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3 years ago

Can someone please help me!

Mathematics

1 answer:

qaws [65]3 years ago

6 0

Answer:

$y = - 16x ^{2} + 281$

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Each sprinkler in Rob’s sprinkler system waters a circular area. Sprinkler A is in an open part of Rob’s lawn and waters up to 1

Jlenok [28]

Answer:

10.17

Step-by-step explanation:

3 0

4 years ago

5y - 25x =10 <br><br><br> I need someone to help me solve this asap

ruslelena [56]

Answer:slope=10.000/2.000=5.000

x-intercept=2/5=minus 0.40000

y-intercept=2/1=2.00000

Step-by-step explanation:

8 0

4 years ago

You and your friends compare how many

Helen [10]

answer:

the mean for this is 27

step-by-step explanation:

all you have to do is add up all of the times

15, 18, 20, 22, 28, 37, 49

which would have given you

189

then you divide that number by how much times were written down

7 would be that number so

189/7 = 27

and this is what it is written in one equation

(15+18+20+22+28+37+49)/7

hope this helps you! :)

3 0

3 years ago

PLS HELP ASAP ILL GIVE BRAINLKEST PLS THANKS ASAP PLS

Eddi Din [679]

Answer: Answer is "D" aka 3.

Step-by-step explanation: the number sequence goes up by three on each bar; leaving the final number to be 3.

6 0

2 years ago

EXAMPLE 2 Prove that 9ex is equal to the sum of its Maclaurin series. SOLUTION If f(x) = 9ex, then f (n + 1)(x) = for all n. If

amm1812

Answer:

To Prove: $9e^x$ is equal to the sum of its Maclaurin series.

Step-by-step explanation:

If $f(x) = 9e^x$ , then $f ^{(n + 1)(x)} =9e^x$ for all n. If d is any positive number and |x| ≤ d, then $|f^{(n + 1)(x)}| = 9e^x\leq 9e^d.$

So Taylor's Inequality, with a = 0 and M = $9e^d$ , says that $|R_n(x)| \leq \dfrac{9e^d}{(n+1)!} |x|^{n + 1} \:for\: |x| \leq d.$

Notice that the same constant $M = 9e^d$ works for every value of n.

But, since $lim_{n\to\infty}\dfrac{x^n}{n!} =0 $ for every real number x$$ ,

We have $lim_{n\to\infty} \dfrac{9e^d}{(n+1)!} |x|^{n + 1} =9e^d lim_{n\to\infty} \dfrac{|x|^{n + 1}}{(n+1)!} =0$

It follows from the Squeeze Theorem that $lim_{n\to\infty} |R_n(x)|=0$ and therefore $lim_{n\to\infty} R_n(x)=0$ for all values of x.

$THEOREM\\If f(x)=T_n(x)+R_n(x), $where $T_n $is the nth degree Taylor Polynomial of f at a and $ lim_{n\to\infty} R_n(x)=0 \: for \: |x-a|$

By this theorem above, $9e^x$ is equal to the sum of its Maclaurin series, that is,

$9e^x=\sum_{n=0}^{\infty}\frac{9x^n}{n!}$ for all x.

6 0

3 years ago