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ANEK [815]
3 years ago
6

Hii please help i’ll give brainliest!!

Mathematics
2 answers:
Greeley [361]3 years ago
7 0

Answer:

your answer should be C.

castortr0y [4]3 years ago
3 0
I am going to guess C but I’m not entirely sure
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19. Ms. Porter had eight parties at her house last year. The number of guests at each party
KatRina [158]

Answer:

17

Step-by-step explanation:

Order the numbers from Smallest to Largest...

9, 10, 14, 15, 19, 21, 21, 27

Then the middle number is your answer.

Since there are 2 Middle Numbers, you add them together, and divide by 2.

Which would be, 17.

6 0
2 years ago
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Glenn rode his bike for 2 hours
Natali5045456 [20]
Cool. I rode mine for 1.
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2 years ago
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If g(x) = x2 + 2, find g(3).
sammy [17]

Answer:

8

Step-by-step explanation:

g(3)=3(2)+2

g(3)=6+2

g(3)=8

8 0
3 years ago
Write an equation that you can use to solve for x.<br><br> Enter your answer in the box.
patriot [66]
X + 60 = 100 (because they are vertical angles, which means that because they are directly opposite of each other, they are of the same measurements).

Solve for x. Isolate the x, subtract 60 from both sides

x + 60 = 100

x + 60 (-60) = 100 (-60)

x = 100 (-60)

x = 40

40° should be your answer

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8 0
3 years ago
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Find the taylor polynomial t3(x) for the function f centered at the number
inysia [295]

Answer:

t_3(x)=\frac{7\pi}{4}+\frac{7}{2}(x-1)-\frac{7}{4}(x-1)^2+\frac{7}{12}(x-1)^3

Step-by-step explanation:

We are given that

f(x)=7tan^{-1}(x)

a=1

T_n(x)=\sum_{r=0}^{n}\frac{f^r(a)(x-a)^r}{r!}

Substitute n=3 and a=1

t_3(x)=f(1)+f'(1)(x-1)+\frac{f''(1)(x-1)^2}{2!}+\frac{f'''(1)(x-1)^3}{3!}

f(x)=7tan^{-1}(x)

f(1)=7tan^{-1}(1)=7\times \frac{\pi}{4}=\frac{7\pi}{4}

Where tan^{-1}(1)=\frac{\pi}{4}

f'(x)=\frac{7}{1+x^2}

Using the formula

\frac{d(tan^{-1}(x))}{dx}=\frac{1}{1+x^2}

f'(1)=\frac{7}{2}

f''(x)=\frac{-14x}{(1+x^2)^2}

f''(1)=-\frac{7}{2}

f''(x)=-14x(x^2+1)^{-2}

f'''(x)=-14((x^2+1)^{-2}-4x^2(x^2+1)^{-3}})

By using the formula

(uv)'=u'v+v'u

f'''(x)=-14(\frac{x^2+1-4x^2}{(1+x^2)^3}

f'''(x)=(-14)\frac{-3x^2+1}{(1+x^2)^3}

f'''(1)=-14(\frac{-3(1)+1}{2^3})=\frac{7}{2}

Substitute the values

t_3(x)=\frac{7\pi}{4}+\frac{7}{2}(x-1)-\frac{7}{4}(x-1)^2+\frac{7}{2\times 3\times 2\times 1}(x-1)^3

t_3(x)=\frac{7\pi}{4}+\frac{7}{2}(x-1)-\frac{7}{4}(x-1)^2+\frac{7}{12}(x-1)^3

7 0
3 years ago
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