine whether the lines given in each box are parallel, perpendicular, or neither
andrey2020 [161]

Answer/Step-by-sep explanation:

To determine whether the lines given in each box are parallel, perpendicular, or neither, take the following simple steps:

1. Ensure the equations for both lines being compared are in the slope-intercept form, y = mx + b. Where m is the slope.

2. If both lines have the same slope value, m, then both lines are parallel.

3. If the slope of one line is the negative reciprocal of the other, then both lines are perpendicular. That is, x = -1/x.

4. If the slope of both lines are not the same, nor the negative reciprocal of each other, then they are neither parallel nor perpendicular.

1. y = 3x - 7 and y = 3x + 1.

Both have the same slope value of 3. Therefore, they are parallel.

2. ⬜ y = -\frac{2}{5}x + 3 and y = \frac{2}{5}x + 8

The slope of both lines are not the same, nor is the slope of one the negative reciprocal of the other. The slope of one is -⅖ and the slope of the other is ⅖. Therefore, they are neither parallel nor perpendicular.⬜

3. y = -\frac{1}{4}x and y = 4x - 5

The slope of the first line, ¼, is the negative reciprocal of the slope of the second line, 4.

Therefore, they are perdendicular.

4. 2x + 7y = 28 and 7x - 2y = 4.

Rewrite both equations in the slope-intercept form, y = mx + b.

2x + 7y = 28

7y = -2x + 28

y = -2x/7 + 28/7

y = -²/7 + 4

And

7x - 2y = 4

-2y = -7x + 4

y = -7x/-2 + 4/-2

y = ⁷/2x - 2

The slope of the first line, -²/7, is the negative reciprocal of the slope of the second line, ⁷/2.

Therefore, they are perdendicular.

5.⬜ y = -5x + 1 and x - 5y = 30.

Rewrite the second line equation in the slope-intercept form.

x - 5y = 30

-5y = -x + 30

y = -2x/-5 + 30/-5

y = ⅖x - 6

The slope of both lines are not the same, nor is the slope of one the negative reciprocal of the other. The slope of one is -5 and the slope of the other is ⅖. therefore, they are neither parallel nor perpendicular.⬜

6.⬜ 3x + 2y = 8 and 2x + 3y = -12.

Rewrite both line equations in the slope-intercept form.

3x + 2y = 8

2y = -3x + 8

y = -3x/2 + 8/2

y = -³/2x + 4

And

2x + 3y = -12

3y = -2x -12

y = -2x/3 - 12/3

y = -⅔x - 4

The slope of both lines are not the same, nor is the slope of one the negative reciprocal of the other. The slope of one is -³/2 and the slope of the other is -⅔ therefore, they are neither parallel nor perpendicular.⬜

7. y = -4x - 1 and 8x + 2y = 14.

Rewrite the equation of the second line in the slope-intercept form.

8x + 2y = 14

2y = -8x + 14

y = -8x/2 + 14/2

y = -4x + 7

Both have the same slope value of -4. Therefore, they are parallel.

8.⬜ x + y = 7 and x - y = 9.

Rewrite the equation of both lines in the slope-intercept form.

x + y = 7

y = -x + 7

And

x - y = 9

-y = -x + 9

y = -x/-1 + 9/-1

y = x - 9

The slope of both lines are not the same, nor is the slope of one the negative reciprocal of the other. The slope of one is -1, and the slope of the other is 1, therefore, they are neither parallel nor perpendicular.⬜

9. y = ⅓x + 9 And x - 3y = 3

Rewrite the equation of the second line.

x - 3y = 3

-3y = -x + 3

y = -x/-3 + 3/-3

y = ⅓x - 1

Both have the same slope value of ⅓. Therefore, they are parallel.

10.⬜ 4x + 9y = 18 and y = 4x + 9

Rewrite the equation of the first line.

4x + 9y = 18

9y = -4x + 18

y = -4x/9 + 18/9

y = -⁴/9x + 2

The slope of both lines are not the same, nor is the slope of one the negative reciprocal of the other. The slope of one is -⁴/9, and the slope of the other is 4, therefore, they are neither parallel nor perpendicular.⬜

11.⬜ 5x - 10y = 20 and y = -2x + 6

Rewrite the equation of the first line.

5x - 10y = 20

-10y = -5x + 20

y = -5x/-10 + 20/-10

y = ²/5x - 2

The slope of both lines are not the same, nor is the slope of one the negative reciprocal of the other. The slope of one is ⅖, and the slope of the other is -2, therefore, they are neither parallel nor perpendicular.⬜

12. -9x + 12y = 24 and y = ¾x - 5

Rewrite the equation of the first line.

-9x + 12y = 24

12y = 9x + 24

y = 9x/12 + 24/12

y = ¾x + 2

Both have the same slope value of ¾. Therefore, they are parallel.

5 0
3 years ago
Suppose quantity s is a length and quantity t is a time. Suppose the quantities v and a are defined by v = ds/dt and a = dv/dt.
finlep [7]

Answer:

a) v = \frac{[L]}{[T]} = LT^{-1}

b) a = \frac{[L}{T}^{-1}]}{{T}}= L T^{-1} T^{-1}= L T^{-2}

c) \int v dt = s(t) = [L]=L

d) \int a dt = v(t) = [L][T]^{-1}=LT^{-1}

e) \frac{da}{dt}= \frac{[L][T]^{-2}}{T} = [L][T]^{-2} [T]^{-1} = LT^{-3}

Step-by-step explanation:

Let define some notation:

[L]= represent longitude , [T] =represent time

And we have defined:

s(t) a position function

v = \frac{ds}{dt}

a= \frac{dv}{dt}

Part a

If we do the dimensional analysis for v we got:

v = \frac{[L]}{[T]} = LT^{-1}

Part b

For the acceleration we can use the result obtained from part a and we got:

a = \frac{[L}{T}^{-1}]}{{T}}= L T^{-1} T^{-1}= L T^{-2}

Part c

From definition if we do the integral of the velocity respect to t we got the position:

\int v dt = s(t)

And the dimensional analysis for the position is:

\int v dt = s(t) = [L]=L

Part d

The integral for the acceleration respect to the time is the velocity:

\int a dt = v(t)

And the dimensional analysis for the position is:

\int a dt = v(t) = [L][T]^{-1}=LT^{-1}

Part e

If we take the derivate respect to the acceleration and we want to find the dimensional analysis for this case we got:

\frac{da}{dt}= \frac{[L][T]^{-2}}{T} = [L][T]^{-2} [T]^{-1} = LT^{-3}

7 0
3 years ago
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Snowcat [4.5K]

i really dont know but i really need to points please

8 0
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