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tester [92]
3 years ago
13

If a = 2, what is the simplified form of the expression below? (x − a)(x^2 +ax + a^2)

Mathematics
2 answers:
dem82 [27]3 years ago
6 0

Answer:x^3+2x^2+4x-ax^2-2ax-4a

Step-by-step explanation:

Ksivusya [100]3 years ago
6 0

Answer:

x³+8

Step-by-step explanation:

(x − a)(x^2 +ax + a^2)

a=2

Substitute a for 2 in the algebraic expression

(x − 2)(x^2 +2x + 2^2)

(x − 2)(x^2 +2x + 4)

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

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

x³+2x²+4x-2x²-4x-8

Collect like terms

x³+2x²-2x²+4x-4x+8

x³+8

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Answer:

11.6 is the answer friend.

Step-by-step explanation:

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3 years ago
If y varies directly as x and y=3 when x=10, find x when y=8<br><br>Please, somebody, I need help
Kryger [21]

Answer:

26.6

Step-by-step explanation

3=10*k

3/10=.3

y=8

8=.3*x

8/.3=26.6

8 0
3 years ago
To rewrite the expression ab+a+b+1, Jozelle first writes (ab+a) + (b+1). What property justifies this?
rodikova [14]

Answer:

I believe it is associative property of addition or commutative

Step-by-step explanation:

7 0
3 years ago
Find the slope of the line containing the points (4, -7) and (6,-8).
masha68 [24]
-1/2 ans. Hope it helps..
6 0
3 years ago
It is possible to get 2 solutions when the system of equations is: (check all that apply)
SVETLANKA909090 [29]

Answer:

A system of the equation of a circle and a linear equation

A system of the equation of a parabola and a linear equation

Step-by-step explanation:

Let us verify our answer

A system of the equation of a circle and a linear equation

Let an equation of a circle as x^2+ y^2 = 1 ..........(1)

Let a liner equation Y = x ............(2)

substitute (2) in (1)

x^2 + x^2 = 1\\2x^2 = 1\\

x^2 = \frac{1}{\sqrt{2} } \\x = +\frac{1}{\sqrt{2} } , -\frac{1}{\sqrt{2} }  so Y = +\frac{1}{\sqrt{2} } , -\frac{1}{\sqrt{2} }

so the two solution are ( (\frac{1}{\sqrt{2} } ,\frac{1}{\sqrt{2} }) (-\frac{1}{\sqrt{2} }, -\frac{1}{\sqrt{2} })

A system of the equation of a parabola and a linear equation

Let equation of Parabola be y^2 = x

and linear equation y = x

substitute

x^2 = x\\x^2 - x= 0\\x(x-1) = \\x = 0 , 1

Y = 0,1

so the two solutions will be (0,0) and (1,1)

3 0
3 years ago
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