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Whitepunk [10]
2 years ago
8

Write an equation in point-slope form of the line that passes through the point (2, -8) and has a slope of 4

Mathematics
1 answer:
ozzi2 years ago
3 0

Answer:

Point slope: y - y1 = m(x - x1)

Plug those in

Solution: y + 8 = 4(x - 2)

Step-by-step explanation:

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3. Find the quotient 20.7.3.<br> A 6.9<br> B 17.7<br> D 62.1
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Answer:

if your trying to say 20 ÷ 73 it is 0.2739 or 73 ÷ 20 it's 3.65

3 0
3 years ago
PLZ HELP!!! Use limits to evaluate the integral.
Marrrta [24]

Split up the interval [0, 2] into <em>n</em> equally spaced subintervals:

\left[0,\dfrac2n\right],\left[\dfrac2n,\dfrac4n\right],\left[\dfrac4n,\dfrac6n\right],\ldots,\left[\dfrac{2(n-1)}n,2\right]

Let's use the right endpoints as our sampling points; they are given by the arithmetic sequence,

r_i=\dfrac{2i}n

where 1\le i\le n. Each interval has length \Delta x_i=\frac{2-0}n=\frac2n.

At these sampling points, the function takes on values of

f(r_i)=7{r_i}^3=7\left(\dfrac{2i}n\right)^3=\dfrac{56i^3}{n^3}

We approximate the integral with the Riemann sum:

\displaystyle\sum_{i=1}^nf(r_i)\Delta x_i=\frac{112}n\sum_{i=1}^ni^3

Recall that

\displaystyle\sum_{i=1}^ni^3=\frac{n^2(n+1)^2}4

so that the sum reduces to

\displaystyle\sum_{i=1}^nf(r_i)\Delta x_i=\frac{28n^2(n+1)^2}{n^4}

Take the limit as <em>n</em> approaches infinity, and the Riemann sum converges to the value of the integral:

\displaystyle\int_0^27x^3\,\mathrm dx=\lim_{n\to\infty}\frac{28n^2(n+1)^2}{n^4}=\boxed{28}

Just to check:

\displaystyle\int_0^27x^3\,\mathrm dx=\frac{7x^4}4\bigg|_0^2=\frac{7\cdot2^4}4=28

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3 years ago
Find the y-intercept (b) of line AB? -5 -6 5 6
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Answer:

Step-by-step explanation:

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Answer: The correct answer is fraction 3 and 3 over 5 cubic feet

Step-by-step explanation:

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3 years ago
How do you solve the following equation for x:
Advocard [28]
You would have to figure out what x would be...  I use trial and error because i don't really know a better way to do it
8 0
3 years ago
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