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erastovalidia [21]
3 years ago
12

How do I solve for v?

Mathematics
1 answer:
sineoko [7]3 years ago
5 0

Answer:

v= 18

Step-by-step explanation:

Subtract 8 from both sides.  This yields 18 = v, or v = 18.

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If 2x+5= 12 for some value of x, then what does 6x+20 equal???
ANEK [815]
1. Solve for x in 2x + 5 = 12
2x + 5 = 12       Subtract 5 from both sides.
2x = 7               Divide 2 from both sides.
x = 3.5

2. Substitute 3.5 in for x in 6x + 20
6(3.5) + 20
21 + 20
41

Your answer is 41.


6 0
3 years ago
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Which set of parametric equations represents the graph of the function shown?
Molodets [167]
Stupid get yo brain to work BOIIIIIJ
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V = Bh for h. <br> A. h = BV<br> B. h=B/V<br> C. h=V/B<br> D. h = V – B
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5 0
3 years ago
The distance between the point (2,1,1) and the plane x-2y=5 is​
Eduardwww [97]

The given plane has normal vector

x-2y=5\implies\mathbf n=\langle1,-2,0\rangle

Scaling <em>n</em> by a real number <em>t</em> gives a set of vectors that span an entire line through the origin. Translating this line by adding the vector <2, 1, 1> makes it so that this line passes through the point (2, 1, 1). So this line has equation

\mathbf r(t)=\langle1,-2,0\rangle t+\langle2,1,1\rangle=\langle 2+t, 1-2t, 1\rangle

This line passes through (2, 1, 1) when <em>t</em> = 0, and the line intersects with the plane when

x-2y=5\implies(2+t)-2(1-2t)=5\implies5t=5\implies t=1

which corresponds the point (3, -1, 1) (simply plug <em>t</em> = 1 into the coordinates of \mathbf r(t)).

So the distance between the plane and the point is the distance between the points (2, 1, 1) and (3, -1, 1):

\sqrt{(2-3)^2+(1-(-1))^2+(1-1)^2}=\boxed{\sqrt5}

7 0
3 years ago
If f(x)=8-10x and g(x)=5x+4 what is the value of (fg)(-2)?
Shalnov [3]
First, calculate f(g(x))==> you plug (5x+4) in the value of x in f(x)

==>f(g(x))= 8-[10(5x+4)===>8-50x-40===>f(g(x))= -50x  + 32

& f(g(-2))= -100+32 =68.

There is a mistake in your answers, it should be 68 & not 78
6 0
3 years ago
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