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rosijanka [135]
3 years ago
13

*HIGH POINT QUESTION* if 6x = 12, and xy = 6, then what is 12y? Please answer!

Mathematics
2 answers:
sasho [114]3 years ago
5 0
Does y = 1 because x = 6 and xy=6 so 6 time 1 =6? not sure just what i think
Vladimir [108]3 years ago
5 0
12y would be the quotient that you get after you multiply 6x=12
6x • Xy =12y

I hope it helps,I’m still kinda new to this topic. ♥️
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A flagpole is at the top of a building. 300 feet from the base of the building, the angle of elevation of the top of the pole is
Grace [21]

Answer: We can use tangent here to find the length of the flagpole

First, find the length of the building + the flag pole

tan 32 = x/300

x=300tan32=187.460805573

(length of building + flagpole)

find length of building

tan 30 = x/300

x=173.205080757

subtract

187.460805573-173.205080757=14.255724816

round, length of flagpole = 14 feet

7 0
3 years ago
Please help! 15 POINTS!<br><br> Questions above
Sonbull [250]
1. is actually 4m
and 2. is actually 44cm^2
5 0
3 years ago
3. *
Sati [7]

Answer:

A first numbers cant be same

5 0
3 years ago
The function y = x2 is transformed to y = (x + 4)2. Which statement is true about the transformed function? It is an even functi
xxMikexx [17]
Even function: f(-x) = f(x). If you replace x by -x you should find the same function.
Odd function: f(-x) = -f(x). If you replace x by -x you  find the same function with opposite sign;
Is f(-x) = f(x)?
f(x) = (x+4)² =  x² + 8x +16
f(-x) = (-x+4)² = x² - 8x + 16, then it's not an even function

Is f(-x) = -f(x)?
f(-x) = (-x+4)² = x² + 8x + 16 , then it's not an odd function
It is neither an even nor an odd function
7 0
4 years ago
Consider a normal distribution curve where 90-th percentile is at 20 and the 25-th percentile is at 10. use this information to
ivann1987 [24]
The 25th percentile corresponds to z ≈ -0.67.
The 90th percentile corresponds to z ≈ 1.28.
So, you can write two equations in μ and σ as here:
.. μ -0.67σ = 10
.. μ +1.28σ = 20

The solution is
.. μ ≈ 13.45
.. σ ≈ 5.11

5 0
3 years ago
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