Answer:
final answer:
y-1= 3x_6
To solve our given problem, we must find the equation of the tangent line of the function f(x)=(x−1)3, at x=2.
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Answer:
see attached diagram
Step-by-step explanation:
First, draw the dashed line 50x+150y=1500 (dashed because the inequality is without notion "or equal to"). You can do it finding x and y intercepts.
When x=0, then 150y=1500, y=10.
When y=0, then 50x=1500, x=30.
Connect points (0,10) and (30,0) to get needed dashed line.
Then determine which region (semiplane) you have to choose. Note that origin's coordinates (0,0) do not satisfy the inequality 50x + 150y>1500, because

This means that origin lies outside the needed region, so you have to choose the semiplane that do not contain origin (see attached diagram).
Given:
The function is:

To find:
All the possible rational zeros for the given function by using the Rational Zero Theorem.
Solution:
According to the rational root theorem, all the rational roots are of the form
, where p is a factor of constant term and q is a factor of leading coefficient.
We have,

Here,
Constant term = -2
Leading coefficient = 10
Factors of -2 are ±1, ±2.
Factors of 10 are ±1, ±2, ±5, ±10.
Using the rational root theorem, all the possible rational roots are:
.
Therefore, all the possible rational roots of the given function are
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Answer:
28 < x < 114
Step-by-step explanation:
- x cannot be larger than or even equal to 71 + 43 = 114. So it must be less than 114.
- But there are limits on the other end as well. 43 +x > 71, otherwise you won't have a triangle. so you have to solve that inequality as well
- 43 + x > 71 Subtract 43 from both sides.
- 43-43 + x > 71 - 43
- x > 28
- To put this in the form of the answer you need it looks like this.
- 28 < x < 114