Answer:
x=34
Step-by-step explanation:
6 - ( x-7) ^ 1/3 = 3
Subtract 6 from each side
6-6 - ( x-7) ^ 1/3 = 3-6
- ( x-7) ^ 1/3 = -3
Divide each side by a negative
( x-7) ^ 1/3 = 3
Cube each side
( x-7) ^ 1/3 ^3 = (3)^3
x-7 = 27
Add 7 to each side
x-7+7 = 27+7
x = 34
Check
6 - ( 34-7) ^ 1/3 = 3
6 - (27^1/3 = 3
6 -3 =3
3=3
Good solution
2x-2 must be zero or greater, since we cannot have a negative quantity under the radical sign (unless we allow for imaginary roots).
Solving 2x-2≥0, we get x-1≥0, or x≥1. x must be equal to or greater than 1.
16 - 16, so the answer to the 2nd problem is the fourth one: x=4.
Answer:
(3,0.5)
Step-by-step explanation:
The given system has equations:
![7x + 4y = 23 \\ 8x - 10y = 19](https://tex.z-dn.net/?f=7x%20%2B%204y%20%3D%2023%20%5C%5C%208x%20-%2010y%20%3D%2019)
Let us multiply the second equation by 4 and the first equation by 10.
![70x + 40y = 230 \\ 32x - 40y = 76](https://tex.z-dn.net/?f=70x%20%2B%2040y%20%3D%20230%20%5C%5C%2032x%20-%2040y%20%3D%2076)
We add both new equations to get:
![102x = 306 \\ x = \frac{306}{102} = 3](https://tex.z-dn.net/?f=102x%20%3D%20306%20%5C%5C%20x%20%3D%20%20%5Cfrac%7B306%7D%7B102%7D%20%20%3D%203)
Put x=3 into the first equation and solve for y.
![7(3) + 4y = 23 \\ 21 + 4y = 23 \\ 4y = 23 - 21 \\ 4y = 2 \\ y = \frac{1}{2}](https://tex.z-dn.net/?f=7%283%29%20%2B%204y%20%3D%2023%20%5C%5C%2021%20%2B%204y%20%3D%2023%20%5C%5C%204y%20%3D%2023%20-%2021%20%5C%5C%204y%20%3D%202%20%5C%5C%20%20y%20%3D%20%20%5Cfrac%7B1%7D%7B2%7D%20)
The point of intersection is the solution to the system of equation which is (3,0.5)
Answer:
- (-1, -32) absolute minimum
- (0, 0) relative maximum
- (2, -32) absolute minimum
- (+∞, +∞) absolute maximum (or "no absolute maximum")
Step-by-step explanation:
There will be extremes at the ends of the domain interval, and at turning points where the first derivative is zero.
The derivative is ...
h'(t) = 24t^2 -48t = 24t(t -2)
This has zeros at t=0 and t=2, so that is where extremes will be located.
We can determine relative and absolute extrema by evaluating the function at the interval ends and at the turning points.
h(-1) = 8(-1)²(-1-3) = -32
h(0) = 8(0)(0-3) = 0
h(2) = 8(2²)(2 -3) = -32
h(∞) = 8(∞)³ = ∞
The absolute minimum is -32, found at t=-1 and at t=2. The absolute maximum is ∞, found at t→∞. The relative maximum is 0, found at t=0.
The extrema are ...
- (-1, -32) absolute minimum
- (0, 0) relative maximum
- (2, -32) absolute minimum
- (+∞, +∞) absolute maximum
_____
Normally, we would not list (∞, ∞) as being an absolute maximum, because it is not a specific value at a specific point. Rather, we might say there is no absolute maximum.