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2 years ago

Can you please help me fill this out thanks

Mathematics

1 answer:

swat322 years ago

7 0

Answer: $\sqrt{2}$

Step-by-step explanation:

In a 45 45 90 triangle the two sides are 1 wile the hypothenuse is $\sqrt{2}$

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Solve for the indicated variable.

marta [7]

Answer:

Option (c) " $r=\dfrac{q}{4c}-h$ "

Step-by-step explanation:

The given equation is :

q = c4(h + r)

We need to solve it for r.

Dividing both sides of the given equation by 4c.

$\dfrac{q}{4c}=\dfrac{4c}{4c}(h+r)\\\\(h+r)=\dfrac{q}{4c}$

Now, subtract h both sides of the equation.

$h+r-h=\dfrac{q}{4c}-h\\\\r=\dfrac{q}{4c}-h$

Hence, this is the required solution.

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2 years ago

Enter the number in standard form.<br> 4.78 · 10-1

kogti [31]

The answer is 46.8
Plz mark me Brainly

8 0

2 years ago

Read 2 more answers

4. In the triangle ABC, angle A = 90° and

vesna_86 [32]

Step-by-step explanation:

here the answers is it helpful

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1 year ago

Determine the amplitude or period as requested. Period of a y = - 1/3 * sin x - 1/3 pi/3 1/3 d 2pi

zysi [14]

Answer:

$Period = 2\pi$

Step-by-step explanation:

Given

$y = -\frac{1}{3} * \sin(x - \frac{1}{3})$

Required

Determine the period

A sine function is represented as:

$y =A \sin(Bx + C) + D$

Where

$Period = \frac{2\pi}{B}$

By comparing:

$y =A \sin(Bx + C) + D$ and $y = -\frac{1}{3} * \sin(x - \frac{1}{3})$

$Bx = x$

So:

$B = 1$

So, we have:

$Period = \frac{2\pi}{B}$

$Period = \frac{2\pi}{1}$

$Period = 2\pi$

Hence, the period of the function is $2\pi$

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2 years ago

What is the horizontal asymptote??? Math 3!! 10 points HELP NEEDED

Ymorist [56]

Answer:

no horizontal asymptote

Step-by-step explanation:

When the degree of the numerator > degree of the denominator there is no horizontal asymptote but there is a slant asymptote.

Dividing the numerator by the denominator gives

$\frac{x^2-x-6}{x+4}$ = x - 5 + $\frac{14}{x+4}$

There is a slant asymptote with equation y = x - 5

8 0

2 years ago