Step-by-step explanation:
::
tbh h.c chhbv. yea I can
Take the deritiivive
it is zero at pi/4 and it repeats at every pi
minimum at where slope goesfrom negative to positive
so at 5pi/4 and 13pi/4 so at every 2pi interval
evaluate origial at 5pi/4
=
=
or aprox 1.22509
Answer:
40 square metres
Step-by-step explanation:
The shaded region is of a triangle, whose area is denoted by: A = (1/2) * b * h, where b is the base and h is the height.
Since the left figure is a square with side lengths 10, we know that the height of the triangle is also 10 metres. The right figure is a rectangle with length 4. Since the total base length of the entire figure is 18 and the base of the square is 10, then the width of the rectangle is 18 - 10 = 8 metres.
This width is also the base of the triangle, so b = 8.
Now plug these values into the equation:
A = (1/2) * b * h
A = (1/2) * 8 * 10 = (1/2) * 80 = 40
The area is 40 square metres.
Answer:
x = infinite amount of solutions
General Formulas and Concepts:
<u>Pre-Algebra</u>
- Order of Operations: BPEMDAS
- Equality Properties
<u>Algebra I</u>
- Solving systems of equations using substitution/elimination
- Solving systems of equations by graphing
Step-by-step explanation:
<u>Step 1: Define systems</u>
4x - 3y = 1
-8x + 6y = -2
<u>Step 2: Rewrite systems</u>
-8x + 6y = -2
- Add 8x to both sides: 6y = 8x - 2
- Divide 6 on both sides: y = 4/3x - 1/3
<u>Step 3: Redefine systems</u>
4x - 3y = 1
y = 4/3x - 1/3
<u>Step 4: Solve for </u><em><u>x</u></em>
<em>Substitution</em>
- Substitute in <em>y</em>: 4x - 3(4/3x - 1/3) = 1
- Distribute -3: 4x - 4x + 1 = 1
- Combine like terms: 1 = 1
Here we see that 1 does indeed equal 1.
∴ x = all real numbers/infinite amount of solutions
<u>Step 5: Graph</u>
<em>Check the systems.</em>
We see that the 2 equations are the same line.
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.