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galben [10]
3 years ago
10

I need this problem simplified thanks! :)

Mathematics
1 answer:
m_a_m_a [10]3 years ago
7 0
I dont understand that

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I NEED THIS ASAP
andrey2020 [161]

Answer:

28

Step-by-step explanation:

8 0
3 years ago
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) Identify each graph as a linear function, nonlinear function, or not a function.
suter [353]

Answer:

1.Non linear function

2. Linear function

3. Non linear function

4. Not a Function

Step-by-step explanation:

4 0
3 years ago
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% is the correct percentage conversion for 2/5
denis-greek [22]

Answer:

%40

Step-by-step explanation:

<u>Step 1:  Divide 2/5</u>

2/5 = 0.4

<u>Step 2:  Convert to Decimal by multiplying by 100</u>

0.4 * 100 = %40

Answer:  %40

6 0
3 years ago
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At what point does the curve have maximum curvature? y = 9 ln(x) (x, y) =
Andrews [41]

y = 9ln(x) 
<span>y' = 9x^-1 =9/x</span>
y'' = -9x^-2 =-9/x^2

curvature k = |y''| / (1 + (y')^2)^(3/2) 

<span>= |-9/x^2| / (1 + (9/x)^2)^(3/2) 
= (9/x^2) / (1 + 81/x^2)^(3/2) 
= (9/x^2) / [(1/x^3) (x^2 + 81)^(3/2)] 
= 9x(x^2 + 81)^(-3/2). 

To maximize the curvature, </span>

we find where k' = 0. <span>
k' = 9 * (x^2 + 81)^(-3/2) + 9x * -3x(x^2 + 81)^(-5/2) 
...= 9(x^2 + 81)^(-5/2) [(x^2 + 81) - 3x^2] 
...= 9(81 - 2x^2)/(x^2 + 81)^(5/2) 

Setting k' = 0 yields x = ±9/√2. 

Since k' < 0 for x < -9/√2 and k' > 0 for x > -9/√2 (and less than 9/√2), 
we have a minimum at x = -9/√2. 

Since k' > 0 for x < 9/√2 (and greater than 9/√2) and k' < 0 for x > 9/√2, 
we have a maximum at x = 9/√2. </span>

x=9/√2=6.36

<span>y=9 ln(x)=9ln(6.36)=16.66</span>  

the answer is
(x,y)=(6.36,16.66)
7 0
4 years ago
PLEASE HELP! 19 points!
Blababa [14]

Answer:

3x^{2}-27x-30

=3x^{2}-(30-3)x-30

=3x^{2}-30x+3x-30

=3x(x-10)+3(x-10)

=(3x+3)(x-10)

Step-by-step explanation:

4 0
3 years ago
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