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3 years ago

Much help needed ....MULTIPLE CHOICE!

Mathematics

2 answers:

Marianna [84]3 years ago

8 0

Answer:

I am almost 100% sure that the answer is B. good luck on your test Ma'am

Step-by-step explanation:

agasfer [191]3 years ago

4 0

the answer is B: There was $2500 in the safe when Lori learned the combination.

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MA = 8x – 2, mB = 2x – 8, and mC = 94 – 4x. List the sides of ABC in order from shortest to longest.

LUCKY_DIMON [66]

I hope this helps you

3 0

3 years ago

127.5 is 250% of what?

Inga [223]

Your answer is 127.5 is 250% of 51

4 0

3 years ago

LCM of 14 and 49? Evaluate plz

ICE Princess25 [194]

Answer:

Step-by-step explanation:

Express the numbers in terms of the product of their prime factors

14 = 2 × 7

49 = 7 × 7 = 7²

Choose the factors which occur most often between the 2 numbers

LCM = 2 × 7² = 2 × 49 = 98

5 0

3 years ago

Im new to this app can someone help me please and thank u so much

GaryK [48]

Answer:

$3\frac{1}{4}$

Step-by-step explanation:

we have to divide the given distance by 3.

Then we will be able to find the distance between markers.

Given distance is

$9\frac{3}{4} \\=\frac{36+3}{4}\\ =\frac{39}{4}$

Since, there are three markers,

So, we have to divide this distance by 3

So, the distance between markers is-

$\frac{\frac{39}{4}} {3} \\=\frac{39}{4} * \frac{1}{3} \\=\frac{13}{4}$

So, the distance between marker is

$\frac{13}{4}$

Now, we have to convert this into mixed number.

Then, the distance is-

$\frac{13}{4}\\ =3\frac{1}{4}$

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 $x\mapsto x+1$ .

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

4 0

3 years ago