Simplify the expression (4x-6y^(3))^(2)

Mathematics

1 answer:

Andru [333]3 years ago

4 0

Answer:

16x^2 -48xy^3 +36y^6

Step-by-step explanation:

(4x-6y^(3))^(2)

FOIL

(4x-6y^3) (4x - 6y^3)

first: 4x*4x = 16x^2

outer : 4x (-6y^3) =-24xy^3

inner : 4x (-6y^3) =-24xy^3

last: -6y^3 * - 6y^3 = 36y^6

Add them together

16x^2 -24xy^3-24xy^3+36y^6

Combine like terms

16x^2 -48xy^3 +36y^6

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The table lists how poverty level income cutoffs (in dollars) for a family of four have changed over time. Use the midpoint for

frez [133]

The poverty level cutoff in 1987 to the nearest dollar was $10787.

<h3>How to find a midpoint?</h3>

The midpoint as the point that divides the line segment exactly in half having two equal segments. Therefore, the midpoint presents the same distance between the endpoints for the line segment. The midpoint formula is: $\mathrm{Midpoint\:of\:}\left(x_1,\:y_1\right),\:\left(x_2,\:y_2\right):\quad \left(\frac{x_2+x_1}{2},\:\:\frac{y_2+y_1}{2}\right)$ .

For solving this exercise, first you need plot the points in a chart. See the image.

Your question asks to approximate the poverty level cutoff in 1987 to the nearest dollar using the midpoint formula. Note that the year 1987 is between 1980 and 1990, thus you should apply the midpoint formula from data for this year (1987).

$\mathrm{Midpoint\:of\:}\left(1980,8429),\:\left(1990,13145):\quad \left(\frac{1990+1980}{2},\:\:\frac{13145+8429}{2}\right)\\\\ \\$

$\mathrm{Midpoint\:of\:}\left(1980,8429)\:=\left(\frac{1990+1987}{2},\:\frac{13145+3843}{2}\right)\\ \\ =\left(\frac{3977}{2},\:\frac{21574}{2}\right)$

$\mathrm{Midpoint\:of\:}\left(1980,8429)\:=\=(\frac{3977}{2},10787)$

The answer for your question will be the value that you calculated for the y-coordinate. Then, the poverty level cutoff in 1987 to the nearest dollar was $10787.

Read more about the midpoint segment here:

brainly.com/question/11408596