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Fofino [41]
3 years ago
13

Solve your equation using trail and improvement. A number plus ten times its square root is is equal to one thousand

Mathematics
1 answer:
Alex3 years ago
5 0

Answer:

x \approx 729

Step-by-step explanation:

Given

Let the number be x

So:

x + 10\sqrt x = 1000

Required

Using trial and improvement, find x

This implies that we keep guessing the values of x until we get a value that match the equation.

Let

x = 625

So, we have:

x + 10\sqrt x = 1000

625 + 10 *\sqrt{625}

Take the square root of 625

625 + 10 *\sqrt{625} = 625 + 10 *25

625 + 10 *\sqrt{625} = 625 + 250

625 + 10 *\sqrt{625} = 875

Let

x = 729

So, we have:

x + 10\sqrt x = 1000

729 + 10 * \sqrt{729}

Take the square root of 729

729 + 10 * \sqrt{729} = 729 + 10 * 27

729 + 10 * \sqrt{729} = 729 + 270

729 + 10 * \sqrt{729} = 999

At this point, we can conclude that x is approximately 729 because

999 \approx 1000

So:

x \approx 729

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PLEASE HELP the formula m=12,000+12,000rt/12t gives keri's monthly loan payment where t is the annual interest rate and t is the
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Answer:

The answer is below

Step-by-step explanation:

The formula m = (12,000 + 12,000rt)/12t gives Keri's monthly loan payment, where r is the annual interest rate and t is the length of the loan, in years. Keri decides that she can afford, at most, a $275 monthly car payment. Give an example of an interest rate greater than 0% and a loan length that would result in a car payment Keri could afford. Provide support for your answer.

Answer: Let us assume an annual interest rate (r) = 10% = 0.1. The maximum monthly payment (m) Keri can afford is $275. i.e. m ≤ $275. Using the monthly loan payment formula, we can calculate a loan length that would result in a car payment Keri could afford.

m=\frac{12000+12000rt}{12t}\\ but\ m\leq275, \ and \ r=10\%=0.1\\275= \frac{12000+12000(0.1)t}{12t}\\275= \frac{12000}{12t} +\frac{12000(0.1)}{12t}\\275= \frac{1000}{t} + 100\\275-100= \frac{1000}{t} \\175= \frac{1000}{t} \\175t = 1000\\t= \frac{1000}{175}\\ t=5.72\ years

The loan must be at least for 5.72 years for an annual interest rate (r) of 10%

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Answer:

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Answer:

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