Answer:
Options A-C-F
Step-by-step explanation:
we know that
<em>The circumference of a circle is equal to</em>
or 
where
D is the diameter and r is the radius
therefore
or 
The number pi is the ratio circumference - diameter or is the ratio circumference - 2 times radius
<em>The area of the circle is equal to</em>
therefore
The number pi is the ratio Area - radius squared
Answer:
-45
-26x+30
Step-by-step explanation:
So first you let's divide this equation into three parts.
Part one -9x(5x+4)
- Step one: you have to distribute the -9x to the numbers in the parentheses. That would leave us with a -9x*5x = -45 and -9x*4 = -36x
- Step two: Put the answers together. That would leave us with -45-36x.
Part two 10(x+3)
- Step one: you have to distribute the 10 to the numbers in the parentheses. That would leave us with a 10*x = 10x and 10*3 = 30
- Step two: Put the answers together. That would leave us with 10x+30
Part three solve
- Step one: combine and put together what you have solved for. That would leave us with a -45-36x+ 10x+30
- Step two: combine like terms. The like terms here are -36x and 10x. When you combine them you would get -26x.
- Step three: Write the equation in standard form. Therefore the answer is -45-26x+30.
Let me know if anything is confusing!
An observation of the Moon was conducted from Friday, November 8, 2013 to Thursday, November 14, 2013. The study of the Moon during this period occurred consistently between the hours of 8 and 9 p.m. EST within the Northern Hemisphere at 37.3346° N, 79.5228° W (Bedford, V.A.). The Moon was noted to be illuminated on the right side and had a dark shadow on the left side indicating a waxing phase. The light region grew over the surface of the Moon with each subsequent night. The first night’s phase was waxing crescent with over 25 percent of the Moon lit up. The next night, the light had grown to cover more of the Moon as it continued through its waxing crescent phase. On November 10th, the Moon exhibited traits of being at first-quarter or …show more content…
An observation of the Moon was conducted from Friday, November 8, 2013 to Thursday, November 14, 2013. The study of the Moon during this period occurred consistently between the hours of 8 and 9 p.m. EST within the Northern Hemisphere at 37.3346° N, 79.5228° W (Bedford, V.A.). The Moon was noted to be illuminated on the right side and had a dark shadow on the left side indicating a waxing phase. The light region grew over the surface of the Moon with each subsequent night. The first night’s phase was waxing crescent with over 25 percent of the Moon lit up. The next night, the light had grown to cover more of the Moon as it continued through its waxing crescent phase. On November 10th, the Moon exhibited traits of being at first-quarter or half-moon status because at least 50 percent of its surface was illuminated. In the following nights, the Moon displayed characteristics of waxing gibbous as the light continued to grow across the moon’s surface from right to left. The Moon was nearing closer to the full moon phase on November 14th as only a very small dark shadow was visible on the left side.
The Moon takes 27.3 days (sidereal month) to complete its actual orbit around the Earth. Like the Sun, the Moon rises in the east and sets in the west each day.
Answer:
+ 8/5
Step-by-step explanation:
Reciprocal is flip
Take the negative flip
- ( - 1 /(5/8))
+ 8/5
Answer:
Correct answer: Fourth answer As = 73.06 m²
Step-by-step explanation:
Given:
Radius of circle R = 16 m
Angle of circular section θ = π/2
The area of a segment is obtained by subtracting from the area of the circular section the area of an right-angled right triangle.
We calculate the circular section area using the formula:
Acs = R²· θ / 2
We calculate the area of an right-angled right triangle using the formula:
Art = R² / 2
The area of a segment is:
As = Acs - Art = R²· θ / 2 - R² / 2 = R² / 2 ( θ - 1)
As = 16² / 2 · ( π/2 - 1) = 256 / 2 · ( 1.570796 - 1) = 128 · 0.570796 = 73.06 m²
As = 73.06 m²
God is with you!!!
