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2 years ago

Instructions: Find the missing side lengths. Leave your answers as radicals in simplest form.

Mathematics

1 answer:

luda_lava [24]2 years ago

7 0

Answer: m=3 radical 2 and n=3

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Use the triangles Name one pair of congruent angles.

Likurg_2 [28]

Your correct answer is option B

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3 years ago

Find the two inequalities that define the region on the grid that is not shaded

hoa [83]

Answer:

y 》 2

y > -x + 3

Step-by-step explanation:

Equation of the bold line:

m = (-3/3) = -1

y = -x + 3

Shaded region is below y=2 (dotted)

So, it's defined by y < 2

Also, it's below the bold line

y 《 - x + 3

Everything outside these is not shaded:

y 》2

and y > -x + 3

3 0

3 years ago

There are 22 events IN ALL at an indoor track and field meet. The ratio of TRACK events to FIELD events is 8:3. How many of the

fenix001 [56]

Answer:

There are 16 track events and 6 field events

8 0

2 years ago

The price of a particular stock is represented by the linear equation y=-0.91x+103.47

Mama L [17]

To find the price of the stock after 12 weeks, you would have to plug in 12 to x

That would make your equation: y = -0.91(12) + 103.47

Then, you solve:

-0.91(12) would be -10.12

Your equation will now look like this:

y = - 10.12 + 103.47

Then you add -10.12 to 103.47

-10.12 + 103.47 = 92.55

y = 92.55

After 12 weeks, the price of the stocks would be $92.55

4 0

3 years ago

Read 2 more answers

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