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san4es73 [151]
3 years ago
13

A sales associate at a computer store receives a bonus of $100 for every computer he sells. He wants to make $2,500 in bonuses n

ext month. Write and solve an inequality to find the minimum number of computers he must sell.

Mathematics
2 answers:
Lostsunrise [7]3 years ago
8 0
$100x should be greater than or equal to $2500. X being the number of computers sold. If you divide 100 by 2500 you will know how many computers he need to sell.
$100x >(or equal to) $2500
rusak2 [61]3 years ago
5 0

Answer:

The inequality that describes this problem is 100x \ge 2500, and the minimum number of computers he must sell is 25.

Step-by-step explanation:

Since there is no restriction on the number of computers he must sell, an inequality can help us describe the situation.

Writing the inequality.

Let x be the number of computers he sells the next month, since he receives a bonus of $100 per computer he sells the total amount he will make for x computers is

100x

And since he wants to make $2,500 in bonuses, the inequality that describes that situation is

100x \ge 2500

Solving the inequality.

In order to find the exact number of computers he must sell, we need to solve for the inequality.

100x \ge 2500

Since 100 is multiplying to x, in order to move that 100 to the other side, we need to apply the inverse operation to multiplication, which is division. We can then divide both sides by 100

\cfrac{100x}{100} \ge \cfrac{2500}{100}

So then we can simplify to get

x \ge 25

Thus the minimum number of computers he must sell is 25.

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3 years ago
Please help having trouble!
zmey [24]

Answer:

Option C is the correct choice that is y

Step-by-step explanation:

As this is a multiple choice question we will reduce the options and work on it with the given points (0,-2),(-2,-3)

Note:We know that \leq ,\geq where there is = sign associated with it have a straight line graph there is no breaking in the line.

And when there is simply we have a dashed line when we plot it on a graph.

So option B and D are discarded.

Now one-by one we will put the values (x,y)\ (-2,3) to know which equation it satisfies.

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