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pychu [463]
3 years ago
11

If you were to translate the coordinate point C using vector(-2,-4), what would the coordinates of the new point be?

Mathematics
1 answer:
uysha [10]3 years ago
6 0

Given:

The point is C(0,3).

The translation vector is \left.

To find:

The coordinates of the new points after translation of point C.

Solution:

If a figure is translated by a vector \left, then

(x,y)\to (x+a,y+b)

The figure is translated by the vector \left, then

(x,y)\to (x+(-2),y+(-4))

(x,y)\to (x-2,y-4)

The point C is at (0,3).

C(0,3)\to C'(0-2,3-4)

C(0,3)\to C'(-2,-1)

Therefore, the coordinates of the new point are C'(-2,-1).

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The perimeter of the triangle is 6a + 3 units. Write an expression in simplest form for the length of Side 3.

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The length of the third side is a-2.

Given that:

Perimeter of triangle: 6a + 3

Length of first side: 2(a+3)

Length of second side: 3a -1

Let the length of the third side of the triangle be x.

Then as perimeter of a triangle is sum of lengths of all 3 sides of the triangle, thus we have:

Thus we have length of the third side as a -2.

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3 years ago
1A. Write the fully expanded expression for 7!
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The number of cents per kilometer it costs to drive a car depends on how fast you drive it. At low speeds the cost is high becau
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Which expression is equivalent to 3x+5+7x+2
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3 years ago
An airline flying to a Midwest destination can sell 20 coach-class tickets per day at a price of $250 and ten business-class tic
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Solution :

Let x = number of coach class tickets sold per day on $ 10 reduced price

y = number of the business class tickets sold per day on $ 50 reduced per price.

Number of the coach class tickets sold = 20 per day at a price of $ 250

Number of the business class tickets sold = 10 per day at a price of $ 750

Decrease in the price of coach class tickets by $ 10 increases the number of tickets sold by 4 per day

Decrease in the price of business class tickets by $ 50 increases the number of tickets sold by 2 per day.

Therefore, the coach price, p(x) = 250 - 10x

Business class price, p(y) = 750 - 50y

Coach tickets, q(x) = 2 + 4x

Business class tickets, q(y) = 6 + 2y  

Revenue R(x, y) = p(x) q(x) + q(y) q(y)

R(x, y) = (250-10x)(20+4x)+(750-50y)(6+2y)

R(x) = (250-10x)(0.4)+(20+4x)(-10)+0 = 0

1000-40x-200-40x=0

80x = 800

x = 10

R(y) = 0 + (750 - 50y)(0.2) + (6+2y)(-50) = 0

1500 - 100y+(-300-100y) = 0

200 y = 1200

y = 6

P(x) = 250 - 10(10) = $ 150

P(y) = 780-50(6) = $450

q(x) = 20 + 4(10) = 60

q(y) = 6 + 2(6) = 18

Maximum revenue = 150 x 60 + 450 x 18

                              = 9000 + 8100

                              = $ 17100

5 0
3 years ago
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