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slavikrds [6]
3 years ago
12

How to do double sided equations

Mathematics
1 answer:
Veronika [31]3 years ago
7 0

Answer:

multiply each side

Step-by-step explanation: this is becasue when numbers are beside eachother they mean multiply


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If there bill was 58.65 and they gave a sever a 15% tip, how much did they pay altogether
masha68 [24]

Answer:

<u>$67.43 altogether</u>

Step-by-step explanation:

A 15% percent tip means 15% OF the bill.

15% of 58.65

0.15*58.65=8.7975

You can't really pay $8.7975, so you would round to the nearest one hundredth which is $8.78.

Now you add the $8.78 to $58.65 which will give you $67.43.

I hope that helps!


7 0
3 years ago
Peter received a trophy at the state math competition. His trophy is in the shape of a square pyramid and is covered in wrapping
lakkis [162]

Answer:315 cm

Step-by-step explanation:

7x7

7x9.5

7x9.5

7x9.5

7x9.5

6 0
3 years ago
What is the area of this partial circle with a radius of 14 inch
AveGali [126]

Step-by-step explanation:

Since this circle is partial we will divide the area of the whole circle by 2

the area of a circle is expressed by \pi r^2 so insert 14 for the radius.

14^2\pi

196\pi

divide the area by 2

98\pi=307.876

Hope that helps :)

6 0
3 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
What is 743x295<br> What do I need to d
Natasha_Volkova [10]

Answer: 219,185

-Multiply the place values to find the answer,

4 0
3 years ago
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