Review the following information about vectors q and r.
What is the value of a?
q=(-4,1)
31
r = (a, 3)
7(q + r) = (-35, 28)
-5

Review The Following Information About Vectors Q And R.What Is The Value Of A?q=(-4,1)31r = (a, 3)7(q

Answers

Answer 1
Answer:  a = -1  (fourth choice)

==========================================

Work Shown:

q = (-4, 1) is one vector

r = (a,3) is another vector

The resultant vector is

q+r = (-4,1)+(a,3)

q+r = (-4+a,1+3)

q+r = (-4+a,4)

Multiply both sides by 7

7(q+r) = 7*(-4+a,4)

7(q+r) = (7*(-4+a),7*4)

7(q+r) = (-28+7a, 28)

---------------

Since 7(q+r) = (-35, 28), we know that,

(-28+7a, 28) =  (-35, 28)

which leads to

-28 + 7a = -35

when we equate the x components of each vector. Let's solve for 'a'

-28 + 7a = -35

7a = -35+28

7a = -7

a = -7/7

a = -1

--------------

Check:

q = (-4,1)

r = (a,3) = (-1,3)

q+r = (-4,1)+(-1,3)

q+r = (-4+(-1), 1+3)

q+r = (-5, 4)

7*(q+r) = 7*(-5, 4)

7*(q+r) = (7*(-5), 7*4)

7*(q+r) = (-35, 28)

The answer is confirmed.

Answer 2

Answer:

D. -1

Step-by-step explanation:

edge test


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Answers

Answer:

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And y is the amount of money collected for the tickets.

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The first table complies with the restriction and has correct values for the variables:

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x = 51, y = 255         ⇒    255 = 5 · 51        ⇒      255 = 255

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Step-by-step explanation:

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Alternative text


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Answer:

No. The input 1.2 has 2 outputs.

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See attachment for complete question

The first circle represents the domain while the second represents the range of the function.

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Answers

It would be 4/12 which is equal to 1/3.

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The fraction of the shaded part of the set is 1/3.

Given is a figure in which there are total 12 parts with 4 parts shaded.

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To find the fraction of the shaded part of the set, we need to determine the ratio of the number of shaded parts to the total number of parts.

In this case, the total number of parts is 12, and the number of shaded parts is 4.

Therefore, the fraction of the shaded part can be calculated as follows:

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Fraction of shaded part = 4 / 12

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k can either be

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