3 years ago

20 Points!!! Pls help me out with this, and please no links.

Mathematics

2 answers:

lara [203]3 years ago

6 0

Answer:

Step-by-step explanation:

KIM [24]3 years ago

5 0

Answer: your answer is B

Step-by-step explanation:

Im so sorry if im wrong, have a good day!

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The cheerleading squad is making posters. They have 3 different colors of poster board and 4 different colors of markers. How ma

satela [25.4K]

12 different boards can be made
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3 years ago

Light travels at a speed of approximately question 2 options:

katrin2010 [14]

Hi,

The answer is B,
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r3t40

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4 years ago

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Use the appropriate substitutions to write down the first four nonzero terms of the Maclaurin series for the binomial (1+3x)^(-1

PolarNik [594]

Answer:

First term=1

Second term=-x

Third term= $2x^2$

Fourth term = $-\frac{28}{3!}x^3$

Step-by-step explanation:

We are given that function

$f(x)=(1+3x)^{-1/3}$

We have to find the first four non zero terms of the Maclaurin series for the binomial.

Maclaurin series of function f(x) is given by

$f(x)=f(0)+f'(0)x+\frac{1}{2!}f''(0)x^2+\frac{1}{3!}f'''(0)x^3+....$

$f(0)=(1+3x)^{\frac{-1}{3}}=1$

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

$f'(0)=-1$

$f''(x)=\frac{4}{3}\times 3 (1+3x)^{-\frac{7}{3}}$

$f''(0)=4$

$f'''(x)=-4\times \frac{7}{3}\times 3(1+3x)^{-\frac{10}{3}}$

$f'''(0)=-28$

Substitute the values we get

$(1+3x)^{-\frac{1}{3}}=1-x+\frac{4}{2!}x^2+\frac{-28}{3!}x^3+...$

$(1+3x)^{-\frac{1}{3}}=1-x+2x^2+\frac{-28}{3!}x^3+...$

First term=1

Second term=-x

Third term= $2x^2$

Fourth term = $-\frac{28}{3!}x^3$

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3 years ago

Find the product -5.5(-1/5)?<br> a. -1.1<br> b. 2.75<br> c. 1.1<br> d. -0.275

stealth61 [152]

The answer is 1.1 brainliest

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3 years ago

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me.norris gave his class 12 minutes to read. Caribe read 5 1/2 pages in that time. at what rate, in pages per hour did she read

Helen [10]

5.5/12 = x/60

One hour/ 12 minutes = 5

5.5•5=27.5

Caribe reads 27.5 pages per hour.

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3 years ago