Help PLEASE!! <br>how do you calculate your monthly mortgage payment when you buy a property?

Who is farthest from the school

postnew [5]

Using the law of cosines:

The grocery store to the school:

Distance = √(7^2 + 10^2 - 2*7*10*cos(100)

Distance = 13.16 miles

The movie store to the school:

Distance = √(7^2 + 10^2 - 2*7*10*cos(120)

Distance = 14.80 miles

The friend is further away.

7 0

3 years ago

The radius of a cone is decreasing at a constant rate of 7 inches per second, and the volume is decreasing at a rate of 948 cubi

inessss [21]

Answer:

The height of cone is decreasing at a rate of 0.085131 inch per second.

Step-by-step explanation:

We are given the following information in the question:

The radius of a cone is decreasing at a constant rate.

$\displaystyle\frac{dr}{dt} = -7\text{ inch per second}$

The volume is decreasing at a constant rate.

$\displaystyle\frac{dV}{dt} = -948\text{ cubic inch per second}$

Instant radius = 99 inch

Instant Volume = 525 cubic inches

We have to find the rate of change of height with respect to time.

Volume of cone =

$V = \displaystyle\frac{1}{3}\pi r^2 h$

Instant volume =

$525 = \displaystyle\frac{1}{3}\pi r^2h = \frac{1}{3}\pi (99)^2h\\\\\text{Instant heigth} = h = \frac{525\times 3}{\pi(99)^2}$

Differentiating with respect to t,

$\displaystyle\frac{dV}{dt} = \frac{1}{3}\pi \bigg(2r\frac{dr}{dt}h + r^2\frac{dh}{dt}\bigg)$

Putting all the values, we get,

$\displaystyle\frac{dV}{dt} = \frac{1}{3}\pi \bigg(2r\frac{dr}{dt}h + r^2\frac{dh}{dt}\bigg)\\\\-948 = \frac{1}{3}\pi\bigg(2(99)(-7)(\frac{525\times 3}{\pi(99)^2}) + (99)(99)\frac{dh}{dt}\bigg)\\\\\frac{-948\times 3}{\pi} + \frac{2\times 7\times 525\times 3}{99\times \pi} = (99)^2\frac{dh}{dt}\\\\\frac{1}{(99)^2}\bigg(\frac{-948\times 3}{\pi} + \frac{2\times 7\times 525\times 3}{99\times \pi}\bigg) = \frac{dh}{dt}\\\\\frac{dh}{dt} = -0.085131$

Thus, the height of cone is decreasing at a rate of 0.085131 inch per second.

3 0

2 years ago

Rectangle ABCD has vertices at (-7,-2); (1, -2); (1, -8); and (-7, -8) respectively. If GHJK is a similar rectangle where G(2, 5

Minchanka [31]

Answer:

J (6,2) K (2,2)

Step-by-step explanation:

Now distance of AB = [(y2-y1)^2 + (x2-x1)^2]^(1/2)

= 8 units

CB = [(y2-y1)^2 + (x2-x1)^2]^(1/2)

= 6 units by using same formula

Now AB/CB must equal to GH/HJ as rectangles are similar

GH = [(y2-y1)^2 + (x2-x1)^2]^(1/2)

= 4 units

so

8/6 = 4/HJ

So,

HJ = 3 units

Now if we see the coordinates given carefully, it is obvious that two perpendicular lines lie perfectly parallel to x and y coordinates in rectangle ABCD. A is (-7,-2) and B is (1,-2) which means distance along y-axis doesn't change. Similarly for C (1,-8) and D (-7,-8), one can see that distance between y-axis doesn't change. So lines AB and CD of rectangle are parallel with x and AD and BC are parallel with y-axis.

In rectangle GHJK one can see that in given coordinates, G(2,5) and H(6,5), y coordinate is same so it is parallel to x axis. Now, HJ is perpendicular to GH so it must be parallel to y axis. It means if we know the lengths of sides we can easily determine unknown coordinates by simple addition and subtraction.

So, we know HJ = 3 units

J is (6,2) since HJ is parallel to y axis so distance on x axis will remain unchanged and length of line HJ will effect distance of y axis.

Similarly K is (2,2) for the same reason.

6 0

2 years ago

a diet plan claims that people who follow the diet lose an average of 2 pounds per week. if this claim is true, what is the aver

ser-zykov [4K]

Step-by-step explanation:

7 0

2 years ago

Read 2 more answers

I need help finding the slope