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ikadub [295]
3 years ago
8

I'm confused on what to do since I day dream during class, I know it's a bad thing but I have a hard time paying attention​

Mathematics
1 answer:
vagabundo [1.1K]3 years ago
4 0

Answer:17/20 answer a

Step-by-step explanation:

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Write the quadratic function in standard form.<br><br> y=2(x - 3)² +9
xz_007 [3.2K]

Answer:

y = 2x^2 - 12x + 27

Step-by-step explanation:

<u>Step 1:  Distribute the power</u>

y = 2(x - 3)² + 9

y = 2(x^2 - 6x + 9) + 9

y = 2x^2 - 12x + 18 + 9

y = 2x^2 - 12x + 27

Answer: y = 2x^2 - 12x + 27

5 0
3 years ago
Prove that (Root of Sec A - 1 / Root of Sec A + 1) + (Root of Sec A + 1 / Root of Sec A - 1) = 2 cosec A
iVinArrow [24]

Answer:

The answer is below

Step-by-step explanation:

We need to prove that:

(Root of Sec A - 1 / Root of Sec A + 1) + (Root of Sec A + 1 / Root of Sec A - 1) = 2 cosec A.

Firstly, 1 / cos A = sec A, 1 / sin A = cosec A and tanA = sinA / cosA.

Also, 1 + tan²A = sec²A; sec²A - 1 = tan²A

\frac{\sqrt{secA-1} }{\sqrt{secA+1} } +\frac{\sqrt{secA+1} }{\sqrt{secA-1} } =\frac{(\sqrt{secA-1)}(\sqrt{secA-1})+(\sqrt{secA+1)}(\sqrt{secA+1}) }{(\sqrt{secA+1})(\sqrt{secA-1}) } \\\\=\frac{secA-1+(secA+1)}{\sqrt{sec^2A-secA+secA-1} } \\\\=\frac{2secA}{\sqrt{sec^2A-1} } \\\\=\frac{2secA}{\sqrt{tan^2A} } \\\\=\frac{2secA}{tanA} \\\\=\frac{2*\frac{1}{cosA} }{\frac{sinA}{cosA} }\\\\= 2*\frac{1}{cosA}*\frac{cosA}{sinA}\\\\=2*\frac{1}{sinAA}\\\\=2cosecA

7 0
3 years ago
I don’t get the second question.
tatyana61 [14]

it should be 3+(-7)  because look at the number closest to the number line, which is positive 3 then the number next to it is negative 7 because it is heading downward towards the opposing side.  

8 0
4 years ago
Michael will be running a 15 mile road race this weekend how many feet will he run
sergij07 [2.7K]

Answer:

79,200

Step-by-step explanation:

There are 5280 feet in 1 mile, then you just multiply 5280 by the number of miles (15). So 5280 x 15 = 79,200

5 0
3 years ago
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