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3 years ago

What is the mean to 24 72 46

Mathematics

2 answers:

Dimas [21]3 years ago

5 0

The answer that best fits the problem is 47. Its 47.3 but you round it to the nearest.

QveST [7]3 years ago

4 0

The answer would be 46.

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K-10/2=7 two Step Equation

aivan3 [116]

I hope this helps you

K-10=7.2

K-10=14

K=14+10

K=24

6 0

3 years ago

5m^{2}\sqrt{25m^{8}+15m^{6}+5m^{2}}

xxTIMURxx [149]

You can square the whole problem to cancel out the square root. Might make things easier.

5 0

3 years ago

Pls answer math<br> need help quick!!<br> look at the attached file below for question

harkovskaia [24]

Remark
I didn't know this was a calculus question. How dumb of me.

f(x + h) - f(x)
==========
x + h - h

50/6 - 51/6 =( -1/6+ h)/h
The h's will disappear when the limit is taken.

answer - 1/6 which is A. If anyone else answers take their answer.

7 0

3 years ago

Coefficiants of (2x+y)^4

sattari [20]

By the binomial theorem,

$(2x+y)^4=\displaystyle\sum_{k=0}^4\binom 4k(2x)^{4-k}y^k=\sum_{k=0}^4\binom 4k2^{4-k}x^{4-k}y^k$

where

$\dbinom nk=\dfrac{n!}{k!(n-k)!}$

Then the coefficients of the $x^{4-k}y^k$ terms in the expansion are, in order from $k=0$ to $k=4$ ,

$\dbinom 402^{4-0}=1\cdot2^4=16$

$\dbinom412^{4-1}=4\cdot2^3=32$

$\dbinom422^{4-2}=6\cdot2^2=24$

$\dbinom432^{4-3}=4\cdot2^1=8$

$\dbinom442^{4-4}=1\cdot2^0=1$

3 0

3 years ago

Leno4ka [110]

Answer:

$\frac{s^2-25}{(s^2+25)^2}$

Step-by-step explanation:

Let's use the definition of the Laplace transform and the identity given: $\mathcal{L}[t \cos 5t]=(-1)F'(s)$ with $F(s)=\mathcal{L}[\cos 5t]$ .

Now, $F(s)=\int_0 ^{+ \infty}e^{-st}\cos(5t) dt$ . Using integration by parts with u=e^(-st) and dv=cos(5t), we obtain that $F(s)=\frac{1}{5}\sin(5t)e^{-st} |_{0}^{+\infty}+\frac{s}{5}\int_0 ^{+ \infty}e^{-st}\sin(5t) dt=\int_0 ^{+ \infty}e^{-st}\sin(5t) dt$ .

Using integration by parts again with u=e^(-st) and dv=sin(5t), we obtain that

$F(s)=\frac{s}{5}(\frac{-1}{5}\cos(5t)e^{-st} |_{0}^{+\infty}-\frac{s}{5}\int_0 ^{+ \infty}e^{-st}\sin(5t) dt)=\frac{s}{5}(\frac{1}{5}-\frac{s}{5}\int_0^{+ \infty}e^{-st}\sin(5t) dt)=\frac{s}{5}-\frac{s^2}{25}F(s)$ .

Solving for F(s) on the last equation, $F(s)=\frac{s}{s^2+25}$ , then the Laplace transform we were searching is $-F'(s)=\frac{s^2-25}{(s^2+25)^2}$

3 0

3 years ago