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3 years ago

How many weeks in total are there in January, February, and March?

Mathematics

2 answers:

dalvyx [7]3 years ago

8 0

Total weeks : 12 weeks

Nuetrik [128]3 years ago

6 0

Here is the math:

January-About 5
Febuary-About 4
March-About 4
In all there is about 13 weeks between those 3 months.
~Deceptiøn

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Ian has the option of purchasing or renting a home. The purchase option requires a loan of $100,000 for a 20-year term at a 4.9%

jenyasd209 [6]

Answer:

The correct answer to the following question will be "70.56".

Step-by-step explanation:

The given values are:

Loan requires, PV = $100,000

Years = 20

Number of months, n = 240

Rate interest = 4.90000%

Monthly rate, r = 0.408333%

Monthly rental payment = $725

As we know,

$PV=PMT\times (\frac{1}{r})\times [1-[\frac{1}{(1+r)^n}]]$

On putting the values in the above formula, we get

⇒ $100000=PMT\times (\frac{1}{0.004083333})\times [1-(\frac{1}{(1+0.004083333^{240})})]$

⇒ $100000=PMT\times 152.8014557$

⇒ $PMT=\frac{100000}{152.8014557}$

⇒ $PMT=654.44$

Now,

$Saving \ Per \ Month =Rent \ per \ month-PMT$

On putting the values, we get

⇒ $=725-654.44$

⇒ $=70.56$

3 0

3 years ago

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Flauer [41]

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3 years ago

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Hoochie [10]

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3 years ago

In the given figure, ΔABC is a right triangle.

LUCKY_DIMON [66]

Answer:

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Step-by-step explanation:

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1 year ago

I NEED HELP!! please show all work

PtichkaEL [24]

Take the logarithm of both sides. The base of the logarithm doesn't matter.

$4^{5x} = 3^{x-2}$

$\implies \log 4^{5x} = \log 3^{x-2}$

Drop the exponents:

$\implies 5x \log 4 = (x-2) \log 3$

Expand the right side:

$\implies 5x \log 4 = x \log 3 - 2 \log 3$

Move the terms containing x to the left side and factor out x :

$\implies 5x \log 4 - x \log 3 = - 2 \log 3$

$\implies x (5 \log 4 - \log 3) = - 2 \log 3$

Solve for x by dividing boths ides by 5 log(4) - log(3) :

$\implies \boxed{x = -\dfrac{ 2 \log 3 }{ 5 \log 4 - \log 3 }}$

You can stop there, or continue simplifying the solution by using properties of logarithms:

$\implies x = -\dfrac{ \log 3^2 }{ \log 4^5 - \log 3 }$

$\implies x = -\dfrac{ \log 9 }{ \log 1024 - \log 3 }$

$\implies \boxed{x = -\dfrac{ \log 9 }{ \log \frac{1024}3 }}$

You can condense the solution further using the change-of-base identity,

$\implies \boxed{x = -\log_{\frac{1024}3}9}$

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2 years ago