Ask question

Ask question

All categories

3 years ago

13

Please help! (will mark as brainliest!)

1 answer:

Aloiza [94]3 years ago

5 0

It’s a constant function because it stays constant the whole time

You might be interested in

I need help with this question

zhuklara [117]

4 times 3.14 times 8 squared
803.8

7 0

3 years ago

Show all steps!!!

Lunna [17]

$-\frac{1}{3}\left|1\cdot \left(\frac{1}{2}\right)^2\right|-\frac{2}{3}-2^3-\frac{2}{3}\left|-\frac{1}{2}\right|^3$
$\frac{1}{3}\left|1\cdot \left(\frac{1}{2}\right)^2\right|=\frac{1}{12}$
$\frac{2}{3}\left|-\frac{1}{2}\right|^3=\frac{1}{12}$
$=-\frac{1}{12}-\frac{2}{3}-2^3-\frac{1}{12}$
$=-\frac{53}{6}$

4 0

3 years ago

How do I add and subtract radicals with no like terms

liq [111]

Most real number arithmetic is pretty vacuous

$\sqrt 2 + \sqrt 3 = \sqrt{2} + \sqrt{3}$

That's about as good a way as possible to write this particular real number. But it's far from the only way.

$(\sqrt 2 + \sqrt 3)^2 = \sqrt{2}^2 + 2\sqrt 2 \sqrt 3 + \sqrt{3}^2 = 5 + 2 \sqrt 6$

so

$\sqrt 2 + \sqrt 3 = \sqrt{5 + 2 \sqrt 6}$

Sometimes you can factor something out and you have a common radical:

$\sqrt{32} + \sqrt{128} = \sqrt{2^5} + \sqrt{2^7} = 4 \sqrt{2} + 8 \sqrt{2} = 12 \sqrt{2}$

But most of them are vacuous,

$3\sqrt{7} + 34 \sqrt{13} = 3\sqrt{7} + 34 \sqrt{13}$

8 0

4 years ago

Please solve for g!!!!

laiz [17]

The answer to the question

3 0

3 years ago

Read 2 more answers

Find the circumference and area of a circle with radius 3 yd. use the value 3.14 for pie

just olya [345]

Answer: circumference = 2Πr²

Where r= 3

Π=3.14

Therefore Circumference = 2×3.14× 3²

= 2×28.26

Circumference=56.52

Formula:

Subtitute in 3 for r.

Solve and the answer is 28.26

4 0

3 years ago

Read 2 more answers