3 years ago

dharmesh has a square garden with a perimeter of 132 feet is the area of the garden greater than 1,000 square feet.

Mathematics

1 answer:

k0ka [10]3 years ago

3 0

The area is 1089 square feet, so the area of the garden is greater than 1,000 square feet.

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Volgvan

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13 – 6x = 2(8 + 3x) x= ???????

ruslelena [56]

Answer:

$13-6x=2\left(8+3x\right)$

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3 years ago

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For brainly and points thank you dont mind the thing i clicked it was by mistake

Inga [223]

Answer:

C) x = 1

Step-by-step explanation:

By adding the like terms you get

-12x + 10 = -10x + 8

then you add 12x to both sides and subtract 8 to both sides and get

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3 years ago

<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:

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

Used:

$a(a+b)=ab+ac\\\\\sqrt[3]{a\cdot b}=\sqrt[3]a\cdot\sqrt[3]b\\\\\sqrt[3]{a^3}=a\\\\(a^n)^m=a^{n\cdot m}$

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2 years ago