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2 years ago

9

Hi everyone, I'm having trouble with these questions and I'm not sure how to do them/where to start. Does anyone have a solution

to it? It will help me with my future problems on this topic!

1 answer:

Iteru [2.4K]2 years ago

3 0

Answer:

sounds simple, but getting to know people can be seriously hard. We've got some tips to help you ease the process and get to know people ...

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Is The polar form of a complex number is unique?

Pie

Answer:

no

Step-by-step explanation:

The angle in the polar form of a complex number can have any multiple of 2π radians added to it, and the number will be the same number. That is, there are an infinite number of representations of a complex number in polar form.

6 0

3 years ago

Find the values of the six trigonometric functions of an angle in standard position if the point with coordinates (6,-5) lies on

Tom [10]

C i guesss because you just said it

4 0

3 years ago

b. Tyler practiced the clarinet for 64% as much time as Han practiced the piano. How long did he practice?

Ksivusya [100]

Answer: 48 minutes

Step-by-step explanation:

Han spent 75 minutes practicing the piano.

Tyler spent only 64% as much as that. How long did Tyler practice:

= 75 * 64%

= 48 minutes

6 0

2 years ago

How do you simplify <img src="https://tex.z-dn.net/?f=%5Cfrac%7B%5Csqrt%7B2%7D%20%2B%5Csqrt%7B6%7D%20%7D%7B%5Csqrt%7B8%7D%20%2B%

blondinia [14]

The trick is to exploit the difference of squares formula,

$a^2-b^2=(a-b)(a+b)$

Set a = √8 and b = √6, so that a + b is the expression in the denominator. Multiply by its conjugate a - b:

$(\sqrt8+\sqrt6)(\sqrt8-\sqrt6)=(\sqrt8)^2-(\sqrt6)^2=8-6=2$

Whatever you do to the denominator, you have to do to the numerator too. So

$\dfrac{\sqrt2+\sqrt6}{\sqrt8+\sqrt6}=\dfrac{(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)}{(\sqrt8+\sqrt6)(\sqrt8-\sqrt6)}=\dfrac{(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)}2$

Expand the numerator:

$(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)=\sqrt{2\cdot8}+\sqrt{6\cdot8}-\sqrt{2\cdot6}-(\sqrt6)^2$

$(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)=\sqrt{16}+\sqrt{48}-\sqrt{12}-6$

$(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)=4+\sqrt{48}-\sqrt{12}-6$

$(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)=-2+\sqrt{12}(\sqrt4-1)$

$(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)=-2+\sqrt{12}(2-1)$

$(\sqrt2+\sqrt6)(\sqrt8-\sqrt6}=-2-\sqrt{12}$

So we have

$\dfrac{\sqrt2+\sqrt6}{\sqrt8+\sqrt6}=-\dfrac{2+\sqrt{12}}2$

But √12 = √(3•4) = 2√3, so

$\dfrac{\sqrt2+\sqrt6}{\sqrt8+\sqrt6}=-\dfrac{2+2\sqrt3}2=\boxed{-1-\sqrt3}$

7 0

3 years ago

Reduce the fraction completely.96/53

Alex777 [14]

1 43/53 is the answer in a mixed fraction.

3 0

3 years ago

Read 2 more answers