Answer:
1 cm
Step-by-step explanation:
To solve this problem we can use the Pythagorean theorem. In fact the diagonal of a rectangle is an hypotenuse of a right triangle, while the length is a leg. The width is the other leg
width = √2^2 - (√3)^2 = √4 - 3 = √1 = 1 cm
If both Pamela and Debbie can shelf the same number of books at the end, the smallest number of books each could have shelved is 182
The number of books that Pamela can shelf at a time = 26
The number of books that Debbie can shelf at a time = 7
If both Pamela and Debbie can shelf the same number of books at the end, the smallest number of books each could have shelved is the least common multiple of 26 and 7
The Least common multiple of 26 and 7 = 182
Therefore, if both Pamela and Debbie can shelf the same number of books at the end, the smallest number of books each could have shelved is 182
Learn more on Least Common Multiple here: brainly.com/question/363238
Answer:
Step-by-step explanation:
I'm going to use calculus to solve this, because it's the simplest way.
The acceleration due to gravity in feet is the second derivative of the position function. We will start with the acceleration and work backwards with antiderivatives to get to the position function.
a(t) = -32. Going backwards and using the fact that the initial vertical velocity is 64 ft/sec, our velocity function is
v(t) = -32t + 64. Going backwards and using the fact that the initial height of the ball is 6 feet, our position function is

The first part of this question asks us the maximum height of the ball. From Physics, we learn that the maximum height of a projectile is reached when the velocity is 0, which happens to be right where the projectile stops for a nanosecond in the air to turn around and come back down. We set the velocity function equal to 0 and solve for t.
0 = -32t + 64 and
0 = -32(t - 2). By the Zero Product Property, either -32 = 0 or t - 2 = 0. It's obvious that -32 does not equal 0, so t - 2 must equal 0. Solving this for t:
t - 2 = 0 so
t = 2 seconds. Since the maximum height is reached at a time of 2 seconds, we plug 2 seconds into the position function to get its position at 2 seconds (which is also the max height of the ball).
and
s(2) = -64 + 128 + 6 so
s(2) = 70 feet
Now we want to know when the ball will hit the ground. "When" is a time value, and we know that the height of the ball on the ground is 0, so we sub in a 0 for s(t) and factor the quadratic.
Using the quadratic formula:
and
which gives us the 2 solutions
and

Plugging into your calculator, the first t = -.0916500 and the second t = 4.091
We all know that time cannot ever be negative, so our t value is 4.09.
Again, from Physics, we know that a projectile reaches it max height at halfway through its travels, so it just goes to follow logically that if it halfway through its travels at 2 seconds, then it will hit the ground at 4 seconds. And it does!! How awesome is that?!
Let's start b writing down coordinates of all points:
A(0,0,0)
B(0,5,0)
C(3,5,0)
D(3,0,0)
E(3,0,4)
F(0,0,4)
G(0,5,4)
H(3,5,4)
a.) When we reflect over xz plane x and z coordinates stay same, y coordinate changes to same numerical value but opposite sign. Moving front-back is moving over x-axis, moving left-right is moving over y-axis, moving up-down is moving over z-axis.
A(0,0,0)
Reflecting
A(0,0,0)
B(0,5,0)
Reflecting
B(0,-5,0)
C(3,5,0)
Reflecting
C(3,-5,0)
D(3,0,0)
Reflecting
D(3,0,0)
b.)
A(0,0,0)
Moving
A(-2,-3,1)
B(0,-5,0)
Moving
B(-2,-8,1)
C(3,-5,0)
Moving
C(1,-8,1)
D(3,0,0)
Moving
D(1,-3,1)
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