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3 years ago

Which number is irrational?

Mathematics

2 answers:

vlada-n [284]3 years ago

6 0

The square root of 7(c) because it never repeats itself and continues and never ends

Ilya [14]3 years ago

5 0

The answer is c hope this helps

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Please help with 14 and 15

olchik [2.2K]

n - a number

14. Nine more than the square of a number: n² + 9.

15. Twice the difference between a number and ten: 2(n - 10)

6 0

3 years ago

Two of your friends go bowling. One friend rents a pair of bowling shoes for $3 and bowls 3 games. The other friend brings his o

Nesterboy [21]

Answer: The cost of each game is $2.50

Step-by-step explanation:

Let's find the equations for each friend:

Friend 1:

Pays $3 for the shoes, and plays 3 games, then if X is the cost of each game, Friend 1 pays a total of:

$3 + 3*X

Friend 2:

This friend buys a soda for $0.50, and he plays 4 games, remember that the cost of each game was X. then this friend pays:

$0.50 + 4*X

And we know that both friends pay the exact same amount, then we can write:

$3 + 3*X = $0.50 + 4*X

And solve this for X.

We need to isolate X, then we can move all the terms with X to the right, and all the terms without X to the left:

$3 - $0.50 = 4*X - 3*X

$2.50 = (4 - 3)*X = X

This means the cost of each game is $2.50

4 0

2 years ago

PLEASE!!! ASAP!!! WILL GIVE BRANLIEST!!! SHOW & EXPLAIN!!

pashok25 [27]

Answer:

365 messages

Step-by-step explanation:

x = # text messages)

16.5 = 10 + .10(x - 300)

6.5 = .10x - 30

36.5 = .10x

x = 365

6 0

3 years ago

Read 2 more answers

I know you want to answer this question.

Alik [6]

Answer:

D. x = 3

Step-by-step explanation:

$\frac{1}{2} ^{x-4}$ - 3 = $4^{x-3}$ - 2

First, convert $4^{x-3}$ to base 2:

$4^{x-3}$ = $(2^{2})^{x-3}$

$\frac{1}{2} ^{x-4}$ - 3 = $(2^{2})^{x-3}$ - 2

Next, convert $\frac{1}{2} ^{x-4}$ to base 2:

$\frac{1}{2} ^{x-4}$ = $(2^{-1})^{x-4}$

$(2^{-1})^{x-4}$ - 3 = $(2^{2})^{x-3}$ - 2

Apply exponent rule: $(a^{b})^{c}$ = $a^{bc}$ :

$(2^{-1})^{x-4}$ = $2^{-1*(x-4)}$

$2^{-1*(x-4)}$ - 3 = $(2^{2})^{x-3}$ - 2

Apply exponent rule: $(a^{b})^{c}$ = $a^{bc}$ :

$(2^{2})^{x-3}$ = $2^{2(x-3)}$

$2^{-1*(x-4)}$ - 3 = $2^{2(x-3)}$ - 2

Apply exponent rule: $a^{b+c}$ = $a^{b}$ $a^{c}$ :

$2^{-1(x-4)}$ = $2^{-1x}$ * $2^{4}$ , $2^{2(x-3)}$ = $2^{2x}$ * $2^{-6}$

$2^{-1 * x}$ * $2^{4}$ - 3 = $2^{2x}$ * $2^{-6}$ - 2

Apply exponent rule: $(a^{b})^{c}$ = $a^{bc}$ :

$2^{-1x}$ = $(2^{x})^{-1}$ , $2^{2x}$ = $(2^{x})^{2}$

$(2^{x})^{-1}$ * $2^{4}$ - 3 = $(2^{x})^{2}$ * $2^{-6}$ - 2

Rewrite the equation with $2^{x} = u$ :

$(u)^{-1}$ * $2^{4}$ - 3 = $(u)^{2}$ * $2^{-6}$ - 2

Solve $u^{-1}$ * $2^{4}$ - 3 = $u^{2}$ * $2^{-6}$ - 2:

$u^{-1}$ * $2^{4}$ - 3 = $u^{2}$ * $2^{-6}$ - 2

Refine:

$\frac{16}{u}$ - 3 = $\frac{1}{64}$ $u^{2}$ - 2

Add 3 to both sides:

$\frac{16}{u}$ - 3 + 3 = $\frac{1}{64}$ $u^{2}$ - 2 + 3

Simplify:

$\frac{16}{u}$ = $\frac{1}{64}$ $u^{2}$ + 1

Multiply by the Least Common Multiplier (64u):

$\frac{16}{u}$ * 64u = $\frac{1}{64}$ $u^{2}$ + 1 * 64u

Simplify:

$\frac{16}{u}$ * 64u = $\frac{1}{64}$ $u^{2}$ + 1 * 64u

Simplify $\frac{16}{u}$ * 64u:

1024

Simplify $\frac{1}{64}$ $u^{2}$ * 64u:

$u^{3}$

Substitute:

1024 = $u^{3}$ + 64u

Solve for u:

u = 8

Substitute back u = $2^{x}$ :

8 = $2^{x}$

Solve for x:

x = 3

4 0

3 years ago

A man is on a diet and goes into a shop to buy some turkey slices. He is given 3 slices which together weigh 1/3 of a pound, but

Ghella [55]

The three slices are each approximately 1/9 pound in weight, and, since 2/9 is less than 1/4, he can eat 2 whole slices and be okay. If he wants to eat partial slices, then he could eat 2 1/4 slices, as each slice weighs 4/36 pounds, so 2 slices would equal 8/36 pounds, leaving 1/36 pound left over in his diet, which is a quarter of the third slice.

2.25

6 0

3 years ago