Step-by-step explanation:
remember, the general formula to solve a quadratic equation ax² + bx + c = 0 is
x = (-b ± sqrt(b² - 4ac))/(2a)
a)
I think there is a typo, and this should be
5x² - 7x = 72
5x² -7x - 72 = 0
x = (7 ± sqrt(49 - -1440))/10 = (7 ± sqrt(1489))/10
x1 = (7 + sqrt(1489))/10
x2 = (7 - sqrt(1489))/10
b)
x(x + 2) = 5
x² + 2x - 5 = 0
x = (-2 ± sqrt(4 - -20))/2 = (-2 ± sqrt(24))/2 =
= (-2 ± sqrt(4×6))/2 = (-2 ± 2×sqrt(6))/2 =
= -1 ± sqrt(6)
x1 = -1 + sqrt(6)
x2 = -1 - sqrt(6)
c)
3x² + 2x - 4 = 0
x = (-2 ± sqrt(4 - -48))/6 = (-2 ± sqrt(52))/6 =
= (-2 ± sqrt(4×13))/6 = (-2 ± 2×sqrt(13))/6 =
= (-1 ± sqrt(13))/3
x1 = (-1 + sqrt(13))/3
x2 = (-1 - sqrt(13))/3
Answer:
The angles and lengths should not change at all the image and pre-image should be the same just at a different place on the graph
Answer:
The values of x are
Step-by-step explanation:
we know that
The formula to solve a quadratic equation of the form is equal to
in this problem we have
so
substitute in the formula
1/6·30=5
so your answer is b)5