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3 years ago

Imagine that a basketball is placed at the red free throw line (as shown above). The distance from the line to the edge of

Mathematics

2 answers:

r-ruslan [8.4K]3 years ago

5 0

Answer:

My answer is D) 12m

This is because half of the court is 14.25m and half of that is 7.1 m. That is where the free throw line is. But the basket is 3 meters tall. So.. the logical math doesn't really add up. So I would also choose C).

Thats all i can do i cant give you the exact answer but i can narrow it down 50/50. Unless i'm completely wrong. Sorry.

Step-by-step explanation:

P.S. since i didn't tell the exact answer, my response will probably be deleted soon.

Assoli18 [71]3 years ago

5 0

Answer:

A) 5 m

Step-by-step explanation:

No ball is shown in the diagram. We assume that the distance of interest is the hypotenuse of a right triangle with legs 3 m and 4 m. Your knowledge of the 3-4-5 Pythagorean triple tells you the hypotenuse is 5 m.

___

If you actually need to use the Pythagorean theorem equation, you are looking for "c" where a=4 and b=3:

c² = a² + b²

c² = 4² + 3² = 16 +9 = 25

c = √25 = 5

The straight-line distance is 5 meters.

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Answer:

Step-by-step explanation:

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3 years ago

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crimeas [40]

Step-by-step explanation:

i) $\overline{AB} = \sqrt{(x_A - x_B)^2 + (y_A - y_B)^2}$

$\:\:\:\:\:\:\:=\sqrt{(2)^2 + (12)^2} = 12.3$

ii) $m = \dfrac{y_A - y_B}{x_A - x_B} = \dfrac{-12}{2} = -6$

iii) $(\overline{x},\:\overline{y}) = \left(\dfrac{x_A + x_B}{2},\:\dfrac{y_A + y_B}{2}\right)$

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3 years ago

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Vinil7 [7]

Answer:

Hope 292.5 is right.

Step-by-step explanation:

First thing, multiply the triangles base and height which will give you 143 then divide it by two which is 71.5. Then to the parallelogram which is base times height which is 117. Moving on to the trapezoid, meanin one-hal times height times first base plus second base giving you 104. Finally, add'em up, giving you a total of 292.5.

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3 years ago

What is the solution to the system of equations?

natka813 [3]

Answer:

Step-by-step explanation:

There is an error in the question. The table does not show two linear functions. y₁ is a linear function, but y₂ is not a straight line. It makes a bend at (-6,1).

Line 1 goes through (-12,-3) and (0,5).

slope = (5-(-3))/(0-(-12)) = 2/3

y-intercept = 5

y₁ = (2/3)x + 5

Line 2 goes through (-12,-2) and (-6,1).

slope = (1-(-2))/(-6-(-12)) = 1/2

y₂ = (1/2)x + 4

(2/3)x + 5 = (1/2)x + 4

x = -6

y = (2/3)x + 5 = 1

Solution: (-6,1)

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