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lbvjy [14]
2 years ago
15

Rose is 60 inches tall. how many feet tall is Rose

Mathematics
1 answer:
nata0808 [166]2 years ago
4 0

Answer:

5 feet

Step-by-step explanation: 60 divided by 12 inches is 5 feet

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Which expression is equivalent to 15a^8b^4/5a^4b?<br> 3a^2b^4<br> 3a^4b^3<br> 10a^4b^3<br> 10a^4b^4
EleoNora [17]

Answer:

3 a^12 b^5

Step-by-step explanation:

Simplify the following:

(15 a^8 b^4 a^4 b)/5

15/5 = (5×3)/5 = 3:

3 a^8 b^4 a^4 b

3 a^8 b^4 a^4 b = 3 a^(8 + 4) b^(4 + 1):

3 a^(8 + 4) b^(4 + 1)

4 + 1 = 5:

3 a^(8 + 4) b^5

8 + 4 = 12:

Answer: 3 a^12 b^5

5 0
3 years ago
Angle A and B are supplementary. Angle a has a measure of 80. What is the measure of angle B? The measure of angle B is
ololo11 [35]
Well when two angles are supplementary they equal 180.
So if angle A is 80 degrees then that would mean that angle B equals 100 degrees. 


Hope This Helped :) 



4 0
3 years ago
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
2 years ago
Please help me with this question, thank you and explain!!!
Novosadov [1.4K]

Total distance from w to z is -10 to 6 = 16

Add the ratios: 2 + 3 + 3 = 8

Divide distance by total ratio: 16/8 = 2

Multiply each ratio by 2 : 2x 2 = 4, 3 x 2 = 6, 3 x 2 = 6

WX is 4 units apart, add 4 to -10 to get x = -6

Answer: A. -6

5 0
2 years ago
A:b = 1:5<br> a:c = 2:1<br> please helppp
joja [24]

Answer:

what is the question I don't understand the question so sorry

5 0
2 years ago
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