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3 years ago

5

Rob peeled 125 oranges in 5 hours. At that rate, how many oranges would he peel in 12 minutes?

2 answers:

marin [14]3 years ago

6 0

Answer:

$\boxed {\tt 5 \ oranges}$

Step-by-step explanation:

Let's set up a proportion.

$\frac{oranges}{minutes} =\frac{oranges}{minutes}$

He can peel 125 oranges in 5 hours. First, let's find the minutes in 5 hours.

There are 60 minutes in 1 hour, so multiply 60 and 5 (60*5=300). There are 300 minutes in 5 hours.

$\frac{125 \ oranges}{300 \ minutes}=\frac{oranges}{minutes}$

We don't know how many oranges he can peel in 12 minutes. So we say he can peel x oranges in 12 minutes.

$\frac{125 \ oranges}{300 \ minutes}=\frac{x \ oranges}{12 \ minutes}$

$\frac{125 \ }{300 \ }=\frac{x \ }{12 }$

Solve for x by isolating it. x is being divided by 12. The inverse of division is multiplication. Multiply both sides of the proportion by 12.

$12*\frac{125 \ }{300 \ }=\frac{x \ }{12 \ }*12$

$12*\frac{125 \ }{300 \ }=x$

$12*0.416666667=x$

$5=x$

Rob can peel 5 oranges in 12 minutes.

vlabodo [156]3 years ago

6 0

Answer:

5 oranges in 12 minutes

Step-by-step explanation:

$125 \div 5 = 25$

$25 \div 60 = 0.41$

$0.41 \times 12 = 5$

what i did here was I took the original amount (125), divided it by 5 which is 25 oranges per hour

1 hr = 60 mins

then I divided 25 oranges by 60 mins, then times it by 12 to get 5 around 5 oranges in 12 minutes.

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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

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