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Damm [24]
3 years ago
14

Ill mark brainliest, question is attached

Mathematics
1 answer:
photoshop1234 [79]3 years ago
5 0

Answer:

5 is the answer!!!

Step-by-step explanation:

if its not im sorry lol

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Simplify the following expression and then write down the coefficient of x²: x² + x² + x² + x²
lapo4ka [179]

Answer:

The expression is 4x² and coefficient is 4.

Step-by-step explanation:

All have the same variables, x², so you add up together :

1{x}^{2}  +  1{x}^{2}  +  1{x}^{2}  +  1{x}^{2}

= 4 {x}^{2}

7 0
3 years ago
Find a formula for the inverse of the function. f(x) = e^6x − 9
STatiana [176]

Answer:

f^{-1}(x)=\frac{1}{6}\ln(x+9)

Step-by-step explanation:

So we have the function:

f(x)=e^{6x}-9

To solve for the inverse of a function, change f(x) and x, change the f(x) to f⁻¹(x), and solve for it. Therefore:

x=e^{6f^{-1}(x)}-9

Add 9 to both sides:

x+9=e^{6f^{-1}(x)}

Take the natural log of both sides:

\ln(x+9)=\ln(e^{6f^{-1}(x)})

The right side cancels:

\ln(x+9)=6f^{-1}(x)

Divide both sides by 6:

f^{-1}(x)=\frac{1}{6}\ln(x+9)

And we're done!

7 0
3 years ago
compare the right-hand and left-hand derivatives to show that the function is not differentiable at the point P. find all points
Softa [21]
f(x)=  \left \{ {{ \sqrt{x} } \atop {2x-1}} \right. 
\\
\\ f'(x)=  \left \{ {{  \frac{1}{2\sqrt{x}} } \atop {2}} \right. 
\\
\\  f'(1)=  \left \{ {{  \frac{1}{2}} \atop {2}} \right. 
\\
\\ f'(1^-) \neq f'(1^+)

Therefore, the function is not differentiable at x = 1.

f(x)= \sqrt{x} 
\\
\\f'(x)= \frac{1}{2 \sqrt{x} } 
\\
\\f'(0)=\frac{1}{2 \sqrt{0} } = \infty

Therefore, the function is not differentiable at x = 0.
3 0
3 years ago
The power P required to do a fixed amount of work varies inversely as the time t. If a power of 15 J/h is required to do a fixed
Ira Lisetskai [31]

Answer:

7.5

Step-by-step explanation:

if 2h =15j/h

then 1h=1/2 *15

=7.5

4 0
3 years ago
Read 2 more answers
Need a quick answer please
Olenka [21]

Step-by-step explanation:

2x + 5 = I / 2 ×16

2x=3

x=3/2

7 0
3 years ago
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