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Makovka662 [10]
2 years ago
9

 what would be the equation of a line that passes through (3,-4) with slope 2/3

Mathematics
1 answer:
Len [333]2 years ago
7 0

Answer:

y = 2/3x - 6

Step-by-step explanation:

Use the slope intercept equation, y = mx + b

Plug in the slope and given point, then solve for b

y = mx + b

-4 = 2/3(3) + b

-4 = 2 + b

-6 = b

Plug in the slope and b into the equation

y = 2/3x - 6

So, the equation of the line is y = 2/3x - 6

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3/54+4/6<br> Add the fractions and give your answer as a proper fraction
Rina8888 [55]

Answer:

13/18

Step-by-step explanation:

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3 years ago
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Find the equation of the line passing through the point (–1, –2) and perpendicular to the line y = –1∕2x + 5. Question options:
tankabanditka [31]

Answer:

<h3>The answer is option C</h3>

Step-by-step explanation:

Equation of a line is y = mx + c

where

m is the slope

c is the y intercept

From the question

y = - 1/2x + 5

Comparing with the general equation above

Slope / m = -1/2

Since the lines are perpendicular to each other the slope of the other line is the negative inverse of the original line

That's

Slope of the perpendicular line = 2

Equation of the line using point (–1, –2) and slope 2 is

y + 2 = 2( x + 1)

y + 2 = 2x + 2

y = 2x + 2 - 2

We have the final answer as

<h3>y = 2x</h3>

Hope this helps you

4 0
2 years ago
Given the function f(x)=x2+2x, the value of f(4) is
Citrus2011 [14]

Answer:

24

Step-by-step explanation:

f(x)=x^2+2x

Let x= 4

f(4)=4^2+2*4

     = 16 + 8

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5 0
3 years ago
a supplier sells 2 1/4 pounds of mulch for every 1 1/3 pounds of gravel. The supplier sells 172 pounds of mulch and gravel combi
astraxan [27]
2 1/4=2 3/12 and 1 1/3= 1 4/12
3 0
3 years ago
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
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