PLEASE HELP MEEE Point J is located at (-2, -5) and Point K is located 5 units right and 9 units up Which measurement is closest to the distance between Point J and Point K?
A. 3.7 units
B. 7.5 units
C. 10.3 units
D. 12 units

This value is approximate. This also means that segment JK is roughly 10.3 units.

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Explanation:

The "5 units right and 9 units up" means we have a right triangle with legs 5 and 9. The hypotenuse is unknown, which I'll call c.

Apply the pythagorean theorem. Solve for c.

a^2 + b^2 = c^2

5^2 + 9^2 = c^2

25 + 81 = c^2

106 = c^2

c^2 = 106

c = sqrt(106)

c = 10.2956301 approximately

c = 10.3 approximately

The distance from J to K is roughly 10.3 units

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A slightly different method:

Start at point J (-2, -5) and move 5 units to the right and 9 units up to arrive at K(3, 4)

Let's use the distance formula to find the distance from (-2,-5) to (3,4)

Define the following

Those values are then plugged into the distance formula

The distance from J to K is roughly 10.3 units

One thing to point out, as you probably already noticed, is that the distance formula and pythagorean theorem effectively are saying the same thing when computing distances like this. The distance formula is based off the pythagorean theorem. Note the x1-x2 portion which represents the horizontal distance from J to K. This is the "move 5 units right" portion. Then the "y1-y2" is the vertical change in distance, which is the "move 9 units up". Also, we can see that the square root of 106 shows up again; a lot of elements are repeated (just phrased in a slightly different way) from the last section.

x=17

step-by-step explanation:

solve 3x-24=x+10

then that makes it 2x=34 because you subtract an x from both sides to get rid of the one on the right, and then you add 24 to both sides to get rid of the 24 on the left

divide by both sides by 2 to get your answer so that you can get your answer

answer: domain and range mean all possible values of x and y that could be substituted to the inequality. a solution means all possible values that make the inequality statement true. so while the range is still all real numbers, the solution is the shaded area, for which y will always be greater than the line of the function.