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maria [59]
1 year ago
9

This question has two parts. Be sure to answer both parts of the question. One inequality in a system is shown below. -12x + 3y

> 9 PART A. Create another inequality so the system has no solution.PART B. Create another inequality so the system has infinite solutions.
Mathematics
1 answer:
fgiga [73]1 year ago
7 0

The inequality is -12x + 3y > 9.

PART A:

The sytem has no solution if inequality does not share a common area. The inequality -12x + 3y > 9 consist the region to left of line -12x + 3y = 9. So for no solution the region to left of equation -12x + 3y = 9 is suitable.

Thus inequality for no solution is, -12x + 3y < 9.

PART B:

For infinite solution the region of both inequality must overlap each other, or the inequality is same with some multiplication of divison factor. So inequality for infinite many solutions is,

\begin{gathered} (-12x+3y>9)\times-1 \\ 12x-3y

Thus inequality for infinite many solution is 12x - 3y < -9.

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One light-year equals 5.9 x 1012 miles. How many light-years are in 6.79
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Option B

The number of light years in 6.79 \times 10^{16} miles is 11508 light years

<em><u>Solution:</u></em>

Given that,

One light-year equals 5.9 x 10^12 miles

Therefore,

1 \text{ light year } = 5.9 \times 10^{12} \text{ miles }

To find: Number of light years in 6.79 \times 10^{16} miles

Let "x" be the number of light years in 6.79 \times 10^{16} miles

Then number of light years in 6.79 \times 10^{16} miles can be found by dividing 6.79 \times 10^{16} miles by miles in 1 light year

\text{Number of light years in } 6.79 \times 10^{16} miles = \frac{6.79 \times 10^{16}}{5.9 \times 10^{12}}\\\\\text{Use the law of exponent }\\\\\frac{a^m}{a^n} = a^{m-n}\\\\\text{Number of light years in } 6.79 \times 10^{16} miles = \frac{6.79}{5.9} \times 10^{16-12}\\\\\text{Number of light years in } 6.79 \times 10^{16} miles = 1.1508 \times 10^4\\\\\text{Number of light years in } 6.79 \times 10^{16} miles = 11508

Thus number of light years in 6.79 \times 10^{16} miles is 11508 light years

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