What is the solution set for 2x+5y>-1 and 4x-3<-3?
1 answer:
PROBLEM ONE
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Solving for x in 2x + 5y > -1.
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Step 1 ) Subtract 5y from both sides.
2x + 5y > -1
2x + 5y - 5y > -1 - 5y
2x > -1 - 5y
Step 2 ) Divide both sides by 2.
2x > -1 - 5y
So, the solution for x in 2x + 5y > -1 is...
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Solving for y in 2x + 5y > -1.
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Step 1 ) Subtract 2x from both sides.
2x + 5y > -1
2x - 2x + 5y > -1 - 2x
5y > -1 - 1x
Step 2 ) Divide both sides by 5.
5y > -1 - 1x
So, the solution for y in 2x + 5y > -1 is...
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PROBLEM TWO
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Solving for x in 4x - 3 < -3.
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Step 1 ) Subtract 3 from both sides.
4x - 3 < -3
4x -3 - 3 < -3 - 3
4x < 0
Step 2 ) Divide both sides by x.
4x < 0
x < 0
So, the solution for x in 4x - 3 < -3 is...
x < 0
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- <em>Marlon Nunez</em>
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Answer:
option A.) y-3 = 1/16 (x-2)^2
Step-by-step explanation:
we know that
If the axis of symmetry is parallel to the y-axis, then we have a vertical parabola
The equation of a vertical parabola is equal to
where
(h,k) is the vertex
In this problem we have
(h,k)=(2,3)
p=4
substitute
<h3>Answer: D) 1692 miles</h3>
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Explanation:
A diagram is very handy for this type of problem. Refer to the diagram below (attached image). Here are the steps used to form the diagram
- Plot point A and make that the home base. Plot point D directly south of point A. I'm skipping B and C because those will be used to form the triangle later.
- Imagine you are standing at point A and looking south at point D. Turn 46 degrees toward the east direction (this is what the notation "bearing of S 46° E" means). Then move 550 miles from A to land on point B. This is the location where the plane makes its first turn.
- Plot the points E and F, which are north and south of point B.
- Imagine we are at point B and looking directly south at point F. Turn 55 degrees toward the west, which is due to the "S 55° W" and then move 483 miles to arrive at point C. This is the plane's location where it turns around to go back to home base.
The diagram below shows all of this visually summarized. We have the segments
- AB = 550 miles
- BC = 483 miles
- AC = x miles
And we have these angles
- Angle DAB = 46 degrees and Angle ABE = 46 degrees (blue)
- Angle ABC = 79 degrees (red)
- Angle ACB = 55 degrees and Angle FBC = 55 degrees (green)
Angle ABC comes from the fact that the blue and green angles add with the red angle to get 180. So we basically need to solve 46+y+55 = 180 which leads to y = 79.
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Once the diagram is set up, we'll use the law of cosines to find x.
Focus on triangle ABC. Ignore points D,E,F. Ignore any extra lines as well.
The interior angles of triangle ABC are
- A = 46 degrees
- B = 79 degrees
- C = 55 degrees
The sides opposite the angles are
- a = 483 (opposite angle A)
- b = x (opposite angle B)
- c = 550 (opposite angle C)
Now we can apply the law of cosines
b^2 = a^2 + c^2 - 2*a*c*cos(B)
x^2 = 483^2 + 550^2 - 2*483*550*cos(79)
x^2 = 434412.180756441
x = sqrt(434412.180756441)
x = 659.099522649229
x = 659
Side AC, or CA, is roughly 659 miles long.
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To find the total distance the plane travels, we add up the three sides of the triangle (ie we find the perimeter)
Distance Traveled = AB + BC + CA
Distance Traveled = 550 + 483 + 659
Distance Traveled = 1692 miles