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cestrela7 [59]
3 years ago
12

What is the slope of a line PARALLEL to the graph? A) -3 B) 0 C) 3 D) -1/3

Mathematics
1 answer:
Neporo4naja [7]3 years ago
7 0
You should choose c because parallel lines have the same slope
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Solve simultaneous equation <br> X-2y=1<br> 2x-y=2
Step2247 [10]

Answer: x=1,y=0

Step-by-step explanation:

X-2y=1.....equation 1

2x-y=2.....equation 2

X=1+2y.......equation 3

Substitute equation 3into 2

2(1+2y)-y=2

2+4y-y=2

2+3y=2

3y=2-2

3y=0

Y=0/3

Y=0

Substitute for y=0 in equation 1

X-2y=1

X-2(0)=1

X-0=1

X=1

X=1 & Y=0

7 0
3 years ago
Read 2 more answers
I need help now its due right now can anyone help me
cestrela7 [59]

Answer:

(2,-3)

Step-by-step explanation:

I thought you'd actually have to calculate it but I guess they just wanted to give you guys a graph lol

7 0
2 years ago
Help me please !!!! (:
riadik2000 [5.3K]
3x=x+40
Subtract x from both sides
2x=40
Divide both sides by 2nto get x by itself
x=2
Plug 2 in for x
m∠4 = (3x)
m∠4 = 3(2)
Solve
m∠4 = 6
8 0
3 years ago
9/10 times a number plus 6 is 51
lys-0071 [83]
N = the number

9/10n + 6 = 51                                       Original Problem.
9/10n = 46                                            Subtract 6 from each side.
n = 51.11111111111111111111                  Multiply each side by (10/9)

Answer: n = 51.111111111111111111111

3 0
3 years ago
Read 2 more answers
Determine whether or not the vector field is conservative. if it is conservative, find a function f such that f = ∇f. (if the ve
Akimi4 [234]
A three-dimensional vector field is conservative if it is also irrotational, i.e. its curl is \mathbf 0. We have

\nabla\times\mathbf f(x,y,z)=-2e^{-x}\,\mathbf k

so this vector field is not conservative.

- - -

Another way of determining the same result: We want to find a scalar function f(x,y,z) such that its gradient is equal to the given vector field, \mathbf f(x,y,z):

\nabla f(x,y,z)=\mathbf f(x,y,z)

For this to happen, we need to satisfy

\begin{cases}f_x=ye^{-x}\\f_y=e^{-x}\\f_z=2z\end{cases}

From the first equation, integrating with respect to x yields

f_x=ye^{-x}\implies f(x,y,z)=-ye^{-x}+g(y,z)

Note that g *must* be a function of y,z only.

Now differentiate with respect to y and we have

f_y=-e^{-x}+g_y=e^{-x}\implies g_y=2e^{-x}\implies g(y,z)=2ye^{-x}+\cdots

but this contradicts the assumption that g(y,z) is independent of x. So, the scalar potential function does not exist, and therefore the vector field is not conservative.
8 0
3 years ago
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