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3 years ago

Someone please help ! write an equation that represents the line using exact numbers

Mathematics

1 answer:

Stels [109]3 years ago

8 0

Answer:

y=(-4/3)x+2 = -1.33x+2

Step-by-step explanation:

we have two points:

(3,-2) and (0,2) given from the graph.

we know the equation of the line is

y=mx+b

m=slope

m=(y2-y1)/(x2-x1)

m=(2-(-2))/(0-3)

so we have the slope is

m=4/-3

m=-4/3

-4/3=-1.3333

we need to find "b"

using the point (0,2)

we have

2=m*0+b

2=b

finally we have

the answer is y=(-4/3)x+2

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You don't need to single out $\cos \theta$ to solve this question.

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2 years ago

Read 2 more answers

In Katy’s garden there are 105 ladybugs. They are increasing at two ladybugs per month. There are currently 175 aphids and the n

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Answer:

They won't ever be the same at the same time.

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Add 2 to 150 and subtract 3 from 175 till you hit a common number at the same month

150, 152, 154, 156, 158, 160, 162

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3 years ago

Find the volume of the region between the planes x plus y plus 2 z equals 2 and 4 x plus 4 y plus z equals 8 in the first octant

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Find the intercepts for both planes.

Plane 1, x + y + 2z = 2:

$y=z=0\implies x=2\implies (2,0,0)$

$x=z=0\implies y=2\implies(0,2,0)$

$x=y=0\implies 2z=2\implies z=1\implies(0,0,1)$

Plane 2, 4x + 4y + z = 8:

$y=z=0\implies4x=8\implies x=2\implies(2,0,0)$

$x=z=0\implies4y=8\impliesy=2\implies(0,2,0)$

$x=y=0\implies z=8\implies(0,0,8)$

Both planes share the same x- and y-intercepts, but the second plane's z-intercept is higher, so Plane 2 acts as the roof of the bounded region.

Meanwhile, in the (x, y)-plane where z = 0, we see the bounded region projects down to the triangle in the first quadrant with legs x = 0, y = 0, and x + y = 2, or y = 2 - x.

So the volume of the region is

$\displaystyle\int_0^2\int_0^{2-x}\int_{\frac{2-x-y}2}^{8-4x-4y}\mathrm dz\,\mathrm dy\,\mathrm dx=\displaystyle\int_0^2\int_0^{2-x}\left(8-4x-4y-\frac{2-x-y}2\right)\,\mathrm dy\,\mathrm dx$

$=\displaystyle\int_0^2\int_0^{2-x}\left(7-\frac72(x+y)\right)\,\mathrm dy\,\mathrm dx=\int_0^2\left(7(2-x)-\frac72x(2-x)-\frac74(2-x)^2\right)\,\mathrm dx$

$=\displaystyle\int_0^2\left(7-7x+\frac74 x^2\right)\,\mathrm dx=\boxed{\frac{14}3}$

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