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vlada-n [284]
2 years ago
6

.

Mathematics
1 answer:
allsm [11]2 years ago
8 0

Answer:

wow.

Step-by-step explanation:

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Add using scientific notation.<br> (5.2 x 10') + (3.01 x 10^)
satela [25.4K]

Answer: What's the exponents?

Step-by-step explanation:

5.2+3.01=8.21

but then there's no exponents that you stated. So it would be 8.21x10^?

If the tens do have exponents above them, add them together.

I hope this answer helps you!

7 0
3 years ago
At a university, 6 out of every 13 students live on campus; if there are 156 students enrolled how many live on campus?
Andru [333]

Answer:

72

Step-by-step explanation:

156 divided by 13 times 6.

3 0
2 years ago
Read 2 more answers
Thinking the perimeter of the rectangle is 3x + 2 + x + 3x + 2 + X.
givi [52]
They are both right because the first expression is the perimeter but expanded in terms of length.The second expression is shortened to be a factor expression but they both represent the length.
The benefit of the first one is you can see how it’s expanded and plug the numbers in and the second one is more easy to plug in the numbers with ease.
7 0
3 years ago
19- (1 - 4x) - 9x +30<br>Simplify​
Lapatulllka [165]

19 - (1 - 4x) - 9x + 30 = -5x+48

(Don't forget the negativity symbol before the answer, if it isn't visible. It's not fully visible on my screen.)

6 0
3 years ago
How do you solve this inequality? -x^2-3x+17_&gt;0​
andrew11 [14]

\bf -x^2-3x+14\ge -3 \implies 0\ge x^2+3x-17 \\\\[-0.35em] ~\dotfill\\\\ ~~~~~~~~~~~~\textit{quadratic formula} \\\\ y=\stackrel{\stackrel{a}{\downarrow }}{1}x^2\stackrel{\stackrel{b}{\downarrow }}{+3}x\stackrel{\stackrel{c}{\downarrow }}{-17} \qquad \qquad x= \cfrac{ - b \pm \sqrt { b^2 -4 a c}}{2 a}

\bf x=\cfrac{-3\pm\sqrt{3^2-4(1)(-17)}}{2(1)}\implies x=\cfrac{-3\pm \sqrt{9+68}}{2} \\\\\\ x=\cfrac{-3\pm\sqrt{77}}{2} \\\\[-0.35em] ~\dotfill\\\\ 0\ge \left( x+\cfrac{-3+\sqrt{77}}{2} \right)\left( x+\cfrac{-3-\sqrt{77}}{2} \right) \\\\\\ \cfrac{3-\sqrt{77}}{2}\ge x\qquad and\qquad \cfrac{3+\sqrt{77}}{2}\ge x

8 0
3 years ago
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