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3 years ago

Which choice is equivalent to the quotient below when. The problem is in the photo below.i am scared please

Mathematics

1 answer:

NeTakaya3 years ago

5 0

Hello from MrBillDoesMath!

Answer:

Choice C, sqrt(3-x)

Discussion:

Have no fear, MBill is here!

The quotient is

sqrt (9-x^2) / sqrt( 3+x) = => as sqrt(a)/sqrt(b) = sqrt(a/b), b <> 0

sqrt ( (9-x^2)/( 3+x) ) = => as 9 - x^2 = (3-x) * (3+x)

sqrt ( (3-x)(3+x) /(3+x) ) = => cancel (3+x) from num. and denom.

sqrt( 3-x)

This is Choice C

Note the domain restriction -3 < x < = 3 guarantees that division by zero can't happen.

Thank you,

MrB

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3 years ago

Read 2 more answers

Solve 3>y-2. Graph the solution

Alex787 [66]

Answer:

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Step-by-step explanation:

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2 years ago

Help me please :((: thank u

I am Lyosha [343]

Hey there!

In order for you to find the one that is equivalent to the given expression is to COMBINE YOUR LIKE TERMS then work from there!

12q + q^2 - 8 + q = ?

(q^2) + (12q + q) + (-8)

q^2 = q^2 since it doesn't have any like term(s)

12q + q = 12q + 1q = 13q

-8 = -8 since it doesn't have any like term(s)

Put all the numbers we solved for in An equation and Thats your answer!

If you did it correctly you should have: q^2 + 13q - 8

Answer: q^2 + 13q - 8 ✅

Good luck on your assignment and enjoy your day!

~LoveYourselfFirst:)

6 0

2 years ago

Would appreciate the help !

aleksandr82 [10.1K]

This is one pathway to prove the identity.

Part 1

$\frac{\sin(\theta)}{1-\cos(\theta)}-\frac{1}{\tan(\theta)} = \frac{1}{\sin(\theta)}\\\\\frac{\sin(\theta)}{1-\cos(\theta)}-\cot(\theta) = \frac{1}{\sin(\theta)}\\\\\frac{\sin(\theta)}{1-\cos(\theta)}-\frac{\cos(\theta)}{\sin(\theta)} = \frac{1}{\sin(\theta)}\\\\\frac{\sin(\theta)*\sin(\theta)}{\sin(\theta)(1-\cos(\theta))}-\frac{\cos(\theta)(1-\cos(\theta))}{\sin(\theta)(1-\cos(\theta))} = \frac{1}{\sin(\theta)}\\\\$

Part 2

$\frac{\sin^2(\theta)}{\sin(\theta)(1-\cos(\theta))}-\frac{\cos(\theta)-\cos^2(\theta)}{\sin(\theta)(1-\cos(\theta))} = \frac{1}{\sin(\theta)}\\\\\frac{\sin^2(\theta)-(\cos(\theta)-\cos^2(\theta))}{\sin(\theta)(1-\cos(\theta))} = \frac{1}{\sin(\theta)}\\\\\frac{\sin^2(\theta)-\cos(\theta)+\cos^2(\theta)}{\sin(\theta)(1-\cos(\theta))} = \frac{1}{\sin(\theta)}\\\\$

Part 3

$\frac{\sin^2(\theta)+\cos^2(\theta)-\cos(\theta)}{\sin(\theta)(1-\cos(\theta))} = \frac{1}{\sin(\theta)}\\\\\frac{1-\cos(\theta)}{\sin(\theta)(1-\cos(\theta))} = \frac{1}{\sin(\theta)}\\\\\frac{1}{\sin(\theta)} = \frac{1}{\sin(\theta)} \ \ {\checkmark}\\\\$

As the steps above show, the goal is to get both sides be the same identical expression. You should only work with one side to transform it into the other. In this case, the left side transforms while the right side stays fixed the entire time. The general rule is that you should convert the more complicated expression into a simpler form.

We use other previously established or proven trig identities to work through the steps. For example, I used the pythagorean identity $\sin^2(\theta)+\cos^2(\theta) = 1$ in the second to last step. I broke the steps into three parts to hopefully make it more manageable.

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