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2 years ago

PLS HELP...IM DESPERATE

Mathematics

2 answers:

tatuchka [14]2 years ago

5 0

The answer is 9.

The median is the central number of a data set. Arrange data points from smallest to largest and locate the central number. This is the median.

postnew [5]2 years ago

4 0

Answer:

Step-by-step explanation:

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Fwe points whats x? x+21=99??<br> injoy <br> anyone wan cht

umka21 [38]

Answer:

The answer is 78

Step-by-step explanation:

99-78=21

5 0

3 years ago

Read 2 more answers

Pplllleeaaassseeeee hellppppp meeee<br> this is easy but i dont get it

AURORKA [14]

Answer:

choice 1) -6 and choice 2) -5

Step-by-step explanation:

choice 1)

-(1/2)-6)+6 > 8

9 > 8

choice 2)

-(1/2)(-5)+6 > 8

8 1/2 > 8

4 0

2 years ago

A scale model of a human heart is 16ft long. The scale is 32:1. How many inches long is the actual heart it represents?

melisa1 [442]

0.5 or 6 inches
The model=32
Actual size=1

32/1= 16/?
If 32 is 1 foot which equals 12 inches, and 16 is half of 32, then 16 is 0.5 inches or 6 inches

4 0

3 years ago

If it's 8:23 pm now... What time would it be in 45 min? ( I want to see what everyone comes up with)

andreyandreev [35.5K]

8:23, after 45 minutes is technically 8: 68

however, the hour ends at 60 minutes, so your true answer is 9:08 pm

hope this helps

6 0

3 years ago

This is a question on my partial fractions homework, but no matter what I try I can't figure it out..

Ierofanga [76]

$\dfrac{x^2+x+1}{(x+1)^2(x+2)}=\dfrac{a_1x+a_0}{(x+1)^2}+\dfrac b{x+2}$
$\implies\dfrac{x^2+x+1}{(x+1)^2(x+2)}=\dfrac{(a_1x+a_0)(x+2)+b(x+1)^2}{(x+1)^2(x+2)}$
$\implies x^2+x+1=(a_1+b)x^2+(2a_1+a_0+2b)x+(2a_0+b)$
$\implies\begin{cases}a_1+b=1\\2a_1+a_0+2b=1\\2a_0+b=1\end{cases}\implies a_1=-2,a_0=-1,b=3$

So you have

$\displaystyle\int_0^2\frac{x^2+x+1}{(x+1)^2(x+2)}\,\mathrm dx=-2\int_0^2\frac x{(x+1)^2}\,\mathrm dx-\int_0^2\frac{\mathrm dx}{(x+1)^2}+3\int_0^2\frac{\mathrm dx}{x+2}$
$=\displaystyle-2\int_1^3\dfrac{x-1}{x^2}\,\mathrm dx-\int_0^2\frac{\mathrm dx}{(x+1)^2}+3\int_0^2\frac{\mathrm dx}{x+2}$

where in the first integral we substitute $x\mapsto x+1$ .

$=\displaystyle-2\int_1^3\left(\frac1x-\frac1{x^2}\right)\,\mathrm dx-\frac1{1+x}\bigg|_{x=0}^{x=2}+3\ln|x+2|\bigg|_{x=0}^{x=2}$
$=-2\left(\ln|x|+\dfrac1x\right)\bigg|_{x=1}^{x=3}-\dfrac23+3(\ln4-\ln2)$
$=-2\left(\ln3+\dfrac13-1\right)-\dfrac23+3\ln2$
$=\dfrac23+\ln\dfrac89$

4 0

3 years ago