Answer:
The coordinates of EF are E(5,-4) and F(1,-4).
The line segment EF is in QIV
Step-by-step explanation:
The line segment AB has vertices at: A(-4,5) and B(-4,1).
We apply the rule
to reflect AB in the y-axis to obtain CD.


We apply the rule
to rotate CD 90 degrees clockwise about the origin to obtain EF.


The coordinates of EF are E(5,-4) and F(1,-4).
See attachment
Answer:
- 3.28 should be plotted between 3.2 and 3.4
- 3.28 is closer to 3.0 than 4.0.
- 3.28 is closer to 3.2 than 3.4.
Step-by-step explanation:
3.28 is located towards the positive side of the number line being a positive value. Since the value is located between 3.2 and 3.4, therefore it can be plotted between this two points.
Also 3.28 is known to be closer to 3.0 than 4.0 because the difference between 3.28 and 3.0 is lower than the difference between 3.28 and 4.
4-3.28 = 0.72(larger value)
3.28-3.0 = 0.28 (smaller value)
The smaller the difference, the closer the value of 3.28 to the value in consideration.
Similarly, 3.28 is closer to 3.2 than 3.4, due to their differences. The difference between 3.28 and 3.2 is lower than the difference between 3.28 and 3.4 as shown:
3.28 - 3.2 = 0.08(smaller)
3.4-3.28 = 0.12(larger)
Percent increase
find increase first
10500 to 11300
11300-10500=800
so
percent increase
change/original
origianal=10500
change=800
800/10500=8/105=0.0761
percent means parts out of 100
0.0761/1 times 100/100=7.61/100=7.61%
rond 7.61% to tenth or to 7.6%
7.6%
Answer:
7, 4, 3
Step-by-step explanation:
Using Descartes' rule of signs, you can determine there are zero negative real roots. The above answer choice is the only one with no negative zeros listed.
_____
The count of negative real roots is based on the number of sign changes when the signs of odd-degree terms are reversed. Doing that gives the signs of the terms as - - - -, so there are no changes. That means there are no negative real roots.
A graphing calculator confirms this choice.
Answer:
Part 1)
------> 
Part 2)
------> 
Part 3)
------> 
Part 4)
------> 
Step-by-step explanation:
we know that
The largest cross sectional area of that sphere is equal to the area of a circle with the same radius of the sphere
Part 1) we have

The area of the circle is equal to

so

Solve for r


Find the volume of the sphere
The volume of the sphere is

For 
substitute


Part 2) we have

The area of the circle is equal to

so

Solve for r


Find the volume of the sphere
The volume of the sphere is

For 
substitute


Part 3) we have

The area of the circle is equal to

so

Solve for r


Find the volume of the sphere
The volume of the sphere is

For 
substitute


Part 4) we have

The area of the circle is equal to

so

Solve for r


Find the volume of the sphere
The volume of the sphere is

For 
substitute

