3 years ago

What is the simplest form of 4(y+2)-2

Mathematics

1 answer:

ivanzaharov [21]3 years ago

4 0

2 • (2y + 3)

 step 1 4 • (y + 2) - 2

final step 4y + 6 = 2 • (2y + 3)

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Plz help me get the answer ASAP

lakkis [162]

solve the inequality:

1/2c>-1

divide 1/2 from both sides

c>-2 is the answer

6 0

4 years ago

Read 2 more answers

PLEASE HELP WILL MARK BRAINLIEST!!!

saul85 [17]

Answer:

The answer would be c i did the test on plato

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

<img src="https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7By%7D%20%287%5Csqrt%5B3%5D%7B8y%5E2%7D-%5Csqrt%5B3%5D%7By%5E5%7D%20-4y%5Csqrt%

noname [10]

Answer:

$\huge\boxed{-y^2+2y}$

Step-by-step explanation:

$\sqrt[3]y\cdot\left(7\sqrt[3]{8y^2}-\sqrt[3]{y^5}-4y\sqrt[3]{27y^2}\right)\\\\=(\sqrt[3]y)(7\sqrt[3]{8y^2})-(\sqrt[3]y)(\sqrt[3]{y^5})-(\sqrt[3]y)(4y\sqrt[3]{27y^2})\\\\=7\sqrt[3]{(y)(8y^2)}}-\sqrt[3]{(y)(y^5)}-4y\sqrt[3]{(y)(27y^2)}\\\\=7\sqrt{8y^3}-\sqrt{y^6}-4\sqrt{27y^3}\\\\=7\sqrt[3]{2^3y^3}-\sqrt{y^{2\cdot3}}-4\sqrt{3^3y^3}\\\\=7\sqrt[3]{(2y)^3}-\sqrt{(y^2)^3}-4\sqrt{(3y)^3}\\\\=7\cdot2y-y^2-4\cdot3y\\\\=14y-y^2-12y\\\\=-y^2+2y$