9514 1404 393
Answer:
see attached
Step-by-step explanation:
Polynomial long division is done the way any long division is done. Find a "partial quotient", subtract from the dividend the product of that partial quotient and the divisor. The result is a new dividend. Repeat until the degree of the dividend is less than that of the divisor.
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In the attached, the "Hints" show you how the partial quotient is found, and they show you how the product of the partial quotient and divisor is found.
The partial quotient term is simply the ratio of the highest degree terms of dividend and divisor. (Unlike numerical long division, there is no guessing.)
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The remainder is the dividend of lower degree than the divisor. As in numerical long division, the full quotient expresses the remainder over the divisor.
For example, 5 ÷ 3 = 1 r 2 = 1 + 2/3.
Your full quotient is (n+5) +1/(n-6).
Answer:
the markup percentage is 66.67%
Step-by-step explanation:
The computation of the percent of markup based on cost is shown below:
= (Selling price - paid price) ÷ (paid price)
= ($15 - $9) ÷ ($9)
= 66.67%
By taking the difference of the selling price & paid price and then divided it by paid price we can get the percentage of markup
Hence, the markup percentage is 66.67%
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<em>hope</em><em> </em><em>it</em><em> </em><em>will</em><em> </em><em>help</em><em> </em><em>u</em><em> </em><em>.</em><em>.</em><em>.</em><em>.</em>
Answer:
We conclude that:
h(f(-1)) = -2
∴ option D i.e. -2 is correct.
Step-by-step explanation:
Given
f(x) = 4x² - 1
g(x) = 1/2x + 5
h(x) = 2(x - 4)³
To determine
h(f(-1)) = ?
In order to determine h(f(-1)) first we need to determine f(-1).
substitute x = -1 in the function f(x) = 4x² - 1
f(-1) = 4(-1)² - 1
f(-1) = 4(1) - 1
f(-1) = 4-1
f(-1) = 3
so
h(f(-1)) = h(3)
now substitute h = 3 in the function h(x) = 2(x - 4)³
h(x) = 2(x - 4)³
h(3) = 2(3 - 4)³
h(3) = 2(-1)³
h(3) = 2(-1)
h(3) = -2
Thus,
h(f(-1)) = h(3) = -2
Hence, we conclude that:
h(f(-1)) = -2
∴ option D i.e. -2 is correct.
Answer:
3.33 and 1/3
Step-by-step explanation:
"Dense" here means that there are infinite irrational numbers between two rational numbers. Also, there are infinite rational numbers between two rational numbers. That's the meaning of dense. Actually, that can be apply to all real numbers, there always is gonna be a number between other two.
But, to demonstrate that irrationals are dense, we have to based on an interval with rational limits, because the theorem about dense sets is about rationals, and the dense irrational set is a deduction from it. That's why the best option is 2, because that's an interval with rational limits.
