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Galina-37 [17]
2 years ago
6

Write two numbers that when rounded to the greatest place value have an estimated sum of 11,000 and an estimated difference of 3

,000. Then find the exact sum and the exact difference. Provide a detailed explanation as to how you reached your answers and why they will work.​
Mathematics
1 answer:
lianna [129]2 years ago
8 0

Answer:

x + y = 10,999.5 (rounded up to 11000)

x - y =  3,000.3 (this would be rounded down to 3000)

Step-by-step explanation:

Assume;

Two numbers are x, y

So,

x + y = 11,000 .......eq1

x - y = 3,000 .........eq2

eq1 + eq2

So,

2x =14,000

x = 7,000

So y = 4,000

For rounding number

x = 6,999.9 (rounded up to 7,000)

y = 3,999.6 (rounded up to 4,000)

Sum;

x + y = 10999.5 (rounded up to 11000)

x - y =  3000.3 (this would be rounded down to 3000)

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Answer:

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y = 2(2x - 1)

Zero solutions.

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y = 3x - 4

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One solution

3x - 4 = 2x + 2 has one solution

Step-by-step explanation:

* Lets explain how to solve the problem

- The system of equation has zero number of solution if the coefficients

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- The system of equation has infinity many solutions if the

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* Lets solve the problem

∵ y = 4x + 2 ⇒ (1)

∵ y = 2(2x - 1) ⇒ (2)

- Lets simplify equation (2) by multiplying the bracket by 2

∴ y = 4x - 2

- The two equations have same coefficient of y and x and different

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∴ They have zero equation

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y = 2(2x - 1)

Zero solutions.

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∵ y = 3x - 4 ⇒ (1)

∵ y = 2x + 2 ⇒ (2)

- The coefficients of x and y are different, then there is one solution

- Equate equations (1) and (2)

∴ 3x - 4 = 2x + 2

- Subtract 2x from both sides

∴ x - 4 = 2

- Add 4 to both sides

∴ x = 6

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∴ y = 2(6) + 2

∴ y = 12 + 2 = 14

∴ y = 14

∴ The solution is (6 , 14)

y = 3x - 4

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One solution

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So... if you notice the picture below

each circle, has their central angle at the vertex of the triangle
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so    \bf \textit{area of an equilateral triangle}\\\\
A=\cfrac{s^2\sqrt{3}}{4}\qquad s=\textit{length of one side}\\\\
-----------------------------\\\\
\textit{area of a circle}\\\\
A=\pi r^2\qquad r=radius\\\\
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\bf \textit{now, let us use the side of 6, and radius of 3}
\\\\\\

\begin{array}{clclll}
\cfrac{6^2\sqrt{3}}{4}&-&\cfrac{\pi 3^2}{2}\\
\uparrow &&\uparrow \\
triangle's&&semi-circle's
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