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SVETLANKA909090 [29]
2 years ago
14

A college is currently accepting students that are both in-state and out-of-state. They plan to accept three times as many in-st

ates students as out-of-state, and they only have space to accept 100 out-of-state students. Let x= the number of out=of=state students and y= the number in-state students. Write the constrains to represent the incoming students at the college.
Mathematics
1 answer:
ziro4ka [17]2 years ago
8 0

Answer:

0 < x ≤ 100 and 0 < y ≤ 300

Step-by-step explanation:

THIS IS THE COMPLETE QUESTION BELOW;

college is currently accepting students that are both in-state and out-of-state. They plan to accept three times as many in-state students as out-of-state, and they only have space to accept 100 out-of-state students. Let x = the number of out-of-state students and y = the number in-state students. Write the constraints to represent the incoming students at the college.

0 < x ≤ 100 and 0 < y ≤ 300

x > 0 and y > 0

0 < x ≤ 100 and y > 300

0 < x and y < 100

SOLUTION

they only have space to accept 100 out-of-state students,which means that the Maximum number of out-of-state students that can be accepted is 100

Then x= 100(Maximum number of out-of-state students that can be accepted)

They plan to accept three times as many in-state students as out-of-state which means that

Y = 3x(Maximum number of in-state students)

Then we can deduced that the numbee out-of-state students that can be accepted can lyes between the range of 0 and 100 which means from interval 0 to 100

Which can be written as 0 < x ≤ 100

But we need to know the interval for the Maximum number of in-state students(Y), to do that we need to multiply the equation above by 3 since Y = 3x

0 < x ≤ 100

3× 0 = 0

3× X = X

3× 100= 300

Then 0 < 3x≤ 300

But we know that Y = 3x then substitute into last equation

We have

0 < y ≤ 300

ThenBthe constraints to represent the incoming students at the college is

0 < x ≤ 100 and 0 < y ≤ 300

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1. S(–4, –4), P(4, –2), A(6, 6) and Z(–2, 4) a) Apply the distance formula for each side to determine whether SPAZ is equilatera
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a) SPAZ is equilateral.

b) Diagonals SA and PZ are perpendicular to each other.

c) Diagonals SA and PZ bisect each other.

Step-by-step explanation:

At first we form the triangle with the help of a graphing tool and whose result is attached below. It seems to be a paralellogram.

a) If figure is equilateral, then SP = PA = AZ = ZS:

SP = \sqrt{[4-(-4)]^{2}+[(-2)-(-4)]^{2}}

SP \approx 8.246

PA = \sqrt{(6-4)^{2}+[6-(-2)]^{2}}

PA \approx  8.246

AZ =\sqrt{(-2-6)^{2}+(4-6)^{2}}

AZ \approx 8.246

ZS = \sqrt{[-4-(-2)]^{2}+(-4-4)^{2}}

ZS \approx 8.246

Therefore, SPAZ is equilateral.

b) We use the slope formula to determine the inclination of diagonals SA and PZ:

m_{SA} = \frac{6-(-4)}{6-(-4)}

m_{SA} = 1

m_{PZ} = \frac{4-(-2)}{-2-4}

m_{PZ} = -1

Since m_{SA}\cdot m_{PZ} = -1, diagonals SA and PZ are perpendicular to each other.

c) The diagonals bisect each other if and only if both have the same midpoint. Now we proceed to determine the midpoints of each diagonal:

M_{SA} = \frac{1}{2}\cdot S(x,y) + \frac{1}{2}\cdot A(x,y)

M_{SA} = \frac{1}{2}\cdot (-4,-4)+\frac{1}{2}\cdot (6,6)

M_{SA} = (-2,-2)+(3,3)

M_{SA} = (1,1)

M_{PZ} = \frac{1}{2}\cdot P(x,y) + \frac{1}{2}\cdot Z(x,y)

M_{PZ} = \frac{1}{2}\cdot (4,-2)+\frac{1}{2}\cdot (-2,4)

M_{PZ} = (2,-1)+(-1,2)

M_{PZ} = (1,1)

Then, the diagonals SA and PZ bisect each other.

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