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Stolb23 [73]
3 years ago
9

Multiply. (2x2 + 4x - 3)(x2 - 2x + 5)

Mathematics
2 answers:
BigorU [14]3 years ago
7 0
For a moment, let y=x^2-2x+5. Then by the distributive property,

(2x^2+4x-3)p=2x^2p+4xp-3p


Again by the distributive property,

2x^2p=2x^2(x^2-2x+5)=2x^4-4x^3+10x^2

4xp=4x(x^2-2x+5)=4x^3-8x^2+20x

-3p=-3(x^2-2x+5)=-3x^2+6x-15

Taking everything together, we get

(2x^2+4x-3)(x^2-2x+5)=(2x^4-4x^3+10x^2)+(4x^3-8x^2+20x)+(-3x^2+6x-15)
(2x^2+4x-3)(x^2-2x+5)=2x^4-x^2+26x-15
eduard3 years ago
4 0

Answer:

the answer is C

Step-by-step explanation:

got it right on the assignment

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The perimeter of the rectangle below is 112 units. Find the length of side VW . Write your answer without variables. Top YX: 4z
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Answer:

VW = 33

Step-by-step explanation:

<em>See attachment for complete question</em>

Given

VY = 4x - 1

VW = 5x + 3

Perimeter = 112

Required

Determine the length of VW

Perimeter is calculated as:

Perimeter = 2 * (Width + Length)

This gives:

Perimeter = 2 * (VW + VY)

Substitute values for VW, VY and Perimeter

112 =  2 * (4x - 1 + 5x + 3)

Collect Like Terms

112 = 2 * (4x + 5x -1 +3)

112 = 2 * (9x +2)

Divide through by 2

56 = 9x + 2

Collect Like Terms

9x = 56 - 2

9x = 54

x = 54/9

x = 6

Substitute 6 for x in VW = 5x + 3

VW = 5 * 6 + 3

VW = 30 + 3

VW = 33

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3 years ago
a collection of dimes and quarters is worth $19.85. There are 128 coins in all. How many of each type of coin are in the collect
PolarNik [594]

Number of dimes were 81 and number of quarters were 47

<em><u>Solution:</u></em>

Let "d" be the number of dimes

Let "q" be the number of quarters

We know that,

value of 1 dime = $ 0.10

value of 1 quarter = $ 0.25

<em><u>Given that There are 128 coins in all</u></em>

number of dimes + number of quarters = 128

d + q = 128 ------ eqn 1

<em><u>Also given that collection of dimes and quarters is worth $19.85</u></em>

number of dimes x value of 1 dime + number of quarters x value of 1 quarter = 19.85

d \times 0.10 + q \times 0.25 = 19.85

0.1d + 0.25q = 19.85  -------- eqn 2

<em><u>Let us solve eqn 1 and eqn 2</u></em>

From eqn 1,

d = 128 - q -------- eqn 3

<em><u>Substitute eqn 3 in eqn 2</u></em>

0.1(128 - q) + 0.25q = 19.85

12.8 - 0.1q + 0.25q = 19.85

12.8 + 0.15q = 19.85

0.15q = 7.05

<h3>q = 47</h3>

Therefore from eqn 3,

d = 128 - q

d = 128 - 47

<h3>d = 81</h3>

Thus number of dimes were 81 and number of quarters were 47

4 0
3 years ago
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