All categories

3 years ago

2 to the Third power -(30÷5) +9

Mathematics

1 answer:

OLga [1]3 years ago

8 0

Answer:

Step-by-step explanation:

2^3-(30/5)+9

First, you divide 30/5 because it is in parentheses and we are following this rule for order of operations:

P- parentheses

E- exponents

M- multiplication

D- division

A- addition

S- subtraction

so 30/5 = 6 and plugged in is:

2^3-(6)+9

Next, we do exponents:

2^3 = 2x2x2 = 8

plugged back in:

8-(6)+9

Now, we do the math from left to right so:

8-6=2

2+9=11

11 is your answer

You might be interested in

Could anyone help??????????????

kvasek [131]

Answer:

a. No

b. Yes

c. No

d. Yes

e. Yes

Step-by-step explanation:

3(x+2) simplified is equal to (3)x+(3)2 which is simplified to 3x+6.

a. 3x+2 is not the same as 3x+6 so it is false.

b. 3(2+x) is simplified to (3)2+3(x) which is simplified to 6+3x or 3x+6 which is equal to 3x+6 so it is true.

c. 3x+2x is simplified to 5x which is not equal to 3x+6 so it is false.

d. x+2x+2+4 is simplified to 3x+6 which is equal to 3x+6 so it is true.

e. x+x+x+1+1+1+1+1+1 is simplified to 3x+6 which is equal to 3x+6 so it is true.

If this answer has helped you please mark as brainliest

8 0

2 years ago

Find the HCF of 20,55,90 Pls help it is urgent if i dont get this answer i will fail

Ann [662]

Answer:

Step-by-step explanation:

Factors of 20 = 1 2 4 5 10 20

Factors of 55 = 1 5 11 55

Factors of 90 = 1 2 3 5 6 9 10 15 18 30 45 90

Hcf = 5

5 0

2 years ago

There are 14 juniors and 16 seniors in a chess club. a) From the 30 members, how many ways are there to arrange 5 members of the

Dmitrij [34]

Answer:

a) 17,100,720

b) 4,717,440

c) 10,920

d) 2821

Step-by-step explanation:

14 juniors and 16 seniors = 30 people

a) From the 30 members, how many ways are there to arrange 5 members of the club in a line?

As it is a ordered arrangement

30.29.28.27.26 = 17,100,720

b) How many ways are there to arrange 5 members of the club in a line if there must be a senior at the beginning of the line and at the end of the line?

16.28.27.26.15 = 4,717,440

c) If the club sends 2 juniors and 2 seniors to the tournament, how many possible groupings are there?

Not ordered arrangement. And means that we need to multiply the results.

C₁₄,₂ * C₁₆,₂

C₁₄,₂ = 14.13.12! = 14.13 = 91

12! 2! 2

C₁₆,₂ = 16.15.14! = 16.15 = 120

14! 2! 2

C₁₄,₂ * C₁₆,₂ = 91.120 = 10,920

d) If the club sends either 4 juniors or 4 seniors, how many possible groupings are there?

Or means that we need to sum the results.

C₁₄,₄ + C₁₆,₄

C₁₄,₄ = 14.13.12.11.10! = 14.13.12.11 = 1001

10! 4! 4.3.2.1

C₁₆,₄ = 16.15.14.13.12! = 16.15.14.13 = 1820

12! 4! 4.3.2.1

C₁₄,₄ + C₁₆,₄ = 1001 + 1820 = 2821

7 0

3 years ago

Can SOMEBODY PLS HELP ME

kkurt [141]

G= -36
If you need a step by step please message me

8 0

3 years ago

Read 2 more answers

An algebra tile configuration. Only the Product spot is shown. 9 tiles are in the Product spot in 5 columns with 5 rows. First r

andrezito [222]

Answer:

The correct answer is 16 c.

Step-by-step explanation:

because i know

8 0

3 years ago

Read 2 more answers