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3 years ago

Need help please and fast thank you

Mathematics

1 answer:

shtirl [24]3 years ago

7 0

Answer:

The second batch

Step-by-step explanation:

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Calculating area of rectangle with a diagonal of 5m height of 3m and length of 4m

bagirrra123 [75]

Answer:

area = 12 m²

Step-by-step explanation:

area = L x W

area = 4 x 3 = 12 m²

8 0

3 years ago

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What is p equal to? Also if you could, can you tell me how you do it so I can do the rest by myself.

svp [43]

Answer:

p=-54

Step-by-step explanation:

$- \frac{p}{6} = 9 \\ \frac{ - p}{6} = 9 \\ \frac{ - p}{6} \times 6 = 9 \times 6 \\ - p = 54 \\ p = - 54$

6 0

3 years ago

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HELP ASAP IM STUPIDDDDD <br><br><br> Write the power as a product of factors: 9^7

algol [13]

Answer:

9x9x9x9x9x9x9, theres your answer

7 0

3 years ago

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Y = 60x + 20<br> y = 65x

dalvyx [7]

Answer:

4=x y=325

Step-by-step explanation:

60x + 20 = 65x

group the variables

20=5x

because you subtracted 60x from both

4=x

because you divided 5 from both

now substitute 5 for x

65×5 is 325

y=325

8 0

2 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