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Firdavs [7]
3 years ago
12

3ab+4ab^3 I really need help please help meeeee!!!!

Mathematics
1 answer:
Advocard [28]3 years ago
3 0

Answer:

ab times (3+4b^2)

Step-by-step explanation:

You would factor the equation.

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Brad is installing a hot tub in his 30 ft by 20 ft back yard. The hot tub has a radius of 5 feet. If he randomly shoots
ddd [48]

Answer:

13% chance

Step-by-step explanation:

Area of backyard = length * width

A = 30 ft * 20 ft

A = 600 ft^{2}

Area of bathtub = π r^{2}

A = π * 5^{2}

A = 78.54 ft^{2}

Probability of arrow landing in tub can be found by dividing Area of the bathtub by the area of the backyard, and then multiplying by 100 for a percentage

Prob. = 78.54/600

Prob. = 0.13

Prob. = 0.13 * 100

Prob. = 13%

7 0
4 years ago
4. Angelina braided b inches of her hair. Brooke braided 4 less than 1/3 the number of inches Angelina braided.
Marta_Voda [28]

Answer:

E

Step-by-step explanation:

3 0
3 years ago
What is t÷5-4=9 what is t ??
Sophie [7]
Add 4 to boh sides
t/5+4-4=9+4
t/5+0=13
t/5=13
times 5/1 both sides
5t/5=5*13
1t=65

t=65
7 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
Which one is it I’m so confused?
Ad libitum [116K]

Answer:

I think its A.) Angle Bisector , I hope this helped!!

Step-by-step explanation:

5 0
3 years ago
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