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kolbaska11 [484]
3 years ago
9

What is the slope of a line that is perpendicular to the line whose equation is 2y = 3x - 1? -3 - -

Mathematics
1 answer:
Shalnov [3]3 years ago
8 0

Answer: 2/3x

Step-by-step explanation:

First, you need to simplify your original line to point slope formula. Dividing the whole thing by 2 to get y by itself leaves you with

y = 3/2x - 1/2

Any perpendicular line has the opposite slope to the line it is perpendicular to, so you flip the slope to get 2/3x

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6 2/3 * -4 1/5 i really need help
deff fn [24]

Answer:

-28

Step-by-step explanation:

6 2/3*-4 1/5=

20/3*-21/5=

4/3*-21/1=

4/1*-7/1=

-28

3 0
3 years ago
Consider the following system of equations 3x+y=6 3x-y=6
marta [7]
3x+y=6
3x-y=6
Cancel out the Y and add the terms
6x=12
X=2
Plug 2 into one of the original equations to find Y
3(6)+y=6
18+y=6
Y=-12
Write as ordered pair: (6,-12)
4 0
3 years ago
Is this the right answer?
iren2701 [21]

Hello from MrBillDoesMath!

Answer:

Yes!


Discussion:

a + b= 2a                =>  subtract "a" from both sides

a - a + b = 2a - a    =>  as (a-a) = 0 and (2a-a) = a

b = a                      (*)

Then  from (*) above

      b-a = b - b       => b-b = 0

      b -a = 0



Thank you,

MrB

8 0
3 years ago
NEED ASAP hurry plz<br><br>Write a story that would explain the graph below.
pishuonlain [190]

Answer:

A kid leaping happily to home after school

8 0
2 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
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