This table discluding the 18

N^2/3=9 indices equation

Mekhanik [1.2K]

Answer:

Exponential equations

Step-by-step explanation:

8 0

3 years ago

Hi :) Is this a quadratic function?

aniked [119]

Answer:

no this is not a quadratic function

6 0

3 years ago

An acute angle, θ, is in a right triangle such that sin of theta is equal to 3 over 8 period What is the value of cot θ?

Serggg [28]

We want to find the value of cot(θ) given that sin(θ) = 3/8 and θ is an angle in a right triangle, we will get:

cot(θ) = (√55)/3

So we know that θ is an acute angle in a right triangle, and we get:

sin(θ) = 3/8

Remember that:

sin(θ) = (opposite cathetus)/(hypotenuse)
hypotenuse = √( (opposite cathetus)^2 + (adjacent cathetus)^2)

Then we have:

opposite cathetus = 3

hypotenuse = 8 = √(3^2 + (adjacent cathetus)^2)

Now we can solve this for the adjacent cathetus, so we get:

adjacent cathetus = √(8^2 - 3^2) = √55

And we know that:

cot(θ) = (adjacent cathetus)/(opposite cathetus)

Then we get:

cot(θ) = (√55)/3

If you want to learn more, you can read:

brainly.com/question/15345177

8 0

3 years ago

Product of 4h × 5m is what

scoundrel [369]

20hm

That simple!. Good Luck.

4 0

3 years ago

Read 2 more answers

Rationalize the denominator and simplify.

cestrela7 [59]

If you're using the app, try seeing this answer through your browser: brainly.com/question/3151018

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$\mathsf{\dfrac{3\sqrt{6}+5\sqrt{2}}{4\sqrt{6}-3\sqrt{2}}}$

Multiply and divide by the conjugate of the denominator, which is $\mathsf{4\sqrt{6}+3\sqrt{2}:}$

$\mathsf{=\dfrac{\big(3\sqrt{6}+5\sqrt{2}\big)\cdot \big(4\sqrt{6}+3\sqrt{2}\big)}{\big(4\sqrt{6}-3\sqrt{2}\big)\cdot \big(4\sqrt{6}+3\sqrt{2}\big)}}$

Expand those products and eliminate the brackets:

$\mathsf{=\dfrac{\big(3\sqrt{6}+5\sqrt{2}\big)\cdot 4\sqrt{6}+ \big(3\sqrt{6}+5\sqrt{2}\big)\cdot 3\sqrt{2}}{\big(4\sqrt{6}-3\sqrt{2}\big)\cdot 4\sqrt{6}+ \big(4\sqrt{6}-3\sqrt{2}\big)\cdot 3\sqrt{2}}}\\\\\\ \mathsf{=\dfrac{12\cdot \big(\sqrt{6}\big)^2+20\sqrt{2}\cdot \sqrt{6}+9\cdot \sqrt{6}\cdot \sqrt{2}+15\cdot \big(\sqrt{2}\big)^2}{16\cdot \big(\sqrt{6}\big)^2-12\cdot \sqrt{2}\cdot \sqrt{6}+ 12\cdot \sqrt{6}\cdot \sqrt{2}-9\cdot \big(\sqrt{2}\big)^2}}\\\\\\ \mathsf{=\dfrac{12\cdot 6+20\sqrt{2\cdot 6}+9\sqrt{6\cdot 2}+15\cdot 2}{16\cdot 6-12\sqrt{2\cdot 6}+ 12\sqrt{6\cdot 2}-9\cdot 2}}\\\\\\ \mathsf{=\dfrac{72+20\sqrt{12}+9\sqrt{12}+30}{96-12\sqrt{12}+12\sqrt{12}-18}}$

$\mathsf{=\dfrac{72+29\sqrt{12}+30}{96-18}}\\\\\\ \mathsf{=\dfrac{102+29\sqrt{2^2\cdot 3}}{78}}\\\\\\ \mathsf{=\dfrac{102+29\cdot \sqrt{2^2}\cdot \sqrt{3}}{78}}\\\\\\ \mathsf{=\dfrac{102+29\cdot 2\sqrt{3}}{78}}$

$\mathsf{=\dfrac{102+58\sqrt{3}}{78}}\\\\\\ \mathsf{=\dfrac{\,\diagup\!\!\!\! 2\cdot \big(51+29\sqrt{3}\big)}{\diagup\!\!\!\! 2\cdot 39}}$

$\mathsf{=\dfrac{51+29\sqrt{3}}{39}}\quad\longleftarrow\quad\textsf{this is the answer.}$

I hope this helps. =)

4 0

3 years ago