QM Q9.) Write the standard form of the equation of the circle with the given center and radius.

A junior college is going to choose a Homecoming Queen, and the candidates are the Freshman Queen and the Sophomore Queen. If th

KonstantinChe [14]

Count the votes, counting each sophomore ballot as 1.5 votes and each freshmen ballot as 1 vote.
ur doing this because there is 200 more freshmen then sophomores...and if u count each sophomore vote as 1.5, it would make up for the 200 more freshmen

5 0

2 years ago

WILL MAKE BRAINLYEST

Valentin [98]

Answer:

none

Step-by-step explanation:

3−12+223−56=158

216−216=0

56+216=272

216+216=432

216−56=160

Brainliest?

8 0

2 years ago

Read 2 more answers

If a ball is thrown straight up into the air with an initial velocity of 70 ft/s, its height in feet after t seconds is given by

Vitek1552 [10]

Answer:

a) Average velocity at 0.1 s is 696 ft/s.

b) Average velocity at 0.01 s is 7536 ft/s.

c) Average velocity at 0.001 s is 75936 ft/s.

Step-by-step explanation:

Given : If a ball is thrown straight up into the air with an initial velocity of 70 ft/s, its height in feet after t seconds is given by $y = 70t-16t^2$ .

To find : The average velocity for the time period beginning when t = 2 and lasting. a. 0.1 s. , b. 0.01 s. , c. 0.001 s.

Solution :

a) The average velocity for the time period beginning when t = 2 and lasting 0.1 s.

$(\text{Average velocity})_{0.1\ s}=\frac{\text{Change in height}}{0.1}$

$(\text{Average velocity})_{0.1\ s}=\frac{h_{2.1}-h_2}{0.1}$

$(\text{Average velocity})_{0.1\ s}=\frac{(70(2.1)-16(2.1)^2)-(70(0.1)-16(0.1)^2)}{0.1}$

$(\text{Average velocity})_{0.1\ s}=\frac{69.6}{0.1}$

$(\text{Average velocity})_{0.1\ s}=696\ ft/s$

b) The average velocity for the time period beginning when t = 2 and lasting 0.01 s.

$(\text{Average velocity})_{0.01\ s}=\frac{\text{Change in height}}{0.01}$

$(\text{Average velocity})_{0.01\ s}=\frac{h_{2.01}-h_2}{0.01}$

$(\text{Average velocity})_{0.01\ s}=\frac{(70(2.01)-16(2.01)^2)-(70(0.01)-16(0.01)^2)}{0.01}$

$(\text{Average velocity})_{0.01\ s}=\frac{75.36}{0.01}$

$(\text{Average velocity})_{0.01\ s}=7536\ ft/s$

c) The average velocity for the time period beginning when t = 2 and lasting 0.001 s.

$(\text{Average velocity})_{0.001\ s}=\frac{\text{Change in height}}{0.001}$

$(\text{Average velocity})_{0.001\ s}=\frac{h_{2.001}-h_2}{0.001}$

$(\text{Average velocity})_{0.001\ s}=\frac{(70(2.001)-16(2.001)^2)-(70(0.001)-16(0.001)^2)}{0.001}$

$(\text{Average velocity})_{0.001\ s}=\frac{75.936}{0.001}$

$(\text{Average velocity})_{0.001\ s}=75936\ ft/s$

5 0

3 years ago

How can u solve the surface area and volume ?

marin [14]

Use these formulas to get the surface area and volume. Hopes this helps! ❤️

7 0

2 years ago

3x-4y=0 2x+y=110 no solution

weeeeeb [17]

Answer:

(x, y) = (40, 30)

Step-by-step explanation:

A graphing calculator can show you the solution to this system of equations is (x, y) = (40, 30). That is the point of intersection where the two lines cross.

__

An algebraic solution can be found by using the substitution method. An expression for y can be found using the second equation:

y = 110 -2x . . . . . . subtract 2x from both sides

Using this in the first equation gives ...

3x -4(110 -2x) = 0 . . . . substitute for y

11x = 440 . . . . . . . . . simplify, add 440

x = 40 . . . . . . . . . . divide by 11

y = 110 -2(40) = 30

The solution is (x, y) = (40, 30).

5 0

1 year ago