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GuDViN [60]
3 years ago
14

The perimeter of the square

Mathematics
1 answer:
Rama09 [41]3 years ago
4 0

Answer:

x = 20/49 units; P = 8 48/49 units

Step-by-step explanation:

All sides of a square are congruent.

That means that the two given side lengths are equal.

We set the two lengths equal and solve the equation for x.

5.5x = (1/5)(3x + 10)

Multiply both sides by 5.

27.5x = 3x + 10

Subtract 3x from both sides.

24.5x = 10

x = 10/24.5

x = 20/49

The perimeter of a square is 4s, where s is the length of a side.

P = 4s

= 4(5.5x)

= 4 * 5.5 * 20/49

= 22 * 20/49

= 440/49

= 8 48/49

Answer: x = 20/49 units; P = 8 48/49 units

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You deposit 2000 in account A, which pays 2.25% annual interest compounded monthly. You deposit another 2000 in account b, which
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To model this situation, we are going to use the compound interest formula: A=P(1+ \frac{r}{n} )^{nt}
where
A is the final amount after t years 
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t is the time in years 

For account A: 
We know for our problem that P=2000 and r= \frac{2.25}{100} =0.0225. Since the interest is compounded monthly, it is compounded 12 times per year; therefore, n=12. Lets replace those values in our formula:
A=2000(1+ \frac{0.0225}{12} )^{12t}

For account B:
P=2000, r= \frac{3}{100} =0.03, n=12. Lest replace those values in our formula:
A=2000(1+ \frac{0.03}{12} )^{12t}

Since we want to find the time, t, <span>when  the sum of the balance in both accounts is at least 5000, we need to add both accounts and set that sum equal to 5000:
</span>2000(1+ \frac{0.0225}{12} )^{12t}+2000(1+ \frac{0.03}{12} )^{12t}=5000

Now that we have our equation, we just need to solve for t:
2000[(1+ \frac{0.0225}{12} )^{12t}+(1+ \frac{0.03}{12} )^{12t}]=5000
(1+ \frac{0.0225}{12} )^{12t}+(1+ \frac{0.03}{12} )^{12t}= \frac{5000} {2000}
(1.001875)^{12t}+(1.0025 )^{12t}= \frac{5}&#10;{2}
ln(1.001875)^{12t}+ln(1.0025 )^{12t}=ln( \frac{5} {2})
12tln(1.001875)+12tln(1.0025 )=ln( \frac{5} {2})
t[12ln(1.001875)+12ln(1.0025 )]=ln( \frac{5} {2})
t= \frac{ln( \frac{5}{2} )}{12ln(1.001875)+12ln(1.0025 )}
17.47

We can conclude that after 17.47 years <span>the sum of the balance in both accounts will be at least 5000.</span>
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