3 years ago

Write a word phrase for the algebraic expression: z/8-9

Mathematics

1 answer:

netineya [11]3 years ago

5 0

Z divided by eight minus 9

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The length of a rectangle is 10 yd less than three times the width, and the area of the rectangle is 77 yd^2. Find the dimension

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Answer:

width = 7, length = 11

Step-by-step explanation:

area = 77

length = 3w - 10

width = w

w(3w - 10) = 77

3w^2 - 10w - 77 = 0

(3w + 11)(w - 7) = 0

we rule out 3w + 11 = 0 because w would be negative

so we use w - 7 = 0

so the width = 7

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length = 21 - 10

length = 11

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What situation should be used to rewrite 16(x^3+1)^2 -22(x^3+1)-3=0 as a quadratic equation

Oliga [24]

$16(x^3+1)^2-22(x^3+1)-3=0$

Use substitution: $x^3+1=t$

$16t^2-22t-3=0\\\\16t^2-24t+2t-3=0\\\\8t(2t-3)+1(2t-3)=0\\\\(2t-3)(8t+1)=0\iff2t-3=0\ \vee\ 8t+1=0$

$2t-3=0\ \ \ |+3\\\\2t=3\ \ \ |:2\\\\t=\dfrac{3}{2}\\..............................\\8t+1=0\ \ \ \ |-1\\\\8t=-1\ \ \ \ |:8\\\\t=-\dfrac{1}{8}$

we're going back to substitution:

$x^3+1=\dfrac{3}{2}\ \vee\ x^3+1=-\dfrac{1}{8}\ \ \ \ |subtract\ 1\ from\ both\ sides\ of\ the\ equations\\\\x^3=\dfrac{1}{2}\ \vee\ x^3=-\dfrac{9}{8}\\\\x=\sqrt[3]{\dfrac{1}{2}}\ \vee\ x=\sqrt[3]{-\dfrac{9}{8}}$

$x=\dfrac{1}{\sqrt[3]2}\ \vee\ x=-\dfrac{\sqrt[3]9}{2}\\\\\boxed{x=\dfrac{\sqrt[3]4}{2}\ \vee\ x=-\dfrac{\sqrt[3]9}{2}}$

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Step-by-step explanation:

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