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BabaBlast [244]
2 years ago
14

Find the slope of the line passing through the points (7, -2) and (3, - 7).

Mathematics
1 answer:
professor190 [17]2 years ago
3 0

Answer:

m = \frac{5}{4}

Step-by-step explanation:

To find the slope, we have to use the slope formula:

m = \frac{y_2 -y_1}{x_2 - x_1}.

Let (x_1,y_1) = (7, -2) and (x_2,y_2) = (3, -7). Then

m = \frac{(-7)-(-2)}{3-7} = \frac{-5}{-4} = \frac{5}{4}

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In the drawing g>h. which statement about the volumes of the two cylinders is true
Contact [7]

Answer:

The bottom option for ttm

Step-by-step explanation:

The volume of the left-hand cylinder is less than the volume of the right-hand cylinder.

7 0
3 years ago
What are the domain and range of f (x) = log x minus 5? domain: x > 0; range: all real numbers domain: x 5; range: y > 5 d
tresset_1 [31]

Answer:

domain: x>5 Range: y>-5

Step-by-step explanation:

6 0
3 years ago
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HELP PLEASE!!
shutvik [7]

Answer:

6.5 x 10^6 To answer this question, you need to divide the mass of the sun by the mass of mercury. So 2.13525 x 10^30 / 3.285 x 10^23 = ? To do the division, divide the mantissas in the normal fashion 2.13525 / 3.285 = 0.65 And subtract the exponents. 30 - 23 = 7 So you get 0.65 x 10^7 Unless the mantissa is zero, the mantissa must be greater than or equal to 0 and less than 10. So multiply the mantissa by 10 and then subtract 1 from the exponent, giving 6.5 x 10^6 So the sun is 6.5 x 10^6 times as massive as mercury.

:}

8 0
3 years ago
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Convert 13 over 7 into a mixed number
zalisa [80]

Answer:

13/7

7 goes into 13 1 time

13 minus 7 equals 6

Answer 1 6/7

Step-by-step explanation:


5 0
3 years ago
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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
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