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Viktor [21]
3 years ago
12

Aladdin has 12 gold coins in his magic bag. The Genie tells him that for every 100 gold coins he has in his magic bag, he will g

et 25 extra gold coins every year. How many years later will Aladdin have 21 gold coins in his bag?
Mathematics
2 answers:
son4ous [18]3 years ago
7 0

Answer:

There is not enough information to answer this question. He only has 12 in his bag so he wont be able to get the 25 coins

Step-by-step explanation:

Salsk061 [2.6K]3 years ago
5 0

Answer:

7 years.

Step-by-step explanation:

We have been given that Aladdin has 12 gold coins in his magic bag. The Genie tells him that for every 100 gold coins he has in his magic bag, he will get 25 extra gold coins every year.

First of all, we will find increment rate of gold coins by finding 25 is what percent of 100 as:

\frac{25}{100}\times 100\%=25\%

The increase in coins in one year would be 25% of 12 that is \frac{25}{100}\times 12=\frac{1}{4}\times 12=3 coins.

Now we will use proportions to solve our problem as:

\frac{\text{Year}}{\text{Coins}}=\frac{1}{3}

\frac{\text{Year}}{21}=\frac{1}{3}

\frac{\text{Year}}{21}\times 21=\frac{1}{3}\times 21

\text{Year}=7

Therefore, 7 years Aladdin will have 21 gold coins in his bag.

 

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Step-by-step explanation:

Vina has $4.70 in quarters, dimes and nickels in her purse. She has nine more dimes than quarters and four more nickels than quarters. How many of each coin are in her purse?

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4 0
2 years ago
What is the equation of the line passing through the points (4,8) and (-1,-12)
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6 0
3 years ago
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The volume V of an ice cream cone is given by V = 2 3 πR3 + 1 3 πR2h where R is the common radius of the spherical cap and the c
Nuetrik [128]

Answer:

The change in volume is estimated to be 17.20 \rm{in^3}

Step-by-step explanation:

The linearization or linear approximation of a function f(x) is given by:

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For the given function, the linearization is:

V(R_0+dR, h_0+dh) = V(R_0, h_0) + \frac{\partial V(R_0, h_0)}{\partial R}dR + \frac{\partial V(R_0, h_0)}{\partial h}dh

Taking R_0=1.5 inches and h=3 inches and evaluating the partial derivatives we obtain:

V(R_0+dR, h_0+dh) = V(R_0, h_0) + \frac{\partial V(R_0, h_0)}{\partial R}dR + \frac{\partial V(R_0, h_0)}{\partial h}dh\\V(R, h) = V(R_0, h_0) + (\frac{2 h \pi r}{3}  + 2 \pi r^2)dR + (\frac{\pi r^2}{3} )dh

substituting the values and taking dx=0.1 and dh=0.3 inches we have:

V(R_0+dR, h_0+dh) =V(R_0, h_0) + (\frac{2 h \pi r}{3}  + 2 \pi r^2)dR + (\frac{\pi r^2}{3} )dh\\V(1.5+0.1, 3+0.3) =V(1.5, 3) + (\frac{2 \cdot 3 \pi \cdot 1.5}{3}  + 2 \pi 1.5^2)\cdot 0.1 + (\frac{\pi 1.5^2}{3} )\cdot 0.3\\V(1.5+0.1, 3+0.3) = 17.2002\\\boxed{V(1.5+0.1, 3+0.3) \approx 17.20}

Therefore the change in volume is estimated to be 17.20 \rm{in^3}

4 0
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