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Assoli18 [71]
3 years ago
6

Evaluate. Round to the nearest hundredth.

Mathematics
1 answer:
KIM [24]3 years ago
3 0

Answer:

22. 6

23. 8^x

24. log2(16^5) = log2((2^4)^5) = log2(2^20) = 20

25. 2x + 1

27. log4(16^(x - 1)) = log4((4^2)^(x - 1)) = log4(4^(2x - 2)) = 2x - 2

28. 17


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I need help!!! ASAP!!!! !!!!!
GalinKa [24]

Answer:

1.

Volume= 729cm^3

Height= 9cm^2

Area of Base= 81cm^2

2.

Volume=450m^3

Height= 3m^2

Area of Base=150m^2

3.

Volume= 480 cm^2

Area of base=75cm^2

Height= 8cm^2

4.

Volume =120in^3

Area of Base=20in^2

Height= 6in^2

Step-by-step explanation:

Volume = (length) Times (Width) Times (Height)

Area os base = (Length) Times (width)

3 0
3 years ago
"Find x" I managed to get -6/2c, I just want to double check if that is correct
Soloha48 [4]

- 6/2c is the correct solution

collect like terms : 3cx - cx = 2cx

hence 2cx = - 6

dividing both sides by 2c gives x = - 6/2c


7 0
4 years ago
Write in word form 10x10
Arada [10]

Answer:

ten times time

Step-by-step explanation:

ten time ten or ten multiple by ten

give brainliest please I need it to level up

6 0
3 years ago
Read 2 more answers
Prove 2√(x) + 1/√(x + 1) <= 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
If (42)p = 414, what is the value of p? (1 point) Question 5 options: 1) 7 2) 8 3) 12 4) 16
timama [110]
I didn’t get an answer that matches your answer choices, but here are mine:
exact form: p = 69/7
decimal form: p = 9.857142
mixed number form: p = 9 (6/7)
6 0
3 years ago
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