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Minchanka [31]
3 years ago
8

Find a line parallel to the graph of y=3x+6 that passes through the point (3,0) in standard form

Mathematics
1 answer:
Ivanshal [37]3 years ago
4 0

Answer:

The line parallel to this one and going through said point is y = 3x - 9

Step-by-step explanation:

In order to find this, we start by noting that the slope will be 3 because parallel lines have the same slope. Now we can use point-slope form to find the equation of the line.

y - y1 = m(x - x1)

y - 0 = 3(x - 3)

y = 3x - 9

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Step-by-step explanation:

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Help needed! no links please, they're annoying.
denis23 [38]
The area of a rectangle is length*width.

The length of this 3x and the width is 2x-3. This means that, to find the area, you need to multiply 3x and 2x-3.

Start by writing out this equation:
A=l*w

Then, plug in your given values:
A= 3x*2x-3 (you can also write it as A= 3*x*2*x-3)

Then, following the order of operations, you start by multiplying. This makes your equation become A = 6x^2 -3. (^2 means squared). It turns into this because 3 * 2 is 6 and x * x is x^2 and you still haven’t used the 3 yet.

After this, there is nothing more that you can do to simplify the equation. Therefore, the area is 6x^2 - 3.

I hope this helped!
4 0
3 years ago
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jekas [21]
This is for 2.X=22 so the total angle is 90° and you take away 68° and that leaves you with 22°
6 0
3 years ago
Hi, the answer is K but can anyone show why
dlinn [17]

Let's do 51 and 52.

51. The contrapositive has the same truth value as the original statement. That's opposed to the converse, which may or may not be true independent of the original statement.

The contrapositive of IF P THEN Q is IF not Q THEN not P.  They're equivalent.  Here that's If the cat is not female then it is not tricolor.

Answer: C

52.  

(x^3)^{(4-b^2)}=1

x^{3(4-b^2)} = 1

For the statement to be true, the exponent must be zero:

3(4-b^2) = 0

b^2 = 4

b = \pm 2

Both positive 2 and negative 2 have a square of 4.

Answer: K

By the way, usually we assume 0^0=1 so the restriction that x \ne 0 isn't really necessary.  Think of the definition of a polynomial or the binomial expansion:

\displaystyle f(x)=\sum_{k=0}^n a_k x^k

\displaystyle(x+y)^n=\sum_{k=0}^n {n \choose k} x^{k}y^{n-k}

For these common equalities to work when x=0 we need to define 0^0=1


3 0
3 years ago
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