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____ [38]
2 years ago
15

Express the quotient in simplest form. x^2 - 4/ x^3 + 7x^2 divided by x^3 - x^2 - 6x/x^2 + 4x - 21

Mathematics
1 answer:
amm18122 years ago
8 0

You must first transform into a multiplication and then factor to see what you can eliminate.

Explanation:

To transform a fractional division into a multiplication, you have to inverse the numerator and the denominator of the second fraction.

<span><span><span><span>x2</span>−4</span><span><span>x3</span>+7<span>x2</span></span></span>×<span><span><span>x2</span>+4x−21</span><span><span>x3</span>−<span>x2</span>−6x</span></span></span>

<span><span><span><span>(x+2)</span><span>(x−2)</span></span><span><span>x2</span><span>(x+7)</span></span></span>×<span><span><span>(x+7)</span><span>(x−3)</span></span><span>x<span>(x−3)</span><span>(x+2)</span></span></span></span>

Since <span><span>aa</span>=1,a=</span> all real numbers, we can proceed by eliminating factors that appear in the numerator and the denominator.

We are left with <span><span><span>x−2/</span><span>x^3</span></span></span>

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1 year ago
What is the minimum value of C = 7x + 8y, given the constraints: 2x + y ≥ 8, x + y ≥ 6, x ≥ 0, y ≥ 0. A. 32 B. 42 C. 46 D. 64
marshall27 [118]

Answer: The minimum value of C is 46.

Step-by-step explanation:

Since, Here, We have to find out Min C = 7x+8y

Given the constraints are 2x+y\geq 8 -------(1)

x+y \geq 6   ------------- (2)

x \geq 0, y \geq 0  -------- (3)

Since, For equation 1) x-intercept, (4, 0) and y-intercept (0,8)

And, 2\times 0+0\geq 8⇒0\geq 8 ( false)

Therefore the area of line 1) does not contain the origin.

For equation 2) x-intercept, (6, 0) and y-intercept (0,6)

And, 0+0\geq 6⇒0\geq 6 ( false)

Therefore the area of line 2) does not contain the origin.

Thus after plotting the constraints 1) 2) and 3) we get Open Shaded feasible region AEB ( Shown in below graph)

At A≡(0,8) , C= 64

At E≡(2,4),  C= 46

At B≡(6,0),  C= 42

Thus at B, C is minimum, And its minimum value = 42


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3 years ago
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