2 years ago

I need help with this problem

Mathematics

1 answer:

Oksanka [162]2 years ago

3 0

The answers is bellemont

You might be interested in

Evaluate the expression below for x = 5. 5(x + 7)

wariber [46]

Answer:

Step-by-step explanation:

5(x + 7)

when x is equal to 5, then

5(x + 7)
5(5+7)
5(12)
5×12
60

7 0

2 years ago

What are all the shapes that are used to create the net of a cylinder

Ksenya-84 [330]

2 circles and for the part in the middle if you look at it as a net it will be a rectangle just rolled up. That is called a lateral surface or a lateral face

6 0

2 years ago

Read 2 more answers

Which statements about opposites are true?

Naily [24]

Answer:

I think B and C are the only ones true

5 0

2 years ago

Charles rolled a number cube labeled from 1 to 6. What is the probability that he will roll a number greater than 2? in a fracti

arlik [135]

Answer:

4/6 or 2/3

Step-by-step explanation:

There are 4 numbers greater than 2 and less than or equal to 6, meaning that to roll a number greater than six, he has a 4/6 chance.

5 0

2 years ago

Read 2 more answers

How many positive integers can be expressed ad a product of two or more of the prime numbers 5,7,11,and 13 if no one product is

sertanlavr [38]

Answer:

11 positive integers can be expressed.

Step-by-step explanation:

Consider the provided information.

The number of possible prime numbers are 5,7,11,and 13.

There are 4 possible prime numbers.

How many positive integers can be expressed as a product of two or more of the prime numbers, that means there can be product of two numbers, three number or four numbers.

The formula to calculate combinations is: $^nC_r=\frac{n!}{r!(n-r)!}$

The number of ways are:

$^4C_2+^4C_3+^4C_4=\frac{4!}{2!(4-2)!}+\frac{4!}{3!(4-3)!}+\frac{4!}{4!}$

$^4C_2+^4C_3+^4C_4=\frac{4!}{2!2!}+\frac{4!}{3!}+1$

$^4C_2+^4C_3+^4C_4=6+4+1$

$^4C_2+^4C_3+^4C_4=11$

Hence, 11 positive integers can be expressed.

8 0

2 years ago