Evaluate the expression 72/x2-2 for x=3

SELECT ALL MODELS I WILL GIVE BRAINLIEST 6TH GRADE MATH

bonufazy [111]

Answer:

The first, third, and fourth options are less than 50% shaded.

Step-by-step explanation:

Each model has 100 squares, and if you count the squares, the first, third, and fourth models have less than 50 squares shaded in.

4 0

2 years ago

A convex polygon has Choose...<br> sides when the interior angle sum is 1980º.

Viefleur [7K]

Answer:

13 sides.

Step-by-step explanation:

(n - 2) = 1980 / 180 = 11.

n = 13

When the sum of the interior angles is 1980 degrees, the convex polygon will have 13 sides.

i hope this helps and is right! p.s. i really need brainliest :)

3 0

2 years ago

The sum of 9 and twice a number

timofeeve [1]

Answer:

2x + 9

Step-by-step explanation:

Let, "A number", be denoted by the variable, x.

Note the meaning of each part:

"the sum" means addition, or + .

"twice" means multiply by 2

Put the parts together:

The sum of 9 and twice a number

2x + 9 is your amsw34.

3 0

2 years ago

Joanna bought 1 1/2 pounds of apples, 3/4 pound of strawberries, 1/2 pound of grapes, 1 3/4pounds of oranges, and 1 1/4 pounds o

Allushta [10]

The correct answer is 5.75 (5 3/4) pounds of fruit
1 1/2 +3/4 +1/2 + 1 3/4+ 1 1/4= 5 3/4

3 0

2 years ago

Read 2 more answers

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