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FromTheMoon [43]
3 years ago
6

Choose the correct zeros for the following quadratic function

Mathematics
1 answer:
Contact [7]3 years ago
7 0
I need more information
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Solve the equation below<img src="https://tex.z-dn.net/?f=%5Csqrt%5B5%5D%7B27%28x-2%29%7D%3D3" id="TexFormula1" title="\sqrt[5]{
Mariana [72]

Answer:

x = 11

Step-by-step explanation:

Raising both sides to the fifth power, we get:

27(x - 2) = 3⁵

x - 2 = 3⁵ / 27

x - 2 = 3⁵ / 3³

x - 2 = 3⁽⁵⁻³⁾ = 3² = 9

x = 11

8 0
3 years ago
What is 4log1/2^w(2log1/2^u-3log1/2^v written as a single logarithm?
IrinaK [193]
Given:

4log1/2^w (2log1/2^u-3log1/2^v)

Req'd:

Single logarithm = ?

Sol'n:

First remove the parenthesis,

4 log 1/2 (w) + 2 log 1/2 (u) - 3 log 1/2 (v)

Simplify each term,

Simplify the 4 log 1/2 (w) by moving the constant 4 inside the logarithm;
Simplify the 2 log 1/2 (u) by moving the constant 2 inside the logarithm;
Simplify the -3 log 1/2 (v) by moving the constant -3 inside the logarithm:

 log 1/2 (w^4)  + 2 log 1/2 (u) - 3 log 1/2 (v) 
log 1/2 (w^4) + log 1/2 (u^2) - log 1/2 (v^3)

We have to use the product property of logarithms which is log of b (x) + log of b (y) = log of b (xy):

Thus,

Log of 1/2 (w^4 u^2) - log of 1/2 (v^3) 

then use the quotient property of logarithms which is log of b (x)  - log of b (y) = log of b (x/y)

Therefore, 

log of 1/2 (w^4 u^2 / v^3)

and for the final step and answer, reorder or rearrange w^4 and u^2:

log of 1/2 (u^2 w^4 / v^3)  
8 0
3 years ago
Read 2 more answers
Please help me due at 11:59 show how you got it if you can.
xz_007 [3.2K]
No clue if this is right

5 0
3 years ago
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7th grade math help me pleaseee
Rufina [12.5K]

Answer:

A. 6.25x + 7.50 = 26.25

Step-by-step explanation:

Clara paid 7.50 for an admission ticket, each round was 6.25 and she did it for x rounds. 6.25x is the same thing as 6.25 times x rounds.

5 0
3 years ago
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Given the function f(x) = 4x2 – 2x4, determine the derivative of f at
ruslelena [56]

Answer:

I'm bout to roast yo old potatoe head looking a*s

8 0
4 years ago
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