Please help with dilation(please click and help me answer)

Determine the lengths of the sides of the rectangle using the given area. Give answers both exactly and

Degger [83]

Answer:

Part a) Exactly values

The width of the rectangle is $\frac{\sqrt{70}}{2}\ cm$

The length of the rectangle is $2\sqrt{70}\ cm$

Part b) Approximately values

The width of the rectangle is $4.2\ cm$

The length of the rectangle is $16.7\ cm$

Step-by-step explanation:

see the attached figure to better understand the problem

we know that

The area of rectangle is equal to

$A=LW$

we have

$L=4x\ cm\\W=x\ cm$

substitute the values in the formula

$A=4x(x)$

$A=4x^2$

$A=70\ cm^2$

so

$70=4x^2$

Solve for x

Divide by 4 both sides

$\frac{70}{4}=x^2$

square root both sides

$x=(+/-)\frac{\sqrt{70}}{2}$

The solution is the positive value

$x=\frac{\sqrt{70}}{2}$

therefore

<em>Exactly values</em>

$L=4\sqrt{70}}{2}=2\sqrt{70}\ cm\\\\W=\frac{\sqrt{70}}{2}\ cm$

<em>Approximately values</em>

$L=4\sqrt{70}}{2}=16.7\ cm\\\\W=\frac{\sqrt{70}}{2}=4.2\ cm$

6 0

3 years ago

The exact value of cos5pi/12 is:

jonny [76]

Step 1: Convert $\frac{5\pi }{12}$ , which is in radians into degrees. To convert it multiply by $\frac{180}{\pi }$

$\frac{5\pi }{12} =\frac{180}{\pi }$

900/12

75

Step 2: 75 degrees isn't on the unit circle but 45 degrees and 30 degrees is. Since 45 + 30 = 75 you can use the cosine of 45 and 30 to find the exact value

cos45 = $\frac{\sqrt{2} }{2}$

cos30 = $\frac{\sqrt{3} }{2}$

Step 3: Add the cos45 and cos30 to get cos5pi/12

$\frac{\sqrt{2} }{2} +\frac{\sqrt{3} }{2} = \frac{\sqrt{2}+\sqrt{3}}{2}$

Hope this helped!

4 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 $x\mapsto x+1$ .

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