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vova2212 [387]
3 years ago
7

What is 12 over three equals four over one

Mathematics
2 answers:
Phoenix [80]3 years ago
4 0
<h2><em>It's exactly what it says. 12/3 = 4. 4/1 = 4. Thus, 4 = 4. It's already solved.</em></h2>
pochemuha3 years ago
3 0

\text{Hey there!}

\mathsf{\dfrac{12}{3}=\dfrac{4}{1}\ true\ or\ false \ ?}

\text{The answer is\ \bf{TRUE}}

\mathsf{\dfrac{12\div3}{3\div3}}\\\\\mathsf{12\div3=4}\\\\\mathsf{3\div3=1}\\\\\mathsf{Which\ convets\ to: \dfrac{4}{1}}

\boxed{\boxed{\mathsf{Thus,\ this\ makes\ your\ answer\ \boxed{\mathsf{TRUE\ because\ they\ both\ equal\ 4}}}}}\checkmark

\text{Good luck on your assignment and enjoy your day!}

~\dfrac{\frak{LoveYourselfFIrst}}{:)}

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Tom Harding is a proofreader. He is paid $0.23 for every page he proofs. What is Tom's total pay for a week in which he proofrea
natka813 [3]
Alrighty, what I was always taught was to start by writing out what you know from the question.

What I know:
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Therefore, Tom made $458.16.

Hope this helps!

3 0
3 years ago
Each month, a shopkeeper spends 5x+11 dollars on rent and electricity. if he spends 2x-3 dollars on rent, how muck dose he spend
timofeeve [1]
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---------------------------------
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---------------------------------
5 0
3 years ago
Which ordered pair is represented by the red point shown?
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3 years ago
Oil is pumped continuously from a well at a rate proportional to the amount of oil left in the well. Initially there were millio
JulijaS [17]

Answer:

The amount of oil was decreasing at 69300 barrels, yearly

Step-by-step explanation:

Given

Initial =1\ million

6\ years\ later = 500,000

Required

At what rate did oil decrease when 600000 barrels remain

To do this, we make use of the following notations

t = Time

A = Amount left in the well

So:

\frac{dA}{dt} = kA

Where k represents the constant of proportionality

\frac{dA}{dt} = kA

Multiply both sides by dt/A

\frac{dA}{dt} * \frac{dt}{A} = kA * \frac{dt}{A}

\frac{dA}{A}  = k\ dt

Integrate both sides

\int\ {\frac{dA}{A}  = \int\ {k\ dt}

ln\ A = kt + lnC

Make A, the subject

A = Ce^{kt}

t = 0\ when\ A =1\ million i.e. At initial

So, we have:

A = Ce^{kt}

1000000 = Ce^{k*0}

1000000 = Ce^{0}

1000000 = C*1

1000000 = C

C =1000000

Substitute C =1000000 in A = Ce^{kt}

A = 1000000e^{kt}

To solve for k;

6\ years\ later = 500,000

i.e.

t = 6\ A = 500000

So:

500000= 1000000e^{k*6}

Divide both sides by 1000000

0.5= e^{k*6}

Take natural logarithm (ln) of both sides

ln(0.5) = ln(e^{k*6})

ln(0.5) = k*6

Solve for k

k = \frac{ln(0.5)}{6}

k = \frac{-0.693}{6}

k = -0.1155

Recall that:

\frac{dA}{dt} = kA

Where

\frac{dA}{dt} = Rate

So, when

A = 600000

The rate is:

\frac{dA}{dt} = -0.1155 * 600000

\frac{dA}{dt} = -69300

<em>Hence, the amount of oil was decreasing at 69300 barrels, yearly</em>

7 0
2 years ago
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