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adoni [48]
2 years ago
8

Which equation has roots of 13 and -3

Mathematics
1 answer:
madreJ [45]2 years ago
3 0
To get roots of 13 and -3 the simplified equation will have to look like this.
(X-13)(X+3)=0
So if we multiply this out with foil method we get this:

X^2 -10X -39

So the answer is A
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Name a unit of measurement that is about the same size of the weight or mass of 20 staples.
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Answer:

D

Step-by-step explanation:

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HELP ASAP. BRAINLIST IF YOU ANSWERED IT RIGHT
S_A_V [24]

Answer:

The answer is D.

Step-by-step explanation:

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Triangle ABC has vertices A(-2, 3), B(0, 3), and C(-1,-1). Find the coordinates of the image after a reflection over the
motikmotik

<u>Given</u>:

Given that the triangle ABC has vertices A(-2,3), B(0,3) and C(-1,-1).

We need to determine the coordinates of the image after a reflection over the x - axis.

Let A'B'C' denote the coordinates of the triangle after a reflection over the x - axis.

<u>Coordinates of the point A':</u>

The general rule to reflect the coordinate across the x - axis is given by

(x,y)\rightarrow (x,-y)

Substituting the point A(-2,3), we get;

(-2,3)\rightarrow (-2,-3)

Thus, the coordinates of the point A' is (-2,-3)

<u>Coordinates of the point B':</u>

The general rule to reflect the coordinate across the x - axis is given by

(x,y)\rightarrow (x,-y)

Substituting the point B(0,3), we get;

(0,3)\rightarrow (0,-3)

Thus, the coordinates of the point B' is (0,-3)

<u>Coordinates of the point C':</u>

The general rule to reflect the coordinate across the x - axis is given by

(x,y)\rightarrow (x,-y)

Substituting the point C(-1,-1), we get;

(-1,-1)\rightarrow (-1,1)

Thus, the coordinates of the point C' is (-1,1)

Hence, the coordinates of the image after a reflection over the x - axis is A'(-2,-3), B(0,-3) and C(-1,1)

8 0
3 years ago
What is the M.A.D of 1,1,1,1,5?
algol [13]
I think it is 3.6 ....... 
8 0
3 years ago
The expression p/x^2 - 5x +6 simplifies to x+4/x-2. Which expression doees p represent?
gladu [14]
We know that
(ad)/(bd)=d/d time a/b=a/b since d's cancel
also
if a/b=c/d in simplest form, then a=c and b=d

we have
p/(x^2-5x+6)=(x+4)/(x-2)
therefor

p/(x^2-5x+6)=d/d times (x+4)/(x-2)
p/(x^2-5x+6)=d(x+4)/d(x-2)

therefor
p=d(x+4) and
x^2-5x+6=d(x-2)

we can solve last one
factor
(x-6)(x+1)=d(x-2)
divide both sides by (x-2)
[(x-6)(x+1)]/(x-2)=d
sub


p=d(x+4)
p=([(x-6)(x+1)]/(x-2))(x+4)
p= \frac {(x-6)(x+1)(x+4)}{(x-2)}

7 0
3 years ago
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