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Mathematics

2 answers:

Alex73 [517]2 years ago

7 0

I gotchu! So you’re going to use the open circle one with the arrow pointing right, then put it at the point -0.75

guajiro [1.7K]2 years ago

4 0

The anwser is c your welcome

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Please dont steal points only real answers

morpeh [17]

1. 4 1/5
2. 11/24
3. 12 1/4
4. 2
5. 3 5/12

I did the math but I’m sorry if it’s not all correct

8 0

3 years ago

What is the order of the set A={-1,20,3,17,5}

laiz [17]

Hey!

Hope this helps...

~~~~~~~~~~~~~~~~~~~~~~~~~~~

I'm assuming your talking about least to greatest, OR greatest to least. If so, that's what I'll do...

Least to Greatest: -1, 3, 5, 17, 20

Greatest to Least: 20, 17, 5, 3, -1

~~~~~~~~~~~~

Unfortunately I did not completely understand the question, for future reference please give AS MUCH detail as possible so that your fellow brainly members can help you ASAP...

And thanks for using Brainly!

3 0

3 years ago

Just want to check to make sure this is right. I got 81 as the average so D seemed right.

sergiy2304 [10]

Im not 100% sure but im pretty sure you are right brother! God bless...Good luck!

6 0

3 years ago

Simplify the answer (3 to the 4th power and (3 to the negative 6th power)

AlladinOne [14]

3^4 and 3^-6
multiply, add, subtract?

8 0

3 years ago

Can someone solve this with steps cause idk how to solve the radicals

IceJOKER [234]

$\bf 343^{\frac{2}{3}}+36^{\frac{1}{2}}-256^{\frac{3}{4}}\qquad \begin{cases}
343=7\cdot 7\cdot 7\\
\qquad 7^3\\
36=6\cdot 6\\
\qquad 6^2\\
256=4\cdot 4\cdot 4\cdot 4\\
\qquad 4^4
\end{cases}\\\\\\ (7^3)^{\frac{2}{3}}+(6^2)^{\frac{1}{2}}-(4^4)^{\frac{3}{4}}
\\\\\\
\sqrt[3]{(7^3)^2}+\sqrt[2]{(6^2)^1}-\sqrt[4]{(4^4)^3}\implies \sqrt[3]{(7^2)^3}+\sqrt[2]{(6^1)^2}-\sqrt[4]{(4^3)^4}
\\\\\\
7^2+6-4^3\implies 49+6-64\implies -9$

to see what you can take out of the radical, you can always do a quick "prime factoring" of the values, that way you can break it in factors to see who is what.

8 0

3 years ago