I believe the answer is one.
Assuming those lines are parallel, then this means the alternate interior angles are the same measure
3x-26 = 2x
3x-26+26 = 2x+26 ... adding 26 to both sides
3x = 2x+26
3x-2x = 2x+26-2x ... subtracting 2x from both sides
3x-2x = 26
<h3>x = 26 is the answer</h3>
Answer:
Step-by-step explanation:
The inequality will be split into two
It is know that, if a<b<c
Then a<b and b<c
-8<2x-4<4
Apply that to this
Then,
-8<2x-4. Equation 1
Also,
2x-4<4 equation 2
Solving equation 1
-8<2x-4
Add 4 to both side of the equation
-8+4<2x-4+4
-4<2x
Divide both sides by 2
-4/2<2x/2
-2<x
Note, if a is less than b, then, b is greater than a, e.g. 4 is less than 10, this implies 10 is greater than 4
Therefore,
-2<x
Then, x greater than -2
Equation 2
2x-4<4
Add 4 to both side of the inequalities
2x-4+4<4+4
2x<8
Divide both side by 2
Then,
2x/2<8/2
x<4
Therefore x is between -2 and +4.
Check attachment for graphical solution
1 + 1 equals 2
check:
2 - 1 = 1
Answer:
b: See first attached photo
c: V = x²y
d and e: V = x(3 - 2x)²
f: 2 cubic feet
Step-by-step explanation:
a: Sketch several boxes and calculate the volumes.
b: See first attached photo a diagram of this situation
The diagram is a square. We are cutting out squares from the corners. We don't know the size of the square yet. The side lengths were 3, but now they are 3 - 2x (since each corner has one side of the square, there are 2 sides of the cut out square on each side of the larger square)
c: The equation for volume is: V = x²y
The length and width of the box are the x values, the height would be the y value
d and e: It wants the equation for the volume for our situation. The base of the box is (3 - 2x)(3 - 2x) or (3 - 2x)². The height of the box is x, so the volume is
V = x(3 - 2x)²
f: Take the derivative, find the critical values, then plug that into x and solve for the volume. See second attached photo for the work for finding the x value that maximizes the box, and the third attached photo for the evaluation of the maximum volume...