Answer:
z = x^3 +1
Step-by-step explanation:
Noting the squared term, it makes sense to substitute for that term:
z = x^3 +1
gives ...
16z^2 -22z -3 = 0 . . . . the quadratic you want
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<em>Solutions derived from that substitution</em>
Factoring gives ...
16z^2 -24z +2z -3 = 0
8z(2z -3) +1(2z -3) = 0
(8z +1)(2z -3) = 0
z = -1/8 or 3/2
Then we can find x:
x^3 +1 = -1/8
x^3 = -9/8 . . . . . subtract 1
x = (-1/2)∛9 . . . . . one of the real solutions
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x^3 +1 = 3/2
x^3 = 1/2 = 4/8 . . . . . . subtract 1
x = (1/2)∛4 . . . . . . the other real solution
The complex solutions will be the two complex cube roots of -9/8 and the two complex cube roots of 1/2.
Answer:
7 < x < 9
Step-by-step explanation:
x^2 – 16x + 63 < 0
this is a parabola that opens up.
The parabola will be < 0 for values between the roots.
to picture this see the graph below.
Factor
(x - 7)(x - 9) < 0
Roots
x = {7, 9}
solution
7 < x < 9
It Dependa what is your speed
Answer:
Step-by-step explanation:
-14<7x
-14/7<7/7x
2<x
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Answer:
30x +25y = 148
Step-by-step explanation:
We can use substitution to find the point of intersection of ...
Using the second equation, we can write an expression for y:
y = 6 -x
Substituting that into the first equation gives ...
3x +14 = 2(6 -x)
3x +14 = 12 -2x . . . . . eliminate parentheses
5x = -2 . . . . . . . . . . . add 2x-14
x = -0.4 . . . . . . . . . divide by 5
y = 6 -(-0.4) = 6.4
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Now, we want an equation for a line through the point (-0.4, 6.4) that is perpendicular to 5x = 6y+1
The perpendicular line will have the coefficients swapped with one of them negated. The constant will accommodate the given point.
6x = -5y + c
6(-0.4) +5(6.4) = c = 29.6
The perpendicular line can be written ...
6x = -5y +29.6
In standard form, the equation is ...
30x +25y = 148