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zloy xaker [14]

4 years ago

12

Which of the following tables represent a proportional relationship?

2 answers:

otez555 [7]4 years ago

5 0

Answer:

B.

Step-by-step explanation:

If you want to find a table with a proportional relationship, you are looking for one where the x-values and y-values increase at the same linear rate.

In Option A, the x-values increase steadily by 1 and then by 2, but the y-values jump around. The numbers are not proportional.

In Option B, every time an x-value is presented, the y-value is two times the value of x. So, this is a proportional relationship.

In Option C, it is true that for almost every time an x-value is presented, the y-value doubles the x-value, but when x is 0, y is 1. That is not a proportional relationship.

In Option D, it is the same thing as Option A.

So, your answer is Option B.

Hope this helps!

KatRina [158]4 years ago

4 0

The answer is B hope it helped

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what is an equation of the line that is parallel to y equals negative 4x - 5 and passes through the point -2, 6

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y = - 4x - 2

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y = - 4x - 5 is in this form

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y = - 4x + c ← is the partial equation

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Find the distance from the point to the line. (-1,-2,1);x=4+4t, y=3+t, z=6-t .The distance is ____ Typn exact answer, using radi

lawyer [7]

Answer:

The distance is<u> 4.726 </u>

Step-by-step explanation:

we need to find the distance from the point to the line

Given:- point (-1,-2,1) and line ; x=4+4t, y=3+t, z=6-t .

used formula $d=\frac{|a\times b|}{|a|}$

Let point P be (-1,-2,1)

using value t=0 and t=1

The point Q (4 , 3, 6) and R ( 8, 4, 5)

Let a be the vector from Q to R : a = < 8 - 4, 4 - 3, 5 - 6 > = < 4, 1, -1 >

Let b be the vector from Q to P: b = < -1 - 4, -2 - 3, 1 - 6> = < -5, -5, -5 >

The cross product of a and b is:

$a \times b= \begin{vmatrix} i & j & k\\ 4 &1&-1\\-5 &-5&-5\\ \end{vmatrix}$

= -6i+15j-15k

The distance is : $d=\frac{\sqrt{(-6)^{2}+(15)^{2}+(-15)^{2}}}{\sqrt{(4)^{2}+(1)^{2}+(-1)^{2}}}$

$=\frac{\sqrt{36+225+225}}{\sqrt{16+1+1}}$

$=\frac{\sqrt{36+225+225}}{\sqrt{16+1+1}}$

$d=\frac{\sqrt{486}}{\sqrt{18}}$

≈4.726

Therefore, the distance is<u> 4.726 </u>

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