Town B is 40km due North of town A. what is the bearing of A from B<br>

Megan hughes deposits $4300 in an account that pays simple interest. when she withdraws her money 7 months later, she receives

Juliette [100K]

The interest is 3.5 percent or 3 percents of the sum she deposited.

3 0

2 years ago

Perform the indicated operation. (y4 - 1) ÷ (y + 1)

Valentin [98]

$The\ domain:\\D:y\neq-1\\\\(y^4-1)\div(y+1)=\dfrac{(y^2)^2-1^2}{y+1}=\dfrac{(y^2-1)(y^2+1)}{y+1}\\\\=\dfrac{(y-1)(y+1)(y^2+1)}{y+1}=(y+1)(y^2+1)=y^3+y^2+y+1\\\\Used:\\(a^n)^m=a^{nm}\\\\a^2-b^2=(a-b)(a+b)$

7 0

2 years ago

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Adventure Play sells an mp3 player for $18.00 and charges $3.25 per song. King Music sells a player for $23.00 and charges $2.00

Rus_ich [418]

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.

4 0

2 years ago

A farmer can spend no more than $4,000 on fertilizer and seeds. The fertilizer costs $2 per pound and seeds cost $20 per pound.

lilavasa [31]

Answer:

The graph in the attached figure

Step-by-step explanation:

Let

f -----> the number of pounds of fertilizer

s ----> the number of pounds of seeds

we know that

The inequality that represent the situation is

$2f+20s\leq 4,000$

using a graphing tool

see the attached figure

The solution is the triangular shaded area

7 0

2 years ago

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a manufacturer of cat toys produces three different toys filled with catnip the toys are the shape of a cone cylinder sphere eac

OLEGan [10]

Statements A, B, and C are true.

Step-by-step explanation:

Step 1:

A cone, a cylinder and a sphere all have radii of 3 inches. The cone and cylinder have heights of 2 inches.

The volume of the cone, $V=\pi r^{2} \frac{h}{3}.$ Here r = 3 inches and h = 2 inches.

So the volume of the cone $=\pi r^{2} \frac{h}{3} = \pi (3^{2}) \frac{2}{3} = 18.85$ cubic inches.

The volume of the cylinder, $V=\pi r^{2} h.$ Here r = 3 inches and h = 2 inches.

So the volume of the cylinder = $=\pi r^{2} h= \pi (3^{2}) (2) = 56.55$ cubic inches.

The volume of the sphere, $V=\frac{4}{3} \pi r^{3}.$ Here r = 3 inches.

So the volume of the sphere $=\frac{4}{3} \pi (3^{3})= 113.1$ cubic inches.

Step 2:

Now we check to see which statements are true.

A. About 113 cubic inches of catnip will fit inside a sphere-shaped toy so it is true.

B. About 18.8 cubic inches of catnip will fit inside a cone-shaped toy so it true.

C. The sphere-shaped toy holds the most of the three so it is true.

D. No shape holds 169.6 cubic inches, so D is false.

E. The toys shaped like a cone and a cylinder do not hold the same amount of catnip in them. So it is false.

P.S. I couldn't see the options B, D, and E completely.

8 0

2 years ago

Town B is 40km due North of town A. what is the bearing of A from B​

Town B is 40km due North of town A. what is the bearing of A from B