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Alekssandra [29.7K]
3 years ago
14

Somebody help this is algebra

Mathematics
2 answers:
sergejj [24]3 years ago
8 0
The answer is 11

You are meant to plug in 5 for x
So: h(5) = 2(5) + 1
h(5) = 11
jeyben [28]3 years ago
3 0

Answer:

2 is your answer

Step-by-step explanation:

If h(x)= 5 then 5= 2x +1

so 5 = 2x + 1

   -1            -1 on both sides

then 2x = 4

        2      2 divide by 2

which x = 2

Hope my answer has helped you!

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Hurry Please! As you move to the _______ on a number line the values of the numbers _______. Which choice best completes the sen
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right; increase

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3x = 2x + 12<br> What is the answer to this equation?
Strike441 [17]

Answer:

x=12

Step-by-step explanation:

3x=2x+12

-2x from each side

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Find the Taylor series for f(x) centered at the given value of a. [Assume that f has a power series expansion. Do not show that
Helga [31]

Answer:

\sum^\infty_{n=0} -5 (\frac{x+2}{2})^n

Step-by-step explanation:

Rn(x) →0

f(x) = 10/x

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Taylor series for the function <em>f </em>at the number a is:

f(x) =  \sum^\infty_{n=0} \frac{f^{(n)}(a)}{n!} (x - a)^n

f(x) = f(a) + \frac{f'(a)}{1!}(x-a)+\frac{f"(a)}{2!} (x-a)^2 + ... ............ equation (1)

Now we will find the function <em>f </em> and all derivatives of the function <em>f</em> at a = -2

f(x) = 10/x            f(-2) = 10/-2

f'(x) = -10/x²         f'(-2) = -10/(-2)²

f"(x) = -10.2/x³      f"(-2) = -10.2/(-2)³

f"'(x) = -10.2.3/x⁴     f'"(-2) = -10.2.3/(-2)⁴

f""(x) = -10.2.3.4/x⁵    f""(-2) = -10.2.3.4/(-2)⁵

∴ The Taylor series for the function <em>f</em> at a = -4 means that we substitute the value of each function into equation (1)

So, we get \sum^\infty_{n=0} - \frac{10(x+2)^n}{2^{n+1}} Or \sum^\infty_{n=0} -5 (\frac{x+2}{2})^n

4 0
3 years ago
HELP ME PLEASE!!!!!!!
inysia [295]
Here you go :) I sent the work in the picture below

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On a coordinate plane, an exponential function approaches y = 0 in quadrant 2 and curves up and increases in quadrant 1.
shepuryov [24]

Answer:

The domain of the function i.e. the value of x can be any real number and the range of the function is y > 0.

Step-by-step explanation:

On a coordinate plane, an exponential function approaches y = 0 in quadrant 2 and curves up and increases in quadrant 1.

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