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vazorg [7]
3 years ago
8

Please someone help me with this

Mathematics
2 answers:
shtirl [24]3 years ago
7 0

Answer:

2,-1.5

Step-by-step explanation:

To find the midpoint of a line segment, take the average of the x coordinates and y coordinates of the two points that make it up.

\frac{2+2}{2} ,\frac{4+(-7)}{2}

=\frac{4}{2} , \frac{-3}{2}\frac{4}{2} , \frac{-3}{2}

=2,-1.5

Zigmanuir [339]3 years ago
6 0

Here's the answer for you

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3 years ago
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15 POINTS!!!! BRAINLIEST FOR THE FIRST ANSWER!!! Solve 3-x/2≤18
Bumek [7]

Answer:

x\geq -30

Step-by-step explanation:

Work to isolate x on one side of the inequality:

3-\frac{x}{2} \leq 18\\3-18\leq \frac{x}{2} \\-15\leq  \frac{x}{2}\\-30 \leq x

Therefore the answer is all x values larger than or equal to -30

x\geq -30

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3 years ago
Explain how to do this please show your work
madreJ [45]

Answer:

(x-  5)(x - 3)

x(x - 3) -5(x - 3)

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2 years ago
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ivanzaharov [21]
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3 years ago
Write 3x^2-18x-6 in vertex form
Vedmedyk [2.9K]
The standard form of a quadratic equation is \displaystyle{ y=ax^2+bx+c, while the vertex form is:

                      y=a(x-h)^2+k, where (h, k) is the vertex of the parabola.

What we want is to write \displaystyle{ y=3x^2-18x-6 as y=a(x-h)^2+k

First, we note that all the three terms have a factor of 3, so we factorize it and write:

\displaystyle{ y=3(x^2-6x-2).


Second, we notice that x^2-6x are the terms produced by (x-3)^2=x^2-6x+9, without the 9. So we can write:

x^2-6x=(x-3)^2-9, and substituting in \displaystyle{ y=3(x^2-6x-2) we have:

\displaystyle{ y=3(x^2-6x-2)=3[(x-3)^2-9-2]=3[(x-3)^2-11].

Finally, distributing 3 over the two terms in the brackets we have:

y=3[x-3]^2-33.


Answer: y=3(x-3)^2-33
6 0
3 years ago
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