Ask question

Ask question

All categories

2 years ago

15

Harrison High School has 768 students. In 6 years, it is projected to have 1,157 students. What is the projected average rate of

change per year in students over this time period?

1 answer:

Marrrta [24]2 years ago

4 0

Probably 65 students per year

You might be interested in

name the rational numbers that are not integers from the list below -11,0,57,8.07, sqrt 13, sqrt 4, 2 1/6, -8/45, 1.1919919991..

weqwewe [10]

Sqrt 13 and 1.1919919991... are irrational, meaning that they can't be described in a fraction of one integer over another, like 1/3, 45/44 or 57/107, these numbers are rational. Most irrationals are known constants like e or π, endless non-repeating decimals, or roots of non-perfect numbers like 13, 7, 5 or 2.

4 0

2 years ago

2. Wpisz taką liczbę, aby rozwiązaniem równania była liczba 2.

lys-0071 [83]

I didn’t understand but when I understand I will help you

5 0

2 years ago

Tickets for the school concert were 3$ and 2$. If 245 tickets were sold for a total of $630, how much of each kind were sold?

Hatshy [7]

2x + 3y = 630
x + y = 245 then you want to get rid of x so in the second equation × by -2 and get -2x -2y =-490 subtract this equation from the first to get y=$140 substitute 140 in for y and get x= $105 and 2x=210 and 3y=420 210 +420=630

7 0

3 years ago

tresset_1 [31]

A and D because it is basically saying 3:36 so when you simplify it comes down to 1:12

8 0

2 years ago

Read 2 more answers

A horizontal trough is 16 m long, and its end are isosceles trapezoids with an altitude of 4 m, a lower base of 4 m, and an uppe

Ganezh [65]

Answer:

0.28cm/min

Step-by-step explanation:

Given the horizontal trough whose ends are isosceles trapezoid

Volume of the Trough =Base Area X Height

=Area of the Trapezoid X Height of the Trough (H)

The length of the base of the trough is constant but as water leaves the trough, the length of the top of the trough at any height h is 4+2x (See the Diagram)

The Volume of water in the trough at any time

$Volume=\frac{1}{2} (b_{1}+4+2x)h X H$

$Volume=\frac{1}{2} (4+4+2x)h X 16$

=8h(8+2x)

V=64h+16hx

We are not given a value for x, however we can express x in terms of h from Figure 3 using Similar Triangles

x/h=1/4

4x=h

x=h/4

Substituting x=h/4 into the Volume, V

$V=64h+16h(\frac{h}{4})$

$V=64h+4h^2\\\frac{dV}{dt}= 64\frac{dh}{dt}+8h \frac{dh}{dt}$

h=3m,

dV/dt=25cm/min=0.25 m/min

$0.25= (64+8*3) \frac{dh}{dt}\\0.25=88\frac{dh}{dt}\\\frac{dh}{dt}=\frac{0.25}{88}$

=0.002841m/min =0.28cm/min

The rate is the water being drawn from the trough is 0.28cm/min.

3 0

3 years ago