Help me please quick!!!

Write (2b^2)^3 without exponents.

Studentka2010 [4]

Answer:

64 * b * b * b * b * b * b

Step-by-step explanation:

Apply exponent to number first, then expand the variable's exponent into repeated multiplication.

4 0

3 years ago

Read 2 more answers

If f(x)=3x−2 and g(x)=x2−5, what is g(f(2))?

Vikki [24]

Answer:

g(f(2)) = 11

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Algebra I

Function Notation

Step-by-step explanation:

Step 1: Define

f(x) = 3x - 2

g(x) = x² - 5

Step 2: Find g(f(2))

f(2)

Substitute in x: f(2) = 3(2) - 2
Multiply: f(2) = 6 - 2
Subtract: f(2) = 4

g(f(2))

Substitute in x: g(4) = 4² - 5
Exponents: g(4) = 16 - 5
Subtract: g(4) = 11

8 0

3 years ago

A large cube has a volume of 1 cubic unit. A small cube has a volume of 1 27 of a cubic unit. What is the difference between the

ExtremeBDS [4]

Given:

Volume of Large cube = 1 cubic unit

Volume of small cube = $\dfrac{1}{27}$ cubic unit

To find:

The difference between the edge length of the large cube and the edge length of the small cube.

Solution:

Let the edges of large and small cubes are $a_1$ and $a_2$ respectively.

We know that, volume of a cube is

$V=(edge)^3$

Volume of Large cube = 1 cubic unit

$(a_1)^3=1$

Taking cube root on both sides, we get

$a_1=1$

So, edge of large cube is 1 unit.

Volume of small cube = $\dfrac{1}{27}$ cubic unit

$(a_2)^3=\dfrac{1}{27}$

Taking cube root on both sides, we get

$a_2=\dfrac{1}{3}$

So, edge of large cube is $\dfrac{1}{3}$ unit.

Now, difference between them is

$d=a_1-a-2$

$d=1-\dfrac{1}{3}$

$d=\dfrac{3-1}{3}$

$d=\dfrac{2}{3}$

Therefore, the difference between the edge length of the large cube and the edge length of the small cube is $\dfrac{2}{3}$ unit.

6 0

3 years ago

A group consists of 6 men and 5 women. Five people are selected to attend a conference. In how many ways can 5 people be selecte

cluponka [151]

Answer:

462

6

1.2%

Step-by-step explanation:

Since the questions are combinations, we must apply the combination formula, which is as follows:

n C r = (n!) / (r! (n-r)!)

Because there are 6 men and 5 women, there are a total of 11 people.

Thus:

n = 11

In the first question in how many ways can 5 people be selected from this group of 11, r = 5.

Replacing in the formula:

11 C 5 = (11!) / (5! * (11-5)!)

11 C 5 = (11!) / (5! * 6!)

11 C 5 = 462

In the second question, in how many ways can 5 men be selected from the 6 men, here n = 6 and r = 5, replacing we are left with:

6 C 5 = (6!) / (5! * (6-5)!)

6 C 5 = (6!) / (5! * 1!)

6 C 5 = 6

In the last question of what is the probability that the selected group is all men, we have that it is the combination of the two previous questions. Since the total would be part A it would be the total of the combinations of choosing 5 of 11 people and part B of the 6 men that there are the combinations of choosing 5.

Divide the two values from parts A and B to get ...

(result from part B) / (result from part A) = (# of ways to pick 5 men) / (# of ways to pick 5 people)

(result from part B) / (result from part A) = 6/462

(result from part B) / (result from part A) = 0.012

In other words, the probability is 1.2%

7 0

3 years ago

PLEASE PLEASE HELP IF YOU KNOW!! THANK YOU

Vladimir [108]

I'm guessing this is what the question is asking?

5 0

3 years ago