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Kipish [7]
3 years ago
8

How would I solve this The number of months in y years

Mathematics
2 answers:
Natali5045456 [20]3 years ago
8 0

Answer:

YEET

Step-by-step explanation:

Whatever y is multiply it my 12, because there are 12 months in a year.

bazaltina [42]3 years ago
5 0

Answer:

I would solve it by dividing the months by 12 then you will get the year.

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Find a pair of integers with a product of -84 and a sum of 5.
Damm [24]

Answer:

12 x -7 = -84 12 + (-7) = 5

Step-by-step explanation:

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Erin has scanned a 4 × 5-inch postcard, and she wants to reduce it to make a graphic for her Web site. She will use a scale fact
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Just multiply both dimensions by 2 to make it into centimeters.

Therefore,

4*2 x 5*2 (in inches) will be:

8cm x 10cm (in centimeters)

We conclude that the dimensions of her postcard into a graphic on her web page will be 8x10 in centimeter units. 

I hope my answer has come to your help. Thank you for posting your question here in Brainly.
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simplify negative 3 and 1 over 9 − negative 8 and 1 over 3. negative 11 and 1 over 12 negative 5 and 2 over 9 5 and 2 over 9 11
garri49 [273]
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5 0
3 years ago
A sphere with a radius of 4.8 centimeters is carved out of a right cone with a base radius of 8 centimeters and a height of 15 c
emmainna [20.7K]

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4 0
3 years ago
Read 2 more answers
The population of Henderson City was 3,381,000 in 1994, and is growing at an annual rate 1.8%
liq [111]
<h2>In the year 2000, population will be 3,762,979 approximately. Population will double by the year 2033.</h2>

Step-by-step explanation:

   Given that the population grows every year at the same rate( 1.8% ), we can model the population similar to a compound Interest problem.

   From 1994, every subsequent year the new population is obtained by multiplying the previous years' population by \frac{100+1.8}{100} = \frac{101.8}{100}.

   So, the population in the year t can be given by P(t)=3,381,000\textrm{x}(\frac{101.8}{100})^{(t-1994)}

   Population in the year 2000 = 3,381,000\textrm{x}(\frac{101.8}{100})^{6}=3,762,979.38

Population in year 2000 = 3,762,979

   Let us assume population doubles by year y.

2\textrm{x}(3,381,000)=(3,381,000)\textrm{x}(\frac{101.8}{100})^{(y-1994)}

log_{10}2=(y-1994)log_{10}(\frac{101.8}{100})

y-1994=\frac{log_{10}2}{log_{10}1.018}=38.8537

y≈2033

∴ By 2033, the population doubles.

4 0
3 years ago
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