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sattari [20]
3 years ago
7

How do you identify the constant of proportionality when given a graph?

Mathematics
1 answer:
Rzqust [24]3 years ago
5 0
Find two easy points on a graph and go over however many times you need to rise and how many times to you run.
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I'm really bad at geometry. Please help
postnew [5]
The sides of the frame are (12+2x) and (6+2x).
Its area A is 112.

Therefore by equating:
(12+2x)*(6+2x)=112
you can get the answer for x. You will obtain a quadratic equation you have to solve.
4 0
3 years ago
What kind of symmetry does a cylinder have?
olasank [31]
Both sides are symmetrical by the length and the shape
4 0
3 years ago
Let C = C1 + C2 where C1 is the quarter circle x^2+y^2=4, z=0,from (0,2,0) to (2,0,0), and where C2 is the line segment from (2,
trapecia [35]
Not much can be done without knowing what \mathbf F(x,y,z) is, but at the least we can set up the integral.

First parameterize the pieces of the contour:

C_1:\mathbf r_1(t_1)=(2\sin t_1,2\cos t_1,0)
C_2:\mathbf r_2(t_2)=(1-t_2)(2,0,0)+t_2(3,3,3)=(2+t_2, 3t_2, 3t_2)

where 0\le t_1\le\dfrac\pi2 and 0\le t_2\le1. You have

\mathrm d\mathbf r_1=(2\cos t_1,-2\sin t_1,0)\,\mathrm dt_1
\mathrm d\mathbf r_2=(1,3,3)\,\mathrm dt_2

and so the work is given by the integral

\displaystyle\int_C\mathbf F(x,y,z)\cdot\mathrm d\mathbf r
=\displaystyle\int_0^{\pi/2}\mathbf F(2\sin t_1,2\cos t_1,0)\cdot(2\cos t_1,-2\sin t_1,0)\,\mathrm dt_1
{}\displaystyle\,\,\,\,\,\,\,\,+\int_0^1\mathbf F(2+t_2,3t_2,3t_2)\cdot(1,3,3)\,\mathrm dt_2
5 0
3 years ago
HELP!!!! I’LL GIVE THE BRIANLIEST TO WHOEVER GETS IT RIGHT!!!!
Leto [7]

Answer:

13 units

Step-by-step explanation:

D is at (1,6) and C is at (-4,-6)

The distance is found by

d = sqrt(( y2-y1)^2+ (x2-x1)^2))

  = sqrt( ( -6-6)^2 + (-4-1)^2)

  = sqrt(  -12^2 + -5^2)

  = sqrt( 144+ 25)

   = sqrt( 169)

   = 13

6 0
3 years ago
5(6-i)+(6-i)(3-i). show work.
Bumek [7]

47 -14i

You can work this out in the straight-forward way, or you can recognize that (6-i) is a common factor. In the latter case, you have ...

... = (6-i)(5 + 3-i)

... = (6 -i)(8 -i)

This product of binomials is found in the usual way. Each term of one factor is multiplied by each term of the other factor and the results summed. Of course, i = √-1, so i² = -1.

... = 6·8 -6i -8i +i²

... = 48 -14i -1

... =

_____

A suitable graphing calculator will work these complex number problems easily.

4 0
3 years ago
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