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3 years ago

Solve showing all work and simplify completely: 9/18 + 1/3

Mathematics

1 answer:

ss7ja [257]3 years ago

6 0

Answer:

5/6

Step-by-step explanation:

9/18 + 1/3 <Need the same denominator so find the lowest common multiple.

3 6 9 12 15 18 < multiples of 3

if 3 goes into 18 than it will be our LCM

3 goes into 18 6 times so multiply 1 and 3 by 6 and rewrite the equation

9/18 + 6/18 < just add 9 + 6 and place it over 18

15/18 <this number can be simplified so write out their factors and pick the highest one to divide by one they have in common.

1 3 5 15 <Factors of 15

1 2 3 6 9 18 <Factors of 18

3 is the highest one they both have so divide both 15 and 18 by 3

15/3 is 5 and 18/3 is 6

put the 5 and 6 in their proper places and you have 5/6 or .83333333333...

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Find the circumference and area of each. Round to nearest tenth:

patriot [66]

№1. Given: r=8 ft, π≈3.14

C=2×π×r=2×3.14×8=50.24=50.2 ft

A=π×r²=3.14×64=200.96=201 ft²

Answer: 50.2 ft; 201 ft²

№2. Given: D=11 cm, π≈3.14

d=2r or r=2/d, so if d is 11 cm, then r is 11÷2=5.5 cm

C=2×π×r=πD=3.14×11=34.54=34.5 cm

A=π×r²=3.14×(5.5)²=94.985=95 cm²

Answer: 34.5 cm; 95 cm²

8 0

2 years ago

What is the radius of a circle given by equation x^2 + y^2 - 2x + 8y - 47 = 0

Lera25 [3.4K]

X^2 + y^2 - 2x + 8y - 47 = 0
x^2 + y^2 - 2x + 8y = 47
(x^2 - 2x) + (y^2 + 8y) = 47
(x^2 - 2(1)x) + (y^2 + 2(4)y) = 47
(x^2 - 2(1)x + 1^2) + (y^2 + 2(4)y + 4^2) = 47 + 1^2 + 4^2
(x - 1)^2 + (y + 4)^2 = 64 = 8^2
r=8

7 0

2 years ago

Read 2 more answers

Translate the expression One less than twice a number

Alja [10]

Answer:

2x – 1 = 17

Explanation:

Hope it helps you..

Y-your welcome in advance..

(;ŏ﹏ŏ)(ㆁωㆁ)

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

Helpppppppppppppp please

Bingel [31]

Answer:29

so the mini triangle in the corner is a mini model of the entire triangle, all you have to do is take the width or bottom number and find the factor in which it is smaller, then take the height of the smaller triangle and add the factor in which it increases to it. so the factor goes up by 31 1/2--9=22 1/2 and adding the factor in which it increases to the height is 22 1/2+6 1/2=29 because 2 halfs make a whole.

Step-by-step explanation:

mr clean has big brien 0-0

5 0

2 years ago

Read 2 more answers