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lana [24]
3 years ago
13

URGENT!!! Help fast!!!

Mathematics
1 answer:
Mashutka [201]3 years ago
6 0

second graph from the left

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A student answers ​90% of the questions on a math exam correctly. If she answers 27 questions​ correctly, how many questions are
alexandr1967 [171]

Answer:

30 questions

Step-by-step explanation:

You can discover this by doing 27 divided by . 90 which equals 30.

6 0
3 years ago
Plz help meeeeeeee <br>I need to turn this in today<br>plzzz!!!!!!​
Katarina [22]

Answer:



Step-by-step explanation:

Step 1 Answer: Yes, 3

Step 2 Answer: Yes, 5

Step 3 Answer: No

8 0
3 years ago
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10+10= hhihihhhiihihihih
svetoff [14.1K]

Answer:

10+10=1010

Step-by-step explanation:

brainliest?

4 0
2 years ago
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Prove 2√(x) + 1/√(x + 1) &lt;= 2√(x+1) for all x in [0,inf)
EleoNora [17]
Start by multiplying each side of the inequality by \sqrt{x + 1} and simplifying:

2 \sqrt x + \frac{1}{\sqrt{x+1}}  \leq  2 \sqrt{x + 1}
(2 \sqrt x + \frac{1}{\sqrt{x+1}})(\sqrt{x + 1}) \leq (2 \sqrt{x + 1})(\sqrt{x + 1})
2 \sqrt{x(x + 1)} + 1 \leq 2(x + 1)
2 \sqrt{x^2 + x} + 1 \leq 2x + 2
2 \sqrt{x^2 + x} \leq 2x + 1
\sqrt{x^2 + x} \leq x + \frac{1}{2}

From here, we can square both sides to get

x^2 + x \leq (x + \frac{1}{2})^2
x^2 + x \leq x^2 + x + \frac{1}{4}
0  \leq \frac{1}{4}, which is always true.
3 0
3 years ago
I NEED THIS ANSWER FAST !
sleet_krkn [62]

Answer:

12

Step-by-step explanation:

6*2=12

3 0
3 years ago
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