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vladimir2022 [97]
3 years ago
8

Find the slope of a ski run that descends 15 feet for every horizontal change of 24 feet.

Mathematics
2 answers:
frutty [35]3 years ago
7 0
-5/8
rise over run
-15/24 = -5/8
Ivahew [28]3 years ago
5 0
The answer is 360 -15 times -24 equals 360
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What is the value of X?
Elodia [21]

Answer:

Vertical Angles

x = 21

Step-by-step explanation:

Vertical Angles

4x + 3 = 87

4x = 84

x = 21

5 0
3 years ago
If angle adc=134 and adb=5x+4 and bdc=7x-2 what does angle bdc measure
ruslelena [56]
As I've shown on picture that I attached:
∠ADC=∠ADB+∠BDC⇒134=5x+4+7x-2
12x+2=134
12x=132⇒x=132/12
x=11
Angle BDC measure 7x-2=7*11-2⇒∠BDC=75.

8 0
3 years ago
How many groups of 1 2/3 are in 4 1/3
Olegator [25]

Answer:

2. can not be 3 bc it would 5

Step-by-step explanation:

3 0
2 years 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
An Someone plz help me plz :(
zlopas [31]

Answer:

18

Step-by-step explanation:

original 3:4

now 18:24

Multiply by 6

6 0
3 years ago
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