All categories

2 years ago

How many fewer people attended the pacers games than attended the clippers games

Mathematics

1 answer:

ohaa [14]2 years ago

3 0

You subtract 582,295 from 670,063 and you get 87,768 so 87,768 fewer people attended the pacers game.

You might be interested in

Geometry// question #8

MAVERICK [17]

Answer:

see explanation

Step-by-step explanation:

The translation rule

(x, y ) → (x + 3, y - 2 )

Means add 3 to the original x- coordinate and subtract 2 from the original y- coordinate, thus

A(- 4, 1 ) → A'(- 4 + 3, 1 - 2 ) → A'(- 1, - 1 )

B(1, 1 ) → B'(1 + 3, 1 - 2 ) → B'(4, - 1 )

C(1, - 2 ) → C'(1 + 3, - 2 - 2 ) → C'(4, - 4 )

D(- 4, - 2 ) → D'(- 4 + 3, - 2 - 2 ) → D'(- 1, - 4 )

4 0

3 years ago

I need help on #11& #8

Musya8 [376]

100 * pi = pi * r^2
Pi cancels on both sides of the equation
So... 100= r^2
= radius is 10

5 0

3 years ago

What is the recursive rule for the sequence?

adoni [48]

An= -2 (an+1), a1 = 5

5 0

3 years ago

A box of Munchkins contains chocolate and glazed donut holes. If Jacob ate 2 chocolate

neonofarm [45]

Answer:

24 munchkins.

Step-by-step explanation:

Let C be the number of chocolate and D be number of glazed donut holes in the original box.

We are told if Jacob ate 2 chocolate munchkins, then 1/11 of the remaining Munchkins would be chocolate. We can represent this information as:

$C-2=\frac{1}{11}*(C+D-2)...(1)$

We are also told if he instead added 4 glazed Munchkins to the original box, 1/7 of the Munchkins would be chocolate. We can represent this information as:

$C=\frac{1}{7}*(C+D+4)...(2)$

Upon substituting C's value from equation (2) in equation (1) we will get,

$\frac{1}{7}*(C+D+4)-2=\frac{1}{11}*(C+D-2)$

Let us have a common denominator on right side of equation.

$\frac{1}{7}*(C+D+4)-\frac{7*2}{7}=\frac{1}{11}*(C+D-2)$

$\frac{C+D+4-14}{7}=\frac{1}{11}*(C+D-2)$

Multiplying both sides of our equation by 7, we will get,

$7*\frac{C+D-10}{7}=7*\frac{1}{11}*(C+D-2)$

$C+D-10=\frac{7}{11}*(C+D-2)$

Multiplying both sides of our equation by 11, we will get,

$11*(C+D-10)=11*\frac{7}{11}*(C+D-2)$

$11*(C+D-10)=7*(C+D-2)$

$11C+11D-110=7C+7D-14$

$11C-7C+11D-7D=-14+110$

$4C+4D=96$

$4(C+D)=96$

$(C+D)=\frac{96}{4}$

$(C+D)=24$

Therefore, the total number of Munchkins in original box is 24.

3 0

3 years ago

i am not 42 but i am an EVEN number. i am more than 25 but less than 50. i am a multiple of 6 but do not have a digit of 3.

sertanlavr [38]

So easyies way is find all multiplues of 6 that are less than 50

6
12
18
24
30
36
42
48

and not 42
6
12
18
24
30
36
48
more than 25
30
35
48

do not have a 3 in them
48

the answer is 48

5 0

3 years ago

Read 2 more answers