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3 years ago

13

For each expression ,use the distributive property to write an equivalent expression

1 answer:

liq [111]3 years ago

3 0

A. 4•x and 4•2 this would turn in 4x+8
B. x•6 and x•8 this would turn into 6x+8x and this equals to 14x
C. 4•2x and 4•3 this would turn into 8x+12.
D. 6•x, 6•y, and 6•z this would equal 6x+6y+6z.

I’m sorry if it’s not clear this is my first answered question.

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Solve for y: 2x+3y=21

stealth61 [152]

Answer:

y = 7 - 2x/3

Step-by-step explanation:

Make the y alone, so 3y = -2x+21

Divide everything else by 3, so y = -2x/3 +7

You end with the answer of y= 7-2x+3

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4 years ago

5x+2y=-3;x+5y=4<br>plz answer me

Alex17521 [72]

Answer:

x = -1 and y = 1

Step-by-step explanation:

5x + 2y = -3 . . . . . . . (i)

x + 5y = 4 . . . . . . . (ii)

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Placing this value of x in eq. (i) :-

5(4 - 5y) + 2y = -3

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3 years ago

Read 2 more answers

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Daniel [21]

Answer:

19.64

Step-by-step explanation:

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3 years ago

<img src="https://tex.z-dn.net/?f=z%5E%7B7%7D%3D128i" id="TexFormula1" title="z^{7}=128i" alt="z^{7}=128i" align="absmiddle" cla

abruzzese [7]

If <em>z</em> ⁷ = 128<em>i</em>, then there are 7 complex numbers <em>z</em> that satisfy this equation.

$z^7 = 128i = 2^7i = 2^7e^{i\frac\pi2}$

$\implies z=\sqrt[7]{2^7} e^{i\frac17\left(\frac\pi2+2n\pi\right)}$

(where <em>n</em> = 0, 1, 2, …, 6)

$\implies z = 2 e^{i\left(\frac\pi{14}+\frac{2n\pi}7\right)}$

$\displaystyle\implies z = 2 \left(\cos\left(\frac\pi{14}+\frac{2n\pi}7\right)+i\sin\left(\frac\pi{14}+\frac{2n\pi}7\right)\right)$

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3 years ago