Which graph represents the solution set of and ?

Mathematics

1 answer:

Anit [1.1K]3 years ago

8 0

Answer:

<h2>GRAPH III</h2>

Step-by-step explanation:

Look at the picture.

We have

y = x² - 4 → move the graph of x² 4 units down (the vertex is (0, -4))

x + y + 2 = 0 <em>subtracy x and 2 from both sides</em>

y = -x - 2 → move the graph of -x 2 units down (the y-intercept os -2)

Only graph III and IV meets these conditions.

The solution is the points of intersection of the graphs.

Therefore your answer is GRAPH III

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At Noise Pollution Cellular, plan A is $30.00 per month and $5.00 for every gigabyte of data. Plan B is $50.00 per month and $2.

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We need to find an equation for each situation. Let x be the number of GB of data used.

Statement 1: At Noise Pollution Cellular, plan A is $30.00 per month and $5.00 for every gigabyte of data used.

Plan A cost equation $C(x) = 30 + 5x$

Statement 2: At Noise Pollution Cellular, Plan B is $50.00 per month and $2.00 for every gigabyte of data used.

Plan B cost equation $C(x) = 50 + 2x$

Number of GB of data for which both the plan A and plan B cost are <em><u>same</u></em>.

$30+5x=50+2x \\ Subtracting \; 2x \; \& \;30\; on\; both\; sides\; \\ \\30+5x-2x-30=50-30+2x-2x \\Combining \; like \; terms\\\\3x=20\\Dividing \; like \; terms\\\\ \frac{3x}{3}=\frac{20}{3} \\ \\x=6\frac{2}{3}$

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$x=6 \\\\ Plan \; A \; Cost = 30+5(6)=30+30=\$60\\ Plan \; B \; Cost = 50+2(6)=50+12= \$72\\ \\ From \; 5\leq x\leq 6\frac{2}{3} , \; Plan \; A \; is \; Cheaper!$

When

$x = 7\\\\ Plan \; A \; Cost= 30 + 5(7) =30+35 = \$65\\\\Plan \; B \; Cost = 50 + 2(7)=50+14=\$64 \\ \\ From \; 6\frac{2}{3} \leq x \leq 8, \; Plan\; B \; is \; Cheaper !$