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3 years ago

Solve the following system of equations -8x -8y=24 2x + y=-7

Mathematics

2 answers:

tatyana61 [14]3 years ago

8 0

Answer:

x=3.33333

y=0.44444

Step-by-step explanation:

Would equal to

-8x-8y=24 -equation 1

16x+8y = -56 -equation 2 (because of multiplying both sides by 8)

=-24x=80 (minus-ing equation two by equation one, on all sides)

= 24x = -80 (switching around the signs)

x =3.3333

and...

y =0.4444

leonid [27]3 years ago

4 0

Solve the equation using substitution?

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Step-by-step explanation:

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2 years ago

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Butoxors [25]

Answer:

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Step-by-step explanation:

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2 years ago

Read 2 more answers

4. In a scale drawing, the length of a

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Answer:

1ft=.33in

Step-by-step explanation:

For every 1ft in the room its .33 on paper.

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3 years ago

Read 2 more answers

(6-3(cube root of 6)/(cube root of 9)

MatroZZZ [7]

For this case we must simplify the following expression:

$\frac {6-3 \sqrt [3] {6}} {\sqrt [3] {9}}$

Multiplying the numerator and denominator by $(\sqrt [3] {9}) ^ 2$

$\frac {6-3 \sqrt [3] {6}} {\sqrt [3] {9}} * \frac {(\sqrt [3] {9}) ^ 2} {(\sqrt [3] { 9}) ^ 2} =$

We rewrite:

$\frac {\frac {6-3 \sqrt [3] {6}} * (\sqrt [3] {9}) ^ 2} {\sqrt [3] {9} * (\sqrt [3] {9 }) ^ 2} =$

By properties of powers we have that:

$a ^ m * a ^ n = a ^ {m + n}\\\frac {(6-3 \sqrt [3] {6}) * (\sqrt [3] {9}) ^ 2} {(\sqrt [3] {9}) ^ 3} =\\\frac {(6-3 \sqrt [3] {6}) * (\sqrt [3] {9}) ^ 2} {9} =$

We rewrite, moving the exponent within the radical:

$\frac {(6-3 \sqrt [3] {6}) * \sqrt [3] {9 ^ 2}} {9} =\\\frac {(6-3 \sqrt [3] {6}) * \sqrt [3] {81}} {9} =$

We can rewrite $3 * 3 ^ 3 = 81$

$\frac {(6-3 \sqrt [3] {6}) * \sqrt [3] {3 * 3 ^ 3}} {9} =$

We simplify:

$\frac {(6-3 \sqrt [3] {6}) * 3 \sqrt [3] {3}} {9} =$

We apply distributive property:

$\frac {18 \sqrt [3] {3} -9 \sqrt [3] {18}} {9} =$

Simplifying we finally have:

$2 \sqrt [3] {3} - \sqrt [3] {18}$

Answer:

$2 \sqrt [3] {3} - \sqrt [3] {18}$

5 0

3 years ago