Question 8 is attached! Please explain how you got your answer.

What is the point-slope form of a line that has a slope of –4 and passes through point (–3, 1)?

ivolga24 [154]

Answer:

y-1= -4(x-(-3))

y minus 1 = negative 4 left-bracket x minus (negative 3) right-bracket

Step-by-step explanation:

The formula for point slope form is written: $y-y_{1} = m(x-x_{1})$ . Now we fill in our know information. m=-4 (because that is the slope), $y_{1}$ =1 (It is the y value in the ordered pair), $x_{1}$ = -3 (It is the x value in the ordered pair).

y-1=-4(x- (-3))

4 0

2 years ago

Write the first 4 terms of the arithmetic sequence With first term 37 and common difference 9

alexandr402 [8]

Answers:

first term = 37

second term = 46

third term = 55

fourth term = 64

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Explanation:

Start at 37. Add 9 to this term to get the second term. Add 9 to that result to get the next term, and so on.

first term = 37

second term = 37+9 = 46

third term = 46+9 = 55

fourth term = 55+9 = 64

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The nth term formula is

a(n) = 9n+28

It can be found by plugging a = 37 and d = 9 into the general form

a(n) = a + d*(n-1)

4 0

2 years ago

Read 2 more answers

Y = 3 sin2x, y = 0, 0 ≤ x ≤ π; about the x−axis

Dominik [7]

I assume you're revolving the region with those bounds about the x-axis, and supposed to find the volume.

Via the disk method,

$\displaystyle\pi\int_0^\pi(3\sin2x)^2\,\mathrm dx=9\pi\int_0^\pi\sin^22x\,\mathrm dx$

Recall the half-angle identity for sine:

$\sin^2t=\dfrac{1-\cos2t}2$
$\implies\displaystyle\frac{9\pi}2\int_0^\pi(1-\cos4x)\,\mathrm dx$
$=\displaystyle\frac{9\pi}2\left(x-\frac14\sin4x\right)\bigg|_{x=0}^{x=\pi}$
$=\dfrac{9\pi^2}2$

8 0

3 years ago

THANK YOU!!

lilavasa [31]

Step-by-step explanation:

x° + 61°+ 68° + 104° = 180°

x° + 129°+104°=180°

x°+133°=180°

x°=180°-133°

x=47°

8 0

2 years ago