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Andrews [41]
2 years ago
10

100 POINTS multiply: (2x+1)(x^2-3x-2)

Mathematics
2 answers:
denis-greek [22]2 years ago
7 0

Answer:

2x^3-5x^2-7x-2

Step-by-step explanation:

1)  Expand by distributing sum groups.

2x(x^2-3x-2)+x^2-3x-2

2) Expand by distributing terms.

2x^3-6x^2-4x+x^2-3x-2

3) Collect like terms.

2x^3+(-6x^2+x^2)+(-4x-3x)-2

4) Simplify.

2x^3-5x^2-7x-2

Thanks!

- Eddie

Vladimir79 [104]2 years ago
6 0

Step-by-step explanation:

Sorry for writing BTS Army but I do lov* them all...

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One pretzel has 20 calories. Write an algebraic expression for the total number of calories in n pretzels.
vredina [299]

Answer:

20n

Step-by-step explanation:

One pretzel has 20 calories. Write an algebraic expression for the total number of calories in n pretzels.

1 pretzel = 20 calories

n pretzel = x

Cross Multiply

x × 1 pretzel = n pretzel × 20 calories

x = n pretzel × 20 calories/1 pretzel

x = 20 n

Therefore, an algebraic expression for the total number of calories in n pretzels = 20n

4 0
3 years ago
What is midpoint of A(2, 3) and B(-2, 5)?
Elis [28]

Answer:

The midpoint of A and B is (0, 4).

Step-by-step explanation:

Use the midpoint formula, ((x₁ + x₂)/2 , (y₁ + y₂)/2)

First, solve for the x part.

(2 + (-2))/2

0/2

0

So the x-coordinate of the midpoint is 0.

Then, solve for the y part.

(3 + 5)/2

8/2

4

So the y-coordinate of the midpoint is 4.

7 0
3 years ago
The length of each side of a rhombus is 10 and the measure of an angle of the rhombus is 60. Find the length of the longer diago
Svetllana [295]
The diagonals of a rhombus are perpendicular bisectors of each other, and are also angle bisectors. The 4 congruent right triangles the rhombus is cut into by the two diagonals are 30-60-90 degree triangles, the longer leg is √3/2 of the hypotenuse, √3/2 of 10=5√3. so the longer diagonal is twice of 5√3=10√3 
4 0
3 years ago
2.6 X 16 fourth grade answers
Masteriza [31]

Answer:

41.6 lol

Step-by-step explanation:

3 0
3 years ago
Factor completely. <br> <img src="https://tex.z-dn.net/?f=x%5E%7B8%7D-%5Cfrac%7B1%7D%7B81%7D" id="TexFormula1" title="x^{8}-\fra
Eduardwww [97]

We have 3⁴ = 81, so we can factorize this as a difference of squares twice:

x^8 - \dfrac1{81} = \left(x^2\right)^4 - \left(\dfrac13\right)^4 \\\\ x^8 - \dfrac1{81} = \left(\left(x^2\right)^2 - \left(\dfrac13\right)^2\right) \left(\left(x^2\right)^2 + \left(\dfrac13\right)^2\right) \\\\ x^8 - \dfrac1{81} = \left(x^2 - \dfrac13\right) \left(x^2 + \dfrac13\right) \left(\left(x^2\right)^2 + \left(\dfrac13\right)^2\right) \\\\ x^8 - \dfrac1{81} = \left(x^2 - \dfrac13\right) \left(x^2 + \dfrac13\right) \left(x^4 + \dfrac19\right)

Depending on the precise definition of "completely" in this context, you can go a bit further and factorize x^2-\frac13 as yet another difference of squares:

x^2 - \dfrac13 = x^2 - \left(\dfrac1{\sqrt3}\right)^2 = \left(x-\dfrac1{\sqrt3}\right)\left(x+\dfrac1{\sqrt3}\right)

And if you're working over the field of complex numbers, you can go even further. For instance,

x^4 + \dfrac19 = \left(x^2\right)^2 - \left(i\dfrac13\right)^2 = \left(x^2 - i\dfrac13\right) \left(x^2 + i\dfrac13\right)

But I think you'd be fine stopping at the first result,

x^8 - \dfrac1{81} = \boxed{\left(x^2 - \dfrac13\right) \left(x^2 + \dfrac13\right) \left(x^4 + \dfrac19\right)}

6 0
3 years ago
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