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3 years ago

Please help me with this math question!

Mathematics

2 answers:

NARA [144]3 years ago

6 0

Option A would be the correct one difference/product :)

sergiy2304 [10]3 years ago

5 0

The answer is A have a good night

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What is the domain of the following function?

SashulF [63]

Answer:

third option (-3,3,5,9)

Step-by-step explanation:

domain is the starting point of the set (x coordinate )

6 0

3 years ago

49 percent of all people who buy running shoes don't run at all Assuming 340,000 people but running shoes how many will use them

Goryan [66]

173, 400

49% of people do not run; thus, 51 percent of people do.

51% * 340,000 = 173, 400

8 0

4 years ago

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Look at the image. (calculus)

mash [69]

<h3>Answer: Choice H) 2</h3>

=============================================

Explanation:

Recall that the pythagorean trig identity is $\sin^2 x + \cos^2x = 1$

If we were to isolate sine, then,

$\sin^2 x + \cos^2x = 1\\\\\sin^2 x = 1-\cos^2x\\\\\sin x = \sqrt{1-\cos^2x}\\\\$

We don't have to worry about the plus minus because sine is positive when 0 < x < pi/2.

Through similar calculations, $\cos x = \sqrt{1-\sin^2x}\\\\$

Cosine is also positive in this quadrant.

-------------

So,

$\frac{\sqrt{1-\cos^2x}}{\sin x}+\frac{\sqrt{1-\sin^2x}}{\cos x}\\\\\frac{\sin x}{\sin x}+\frac{\cos x}{\cos x}\\\\1+1\\\\2$

Therefore,

$\frac{\sqrt{1-\cos^2x}}{\sin x}+\frac{\sqrt{1-\sin^2x}}{\cos x}=2$

is an identity as long as 0 < x < pi/2

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2 years ago

Read 2 more answers

One bag of Roland fertilizer covers 2346 square feet of lawn. How many total bags of fertilizer must be purchased to cover a law

Ivenika [448]

If I did my math correct it should be 47. If you take 284x386 you get 109,624 (that should be the area in square feet) then take 109,624 and divide it into 2346 and you get 46.7280477408. In order to cover it you should round up to 47 bags.

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4 years ago

Resolve into partial fractions<br><br>

lisabon 2012 [21]

Answer:

See below

Step-by-step explanation:

$\frac{11 + x}{(2 - x)(x - 3)} = \frac{A}{2 - x} + \frac{B}{x - 3} \\ \\ \therefore \: \frac{11 + x}{(2 - x)(x - 3)} = \frac{A(x - 3) +B(2 - x) }{(2 - x)(x - 3)} \\ \\ \therefore \: 11 + x = Ax - 3A + 2B - Bx \\ \\ \therefore \: 11 + x = - 3A + 2B + Ax - Bx\\ \\ \therefore \: 11 + x = - 3A + 2B + (A - B)x \\ \\ equating \: the \: like \: terms \: on \: both \: sides \\ A - B = 1 \\ \therefore \: A = B + 1.....(1) \\ \\ - 3A + 2B = 11....(2) \\ from \: eq \: (1) \: and \: (2) \\ - 3(B + 1) + + 2B = 11 \\ \\ - 3B - 3 + 2B = 11 \\ \\ - B = 11 + 3 \\ \\ B = - 14 \\ A = - 14 + 1 = - 13 \\ \\ \frac{11 + x}{(2 - x)(x - 3)} = \frac{ - 13}{2 - x} + \frac{ - 14}{x - 3} \\ \\ \frac{11 + x}{(2 - x)(x - 3)} = - \frac{ 13}{2 - x} - \frac{ 14}{x - 3}$