Answer:
I am pretty sure the answer is A.
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:
y = 3/4x - 5
Step-by-step explanation:
An equation of a line can be written in slope-intercept form:
y = mx + b
where m is the slope and b is the y-intercept.
Let's plug in what we know.
The slope is 3/4.
y = 3/4x + b
To find b, we want to plug in a value that we know is on this line: in this case, I will use the point (4, -2). Plug in the x and y values into the x and y of the standard equation.
-2 = 3/4(4) + b
To find b, multiply the slope and the input of x(4)
-2 = 3 + b
Now, subtract 3 from both sides to isolate b.
-5 = b
Plug this into your standard equation.
y = 3/4x - 5
This is your equation. It has a slope of 3/4, and passes through (4, -2)
Hope this helps!