1) determine the roots for x^2-x=2
2) What are the roots of the equation 9x^2-81x=0
3) The width of a rectangle is 3 less than twice the length of x, If the area is 43 square feet what is the equation used to find the length
Math question(s) below?
2018-07-25 7:55 am
回答 (4)
2018-07-25 8:06 am
1)
x² - x = 2
x² - x - 2 = 0
(x + 1)(x - 2) = 0
x = -1 or x = 2
====
2)
9x² - 81x = 0
x² - 9x = 0
x(x - 9) = 0
x = 0 or x = 9
====
3)
The length of the rectangle = x
Then, the width of the rectangle = 2x - 3
Area of the rectangle:
x(2x - 3) = 43
2x² - 6x - 43 = 0
x² - x = 2
x² - x - 2 = 0
(x + 1)(x - 2) = 0
x = -1 or x = 2
====
2)
9x² - 81x = 0
x² - 9x = 0
x(x - 9) = 0
x = 0 or x = 9
====
3)
The length of the rectangle = x
Then, the width of the rectangle = 2x - 3
Area of the rectangle:
x(2x - 3) = 43
2x² - 6x - 43 = 0
2018-07-25 8:09 am
1) x^2-x-2=0
(x-2)(x+1) = 0
x = -2 or 2
2) 9x(x-9) = 0
x = 0 or 9
3) L = length
W = width = 2L-3
2L^2-3L-43 = 0
(x-2)(x+1) = 0
x = -2 or 2
2) 9x(x-9) = 0
x = 0 or 9
3) L = length
W = width = 2L-3
2L^2-3L-43 = 0
2018-07-25 8:16 pm
1)
x^2 - x = 2
(x + 1)(x - 2) = 0
Solutions:
x = -1
x = 2
2)
9x^2 - 81x = 0
9x(x - 9) = 0
Solutions:
x = 0
x = 9
3)
The width of a rectangle is 3 less than twice the length of x.
If the area is 43 square feet what is the equation used to find the length?
l(2l - 3) = 43
2l^2 - 3l - 43 = 0
l ≈ 5.4471, w = 7.8942
x^2 - x = 2
(x + 1)(x - 2) = 0
Solutions:
x = -1
x = 2
2)
9x^2 - 81x = 0
9x(x - 9) = 0
Solutions:
x = 0
x = 9
3)
The width of a rectangle is 3 less than twice the length of x.
If the area is 43 square feet what is the equation used to find the length?
l(2l - 3) = 43
2l^2 - 3l - 43 = 0
l ≈ 5.4471, w = 7.8942
2018-07-25 7:56 am
Thanks for reminding me to dye my roots
收錄日期: 2021-05-01 22:24:54
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