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omeli [17]
3 years ago
7

Write each of the equations below in standard form (i.e. y = ax^2+bx+c y=ax ​2 ​​ +bx+c):

Mathematics
1 answer:
ArbitrLikvidat [17]3 years ago
5 0
C . Is the answer
Hope it helps
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If =250, then 50% of n equals
qwelly [4]

If n=250, then 50% of n can also be described as n divided by 2.

50% of n = 250/2 = 125

answer: 125

5 0
3 years ago
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Segment addition with variables
oksian1 [2.3K]

<em>x equals 22</em>

<h2>Explanation:</h2>

____________________________________________

The Segment Postulate states the following:

<em>Given two end points A and B, a third point C lines on the segment AB if and only if the distances between the points satisfy the equation:</em>

\overline{AB}=\overline{AC}+\overline{CB}

____________________________________________

From the figure:

\overline{AC}=8+x \\ \\ \overline{CB}=10 \\ \\ \overline{AB}=40

Our goal is to find x:

\overline{AB}=\overline{AC}+\overline{CB} \\ \\ Substituting \ values: \\ \\ 40=(8+x)+10 \\ \\ 18+x=40 \\ \\ Subtracting \ 18 \ from \ both \ sides: \\ \\ 18-18+x=40-18 \\ \\ \boxed{x=22}

<h2>Learn more:</h2>

Dilation: brainly.com/question/2501119

#LearnWithBrainly

4 0
3 years ago
How many whole numbers are there, whose squares and cubes have the same number of digits?
Naddik [55]

Answer:

there are only 4 whole numbers whose squares and cubes have the same number of digits.

Explanations:

let 0, 1, 2 and 4∈W (where W is a whole number), then

0^2=0, 0^3=0,

1^2=1, 1^3=1,

2^2=4, 2^3=8,

4^2=16, 4^3=64.

You can see from the above that only four whole numbers are there whose squares and cubes have the same number of digits







7 0
3 years ago
The factorization of x2 + 3x – 4 is modeled with algebra tiles. An algebra tile configuration. 2 tiles are in the Factor 1 spot:
arsen [322]

Answer:

The factors of x^2+3x-4 are (x-1)(x+4) ....

Step-by-step explanation:

We have to find the factors of x^2+3x-4

As we know that this is a quadratic equation.

So we have to find the roots first.

The roots are -1 and 4.

Now completing the quadratic formula using the roots we have :

x^2+4x-x-4

Make a pair of first two terms and last two terms:

(x^2+4x)-(x+4)

Now take out the common from each pair:

x(x+4)-1(x+4)

(x-1)(x+4)

Thus the factors of x^2+3x-4 are (x-1)(x+4) ....

5 0
3 years ago
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Select the statement that is false.
defon

Answer:

number 3 is incorrect the rest are correct

7 0
3 years ago
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