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seropon [69]
3 years ago
9

Graph the solution to the following system of inequalities in the coordinate plane.

Mathematics
1 answer:
Debora [2.8K]3 years ago
5 0

Answer:

The solution is any point in the common part of red and blue

The two lines intersect each other at (4 , 7)

Step-by-step explanation:

∵ y > 1/4 x + 6 ⇒ y = 1/4 x + 6

∵ y > 2x - 1 ⇒ y = 2x - 1

∴ 2x - 1 = 1/4 x + 6

∴ 2x - 1/4 x = 6 + 1

∴ 7/4 x = 7 ⇒ x = 7 ÷ 7/4

∴ x = 4

∴ y = 2(4) - 1 = 7

∴ The two lines intersect each other at (4 , 7)

∴ The solution is any point in the common part of red and blue

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I do not understand how to work out this question please help
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3 years ago
A population of 1,750 cheetahs decreases by 10% per year. How many cheetahs will there be in the population after 11 years? Roun
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The answer is C. 546.

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After 1 year:     1750 * 0.89 ≈ 1558

After 2 years:  1558 * 0.89 ≈ 1387

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After 5 years:  1098 * 0.89 ≈ 977

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Step-by-step explanation:

4 0
3 years ago
The hypotenuse of a right triangle has endpoints A(4, 1) and B(–1, –2). On a coordinate plane, line A B has points (4, 1) and (n
GarryVolchara [31]

Answer:

(-1,1),(4,-2)

Step-by-step explanation:

Given: The hypotenuse of a right triangle has endpoints A(4, 1) and B(–1, –2).

To find: coordinates of vertex of the right angle

Solution:

Let C be point (x,y)

Distance between points (x_1,y_1),(x_2,y_2) is given by \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

AC=\sqrt{(x-4)^2+(y-1)^2}\\BC=\sqrt{(x+1)^2+(y+2)^2}\\AB=\sqrt{(4+1)^2+(1+2)^2}=\sqrt{25+9}=\sqrt{34}

ΔABC is a right angled triangle, suing Pythagoras theorem (square of hypotenuse is equal to sum of squares of base and perpendicular)

34=\left [ (x-4)^2+(y-1)^2 \right ]+\left [ (x+1)^2+(y+2)^2 \right ]

Put (x,y)=(-1,1)

34=\left [ (-1-4)^2+(1-1)^2 \right ]+\left [ (-1+1)^2+(1+2)^2 \right ]\\34=25+9\\34=34

which is true. So, (-1,1) can be a vertex

Put (x,y)=(4,-2)

34=\left [ (4-4)^2+(-2-1)^2 \right ]+\left [ (4+1)^2+(-2+2)^2 \right ]\\34=9+25\\34=34

which is true. So, (4,-2) can be a vertex

Put (x,y)=(1,1)

34=\left [ (1-4)^2+(1-1)^2 \right ]+\left [ (1+1)^2+(1+2)^2 \right ]\\34=9+4+9\\34=22

which is not true. So, (1,1) cannot be a vertex

Put (x,y)=(2,-2)

34=\left [ (2-4)^2+(-2-1)^2 \right ]+\left [ (2+1)^2+(-2+2)^2 \right ]\\34=4+9+9\\34=22

which is not true. So, (2,-2) cannot be a vertex

Put (x,y)=(4,-1)

34=\left [ (4-4)^2+(-1-1)^2 \right ]+\left [ (4+1)^2+(-1+2)^2 \right ]\\34=4+25+1\\34=30

which is not true. So, (4,-1) cannot be a vertex

Put (x,y)=(-1,4)

34=\left [ (-1-4)^2+(4-1)^2 \right ]+\left [ (-1+1)^2+(4+2)^2 \right ]\\34=25+9+36\\34=70

which is not true. So, (-1,4) cannot be a vertex

So, possible points for the vertex are (-1,1),(4,-2)

7 0
3 years ago
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