Draw and label parts of a triangle

Mathematics

1 answer:

SCORPION-xisa [38]3 years ago

4 0

Answer:

Step-by-step explanation:

Sides of the triangle: AB , BC , AC

Angles: ∠A , ∠B , ∠C

Vertices: A , B ,C

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FRÉD AND ETHEL WACHED TV FOR SEVERAL HOURS ONE AFTERNOON. THEY BEGAN AT 1:30 P.M AND CONTINUED WACHING TV FOR 3 HOURS AND 45 MIN

QveST [7]

$30 + 45 =75$
$3 + 1 = 4$
There are 60 minutes in one hour, so 75 minutes is equal to 1 hour and 15 minutes.

$4 + 1 = 5$
Their mom turned the TV off at 5:15.

7 0

3 years ago

The distance by road from town A to town B is 257 km. What is 50% of that distance?

Marina86 [1]

Answer: 128.5km

Step-by-step explanation:

Since we are given the information that the distance by road from town A to town B is 257 km. To get 50% of the distance, we simply have to multiply the distance given by 50%. This will be:

= 50% × 257km

= 50/100 × 257km

= 0.5 × 257km

= 128.5km

Therefore, 50% of the distance is 128.5km.

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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 $x\mapsto x+1$ .

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