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Nutka1998 [239]
3 years ago
9

The Party Room at Penny's Pizza rents for an initial fee of $30 and then $5 per hour. Aislyn's bill for her Birthday party was $

50. For how many hours did she rent the room?
*NEED HELP WITH ANSWER*
Mathematics
2 answers:
vampirchik [111]3 years ago
5 0
$50- $30 = $20
$20/ $5 = 4 hours
Annette [7]3 years ago
4 0
Initial fee- $30
$5 per hour
=$50

$5* 4(hours)=$20 
$30+$20=$50

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Will the sum of 2x3 + x2 - 4 and -5x² - 6x? be a polynomial?<br> yes<br> no
Eva8 [605]

Answer:

Yes

Step-by-step explanation:

Since we are adding two polynomials

The sum will also be polynomial

4 0
2 years ago
Solve the compound inequality and graph<br> Please help! :)
frozen [14]

Answer:

Step-by-step explanation:

x+4≥10 and x-6>-15

x≥10-4 and x>-15+6

x≥ 6 and x>-9

combining the two

x>-9

8 0
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Simplify: 3 square root 216<br> A. 8<br> B. 16<br> C. 6<br> D. 2
zysi [14]

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6

Step-by-step explanation:

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4 0
2 years ago
Read 2 more answers
∠CAT and ∠MUD are supplementary. If m∠CAT=11x+16 and m∠MUD=4x+14, find the measure of each angle.
Georgia [21]

Answers:

  • x = 10
  • angle CAT = 126 degrees
  • angle MUD = 54 degrees

=========================================================

Explanation:

∠CAT and ∠MUD are supplementary, which means the angle measures add to 180. They form a straight line.

( m∠CAT ) + ( m∠MUD ) = 180

( 11x+16 ) + ( 4x+14 ) = 180

11x+16 + 4x+14 = 180

(11x+4x) + (16+14) = 180

15x+30 = 180

15x = 180-30

15x = 150

x = 150/15

x = 10

Let's find each angle based on this x value

  • m∠CAT=11x+16 = 11*10+16 = 110+16 = 126 degrees
  • m∠MUD=4x+14 = 4*10+14 = 40+14 = 54 degrees

Those two angles add to 126+54 = 180 to confirm we do indeed have supplementary angles, and confirm the correct answers.

8 0
1 year ago
If n is a positive integer, how many 5-tuples of integers from 1 through n can be formed in which the elements of the 5-tuple ar
Oksana_A [137]

Answer:

n + 4 {n \choose 2} + 6 {n \choose 3} + 4 {n \choose 4} + {n \choose 5}

Step-by-step explanation:

Lets divide it in cases, then sum everything

Case (1): All 5 numbers are different

 In this case, the problem is reduced to count the number of subsets of cardinality 5 from a set of cardinality n. The order doesnt matter because once we have two different sets, we can order them descendently, and we obtain two different 5-tuples in decreasing order.

The total cardinality of this case therefore is the Combinatorial number of n with 5, in other words, the total amount of possibilities to pick 5 elements from a set of n.

{n \choose 5 } = \frac{n!}{5!(n-5)!}

Case (2): 4 numbers are different

We start this case similarly to the previous one, we count how many subsets of 4 elements we can form from a set of n elements. The answer is the combinatorial number of n with 4 {n \choose 4} .

We still have to localize the other element, that forcibly, is one of the four chosen. Therefore, the total amount of possibilities for this case is multiplied by those 4 options.

The total cardinality of this case is 4 * {n \choose 4} .

Case (3): 3 numbers are different

As we did before, we pick 3 elements from a set of n. The amount of possibilities is {n \choose 3} .

Then, we need to define the other 2 numbers. They can be the same number, in which case we have 3 possibilities, or they can be 2 different ones, in which case we have {3 \choose 2 } = 3  possibilities. Therefore, we have a total of 6 possibilities to define the other 2 numbers. That multiplies by 6 the total of cases for this part, giving a total of 6 * {n \choose 3}

Case (4): 2 numbers are different

We pick 2 numbers from a set of n, with a total of {n \choose 2}  possibilities. We have 4 options to define the other 3 numbers, they can all three of them be equal to the biggest number, there can be 2 equal to the biggest number and 1 to the smallest one, there can be 1 equal to the biggest number and 2 to the smallest one, and they can all three of them be equal to the smallest number.

The total amount of possibilities for this case is

4 * {n \choose 2}

Case (5): All numbers are the same

This is easy, he have as many possibilities as numbers the set has. In other words, n

Conclussion

By summing over all 5 cases, the total amount of possibilities to form 5-tuples of integers from 1 through n is

n + 4 {n \choose 2} + 6 {n \choose 3} + 4 {n \choose 4} + {n \choose 5}

I hope that works for you!

4 0
3 years ago
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