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Zanzabum
3 years ago
8

A room has an area of 216 square feet. One dimension is 6 feet more than the other. Find the length of the longer side.

Mathematics
2 answers:
zlopas [31]3 years ago
3 0
The dimensions would be x and x+6. The area is 216.

x(x+6) = 216
x^2 + 6x - 216 = 0
(x-12)(x+18) = 0

x = 12---Yes---
x = -18---No, sides cannot be negative---

If one side is 12, the other side is 18. To check:

12 * 18 = 216
216 = 216

Correct
kicyunya [14]3 years ago
3 0
He correct cause I just did it. ;)
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Find f(x) of -n^2 -3n if x is -5​
dangina [55]

Answer:

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Step-by-step explanation:

f(5)= -(5)^2-3(5)

-25-15=-40

5 0
3 years ago
Add: <br> 3x+9+8x + 12<br> 2x + 6+x² + 6x + 9
Oxana [17]

I'll treat these like they're two seperate problems because  how you set it up they're not together. If it's one whole problem please tell me and I'll solve it that way. :-)

First one: 3x+9+8x + 12

Collect like terms and simplify

(3x+8x)+(9+12)

11x+21

Second one: 2x + 6+x² + 6x + 9

Collect like terms and simplify

(2x+6x)+(6+9)+x^{2}

8x+15+x^{2}

Hope this helps you, have a BLESSED and wonderful day, as well as a safe one!

-Cutiepatutie ☺❀❤

6 0
3 years ago
Can someone help me find the equivalent expressions to the picture below? I’m having trouble
miss Akunina [59]

Answer:

Options (1), (2), (3) and (7)

Step-by-step explanation:

Given expression is \frac{\sqrt[3]{8^{\frac{1}{3}}\times 3} }{3\times2^{\frac{1}{9}}}.

Now we will solve this expression with the help of law of exponents.

\frac{\sqrt[3]{8^{\frac{1}{3}}\times 3} }{3\times2^{\frac{1}{9}}}=\frac{\sqrt[3]{(2^3)^{\frac{1}{3}}\times 3} }{3\times2^{\frac{1}{9}}}

           =\frac{\sqrt[3]{2\times 3} }{3\times2^{\frac{1}{9}}}

           =\frac{2^{\frac{1}{3}}\times 3^{\frac{1}{3}}}{3\times 2^{\frac{1}{9}}}

           =2^{\frac{1}{3}}\times 3^{\frac{1}{3}}\times 2^{-\frac{1}{9}}\times 3^{-1}

           =2^{\frac{1}{3}-\frac{1}{9}}\times 3^{\frac{1}{3}-1}

           =2^{\frac{3-1}{9}}\times 3^{\frac{1-3}{3}}

           =2^{\frac{2}{9}}\times 3^{-\frac{2}{3} } [Option 2]

2^{\frac{2}{9}}\times 3^{-\frac{2}{3} }=(\sqrt[9]{2})^2\times (\sqrt[3]{\frac{1}{3} } )^2 [Option 1]

2^{\frac{2}{9}}\times 3^{-\frac{2}{3} }=(\sqrt[9]{2})^2\times (\sqrt[3]{\frac{1}{3} } )^2

                =(2^2)^{\frac{1}{9}}\times (3^2)^{-\frac{1}{3} }

                =\sqrt[9]{4}\times \sqrt[3]{\frac{1}{9} } [Option 3]

2^{\frac{2}{9}}\times 3^{-\frac{2}{3} }=(2^2)^{\frac{1}{9}}\times (3^{-2})^{\frac{1}{3} }

               =\sqrt[9]{2^2}\times \sqrt[3]{3^{-2}} [Option 7]

Therefore, Options (1), (2), (3) and (7) are the correct options.

6 0
2 years ago
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Mandarinka [93]

Step-by-step explanation:

By Exterior Angle Theorem,

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8 0
2 years ago
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tino4ka555 [31]

Answer:

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Step-by-step explanation:

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