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Nimfa-mama [501]
3 years ago
5

Two positive integers have their Hcf as 12 and their product as 6336. The number of possible pairs for the numbers is

Mathematics
2 answers:
oksano4ka [1.4K]3 years ago
7 0

the answer is 4? I am not sure

melamori03 [73]3 years ago
7 0

Answer:

The number of possible pairs for the number (B) 3

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A theater group made appearances into cities the hotel charge before tax and the second city was 1500 higher than the first the
Svetlanka [38]

Answer:

The hotel charge in each city before tax was <u>$5125</u> of the first city and <u>$3625</u> of the second city.

Step-by-step explanation:

Given:

A theater group made appearances into cities the hotel charge before tax and the second city was 1500 higher than the first.

The tax of the first city was 6% and the tax of the second city was 10%.

Total hotel tax paid for two cities with $670.

Now, to find the hotel charge in each city before tax.

Let the hotel charge in first city before tax be x.

And the hotel charge in second city before tax be y.

<em>So, as the hotel charge of the second city was 1500 higher than the first.</em>

<em>Thus</em>,

y=x-1500   ........(1)

<em>And as given, the tax of the first city was 6% and the tax of the second city was 10%, total hotel tax paid for two cities with $670.</em>

6% of x + 10% of y = $670.

\frac{6x}{100} +\frac{10y}{100} =670

0.06x+0.10y=670

Substituting the value of y from equation (1) we get:

0.06x+0.10(x-1500)=670

0.06x+0.10x-150=670

0.16x-150=670

<em>Adding both sides by 150 we get:</em>

0.16x=820

<em>Dividing both sides by 0.16 we get:</em>

x=5125.

<em>The hotel charge in first city before tax = $5125.</em>

Now, substituting the value of x in equation (1) we get:

y=x-1500

y=5125-1500

y=3625.

<em>The hotel charge in second city before tax = $3625.</em>

Therefore, the hotel charge in each city before tax was $5125 of the first city and $3625 of the second city.

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