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Murrr4er [49]
4 years ago
7

Really need this, it will help alottt

Mathematics
1 answer:
Nuetrik [128]4 years ago
8 0

Answer:

\huge\boxed{m = \frac{y-b}{x}}

Step-by-step explanation:

In order to solve for y=mx+b and we want to solve for m, we want to get m on one side of the equation.

y = mx+b

Let’s subtract b from both sides.

y - b = mx

Now we can divide both sides by x so we have m on one side of the equation.

\frac{y-b}{x} = m

Hope this helped!

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Can somebody please help me figure this out?
Vera_Pavlovna [14]

Answer: You would put the 35° and the 4x+7° equal to each other.

Step-by-step explanation:

    1. 4x+7=35  

    2. 4x+7=35   -> Subtract 7 from the positive 7 and the 35

           -7     -7

    3. 4x=28       -> Divide 4x and 28

    4. x=7

5 0
3 years ago
Multiply and simplify. (x + 3)(x - 8)
Kazeer [188]

Answer:

x^{2} - 5x - 24

Step-by-step explanation:

( x + 3 )( x - 8 ) =

x^{2} - 8x + 3x - 24 = x^{2} - 5x -24

4 0
4 years ago
Write an equation in Standard Form using the correct properties. Convert this equation to Slope-Intercept Form.
Lubov Fominskaja [6]

Answer:

I think what you are asking for is an example.

Step-by-step explanation:

Standard form equation

8x-4y = 20

1. move 8x to the other side

-4y=20-8x

2. divide both sides by -4

\frac{-4y}{-4} = \frac{20-8x}{-4}

3. Solve for this...

y=2x-5

7 0
3 years ago
Where will these 2 lines intersect?<br><br> y = 2x + 7 and x = y + 4
defon
To find the intersecting point, you have to put the two formulas equal to each other
First isolate the same variable, in this case y
So y=x-4

Now we can say
2x + 7 = x - 4

Now solve for x

X= 3

Now fill this in in either one of the formulas

Y= 2 (3) + 7
Y= 13

So they’ll intersect at (3,13)
5 0
3 years ago
Read 2 more answers
(1) Let {v1,v2,v3} be a set of vectors in Rn . If u is Span {v1,v2,v3}, show that 3u is in Span {v1,v2,v3}.
Evgesh-ka [11]

Answer:

(1)

Multiplying by 3 both sides of the equality you get that

3u = 3c_1v_1+3c_2v_2+3c_3v_3

3u  is in the Span of the vectors \{v_1,v_2,v_3\}.

(2)

That's not true, consider the following counter example.

v_1 = (0,0,0,1)\\v_2 = (0,0,1,0)\\v_3 = (0,1,0,0)\\v_4 = (1,0,0,0)\\u = (0,1,1,1)

u is a linear combination of v_1,v_2,v_3 but is NOT a linear combination of v_1,v_2,v_3,v_4.

Step-by-step explanation:

(1)

As the hint indicates, you know that

u = c_1 v_1 + c_2v_2+c_3v_3

Then, if you multiply both sides of the equality by 3, you get that

3u = 3c_1v_1+3c_2v_2+3c_3v_3

And that's it. 3u  is in the Span of the vectors \{v_1,v_2,v_3\}

(2)

That's not true, consider the following counter example.

v_1 = (0,0,0,1)\\v_2 = (0,0,1,0)\\v_3 = (0,1,0,0)\\v_4 = (1,0,0,0)\\u = (0,1,1,1)

u is a linear combination of v_1,v_2,v_3 but is NOT a linear combination of v_1,v_2,v_3,v_4.

4 0
3 years ago
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