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Bumek [7]
3 years ago
6

If the relationship is proportional, what is the missing value from the table

Mathematics
1 answer:
lesya [120]3 years ago
7 0

Given:

Consider the below figure attached with this question.

The table represents a proportional relationship.

To find:

The missing value from the table.

Solution:

If y is proportional to x, then

y\propto x

y=kx                 ...(i)

Where, k is a constant of proportionality.

The relationship passes through the point (-3,-1). Substituting x=-3,y=-1 in (i), we get

-1=k(-3)

\dfrac{-1}{-3}=k

\dfrac{1}{3}=k

Putting k=\dfrac{1}{3} in (i), we get

y=\dfrac{1}{3}x           ...(ii)

We need to find the y-value for x=-12.

Substituting x=-12 in (ii), we get

y=\dfrac{1}{3}(-12)

y=-4

Therefore, the missing value in the table is -4. Hence, option D is correct.

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Answer:

45

Step-by-step explanation:

\frac{x}{3} -6=9

add 6 on both sides to eliminate it(doing this will get us closer to the variable)

x/3=15

multiply both sides by 3

x=45

hope this helps!

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What are their perimeters?
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and

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Seems like thier perimeters are equal.

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From the sum of a2−2ab+b2 and 2a2+2ab+b2 subtract the sum of a2−b2 and a2+ab+3b2
Sedaia [141]

Answer:

a^2-ab

Step-by-step explanation:

We need to find the sum of a^2-2ab+b^2 and 2a^2+2ab+b^2 first.

Adding a^2-2ab+b^2 + 2a^2+2ab+b^2

Combining like terms, we get

a^2+2a^2=3a^2

-2ab+2ab = 0

b^2+b^2=2b^2

Therefore,

a^2-2ab+b^2 +2a^2+2ab+b^2=3a^2+2b^2.

Now, we need to find the sum of a^2-b^2 \ \ and \ \ a^2+ab+3b^2..

Adding a^2−b^2+a^2+ab+3b^2.

Combining like terms, we get

a^2+a^2=2a^2

-b^2+3b^2=2b^2.

Therefore,

a^2−b^2+a^2+ab+3b^2=2a^2+ab+2b^2.

Now, subtracting

3a^2+2b^2 -(2a^2+ab+2b^2).

Distributing minus sign over second parenthesis, we get

3a^2+2b^2-2a^2-ab-2b^2.

Combining like terms,

3a^2-2a^2=a^2

2b^2-2b^2=0

Therefore,

3a^2+2b^2-2a^2-ab-2b^2=a^2-ab.

Therefore, the difference of the sum of a^2-2ab+b^2 + 2a^2+2ab+b^2 and a^2-b^2+a^2+ab+3b^2. is a^2-ab.

5 0
3 years ago
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