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BartSMP [9]
2 years ago
15

Please help! I cannot figure this out, I have no idea what to do.

Mathematics
1 answer:
joja [24]2 years ago
6 0

3. First you should identify which line corresponds to which equation.

x + 2y = -8   ⇒   y = -x/2 - 4

is the line with slope -1/2 and intercept (0, -4), so the first inequality refers to the lower line in the graph.

For the inequality itself, the solution set that satisfies it is either the region above or below it. To decide which, pick any point in the plane and plug its coordinates into the inequality. The origin is a natural choice, since it's not on the line and working with 0s is easy.

In the first inequality, we have

x = 0 and y = 0   ⇒   0 + 2•0 = 0 < -8

which is not true. So (0, 0) is not in the solution set, and it stands to reason that any point in the same region will not be a solution. In other words, the region above the line x + 2y = -8 does not satisfy the inequality, while the one below it does.

In the second inequality,

x = 0 and y = 0   ⇒   2•0 + 0 = 0 ≥ -1

which is true. This means the region containing (0, 0) above the upper line solves the second inequality.

The solution to the system of inequalities is then the intersection of these two regions. (See attached; I've labeled the solution to the first inequality in blue, the solution to the second one in red, and the solution to both in green.)

All this work is kind of overkill for this particular question, though. All you really need to do is check if the given point satisfies the inequalities:

• (-5, 5) :

x = -5 and y = 5   ⇒   -5 + 2•5 = 5 < -8   ⇒   NO

• (2, -5) :

x = 2 and y = -5   ⇒   -2 + 2•5 = 8 < -8   ⇒   NO

• (-4, -6) :

x = -4 and y = -6   ⇒   -4 + 2•(-6) = -16 < -8   ⇒   MAYBE

x = -4 and y = -6   ⇒   2•(-4) + (-6) = -14 < -8   ⇒   YES

• (5, -7) :

x = 5 and y = -7   ⇒   5 + 2•(-7) = -9 < -8   ⇒   MAYBE

x = 5 and y = -7   ⇒   2•5 + (-7) = 3 < -8   ⇒   NO

4. x is the number of candy boxes with chocolates and y is the number of boxes without chocolates. Each of the x boxes cost $27.50, so if you have x boxes, their total cost is $27.5x. Each of the y boxes cost $25.00, so y boxes cost $25y. The sponsor doesn't want to order more than 100 boxes total, so

x + y < 100

but wants to raise at least $2000, so

27.5x + 25y ≤ 2000

Now do the same thing as before; plug in the listed x- and y-coordinates and pick the point that satisfies both inequalities. You will find that (50, 30) is the correct choice.

5. Consult the "overkill" part of problem 3. The line y = -3x + 6 has a negative slope, so it's the downward sloping one. Check if the origin satisfies the inequality:

x = 0 and y = 0   ⇒   0 ≤ -3•0 + 6   ⇒   0 ≤ 6   ⇒   YES

This means Regions II and III solve the first inequality.

Do the same with the other inequality:

x = 0 and y = 0   ⇒   0 ≥ 1/2•0 + 1   ⇒   0 ≥ 1   ⇒   NO

This tells us that Regions I and II solve the second inequality.

The intersection of these regions is of course Region II.

6. One of the plotted lines is apparently y = x + 4, which has a positive slope, so this must be the upper line. The shaded region below it corresponds to the solution of either y < x + 4 or y ≤ x + 4 and we eliminate B.

The other line has negative slope, which eliminates A and D since

3x - 4y = 20   ⇒   y = 3/4x - 5

has positive slope. This leaves D.

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SOVA2 [1]
A linear equation would be the best fit, but the last point (-1,-7) kinda messes it up. If the -7 would have been a -6 the line y=-2x-8 would fit perfectly.
7 0
3 years ago
Round 4.675 to the nearest cent
Viktor [21]
Money is written out to the hundredths place, so when rounding to the nearest cent, round to the hundredths place. In 4.675, the 7 is in the hundredths place. The value to its right, 5, is greater than 4, so the 7 is rounded up to 8. So 4.675 to the nearest cent is 4.68
3 0
3 years ago
A computer training institute has 625 students that are paying a course fee of $400. Their research shows that for every $20 red
Anastasy [175]
1.
If no changes are made, the school has a revenue of :

625*400$/student=250,000$

2.
Assume that the school decides to reduce n*20$.

This means that there will be an increase of 50n students.

Thus there are 625 + 50n students, each paying 400-20n dollars.

The revenue is: 

(625 + 50n)*(400-20n)=12.5(50+n)*20(20-n)=250(n+50)(20-n)

3.

check the options that we have, 

a fee of $380 means that n=1, thus 

250(n+50)(20-n)=250(1+50)(20-1)=242,250   ($)


a fee of $320 means that n=4, thus

 250(n+50)(20-n)=250(4+50)(20-4)=216,000    ($)


the other options cannot be considered since neither 400-275, nor 400-325 are multiples of 20.

Conclusion, neither of the possible choices should be applied, since they will reduce the revenue.
6 0
3 years ago
Joexreader where are you Joexreader​
Anit [1.1K]

Answer:

this seems like an interesting story

4 0
3 years ago
Read 2 more answers
The both that are not in green I don't know how to do them . I need some help.
attashe74 [19]

Answer:

Step-by-step explanation:

B. 0.00604

D.1.47 ×10^{-3}

3 0
3 years ago
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