All categories

3 years ago

Mary is three quarters of Cameron's age. Mary is 24 years old. How old is Cameron?

2 answers:

6 0

<em> $answer \\ 32 \: years \: old \\ solution \\ mary's \: age = 24 \\ let \: cameron's \: age \: be \: x \\ given \\ \frac{3}{4} x = 24 \\ or \: x = 24 \times \frac{4}{3} \\ x = 32 \\ hope \: it \: helps$ </em>

Bumek [7]3 years ago

3 0

Answer:

<h2>32 years old</h2>

Step-by-step explanation:

3/4=24 so 1/4= 24÷3= 8

1/4=8

So to get 4/4 or Cameron's age it is 8×4=32yrs

You might be interested in

Solve. 3(x-6)-2(2x+3)=14

coldgirl [10]

Answer:

x=-38

Step-by-step explanation:

Step 1: Simplify both sides of the equation.

3(x−6)−2(2x+3)=14

(3)(x)+(3)(−6)+(−2)(2x)+(−2)(3)=14(Distribute)

3x+−18+−4x+−6=14

(3x+−4x)+(−18+−6)=14(Combine Like Terms)

−x+−24=14

−x−24=14

Step 2: Add 24 to both sides.

−x−24+24=14+24

−x=38

Step 3: Divide both sides by -1.

−x

−1

x=−38

5 0

3 years ago

Read 2 more answers

Round 0.9998 to 3 decimal places

mojhsa [17]

The answer would be one because .9998 would round the 9 to a 10 which would round the second 9 and then the third nine to make 1

3 0

2 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

(15pts) factor each completly

Natali [406]

Your answers:

10) (m+7)(m+8) 11) (b-10)(b+2)

12) (k-4)(k+10) 13) (n-9)(n-8)

14) (7r+3)(r-2) 15) (5a+6)(a+2)

Hope it helps

5 0

3 years ago

Read 2 more answers

Which statements must be true it says check alll that apply ?

Zinaida [17]

The statements that must be true are c,e,f, but these are the statements that I think are correct

4 0

3 years ago