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4 years ago

What fraction makes the equation true? Enter the answer in the boxes. 3/5____=15/25

Mathematics

2 answers:

Yuki888 [10]4 years ago

8 0

Answer:

times 5

Step-by-step explanation:

Brums [2.3K]4 years ago

6 0

Answer:

5/5

Step-by-step explanation:

If you multiply the top by 5 you get 15 and you multiply the bottom by 5 you get 25.

Hope this helps.

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A BASKETBALL PLAYER MAKES A BASKET THROUGH A TEN FOOT HIGH BASKET FROM 20 FEET AWAY. IF THE SHOOTER RELEASES THE BALL FROM 8 FEE

mixas84 [53]

Answer:

The vertical velocity of the ball when released = 10.81 m/s

The vertical velocity of the ball in the basket = 8.81 m/s

Step-by-step explanation:

The given parameters are;

The height of the basket = 10 feet

The horizontal location of the shooter from the basket = 20 feet

The height from which the shooter releases the ball = 8 feet

The horizontal velocity of the ball = 10 ft/s

Therefore, we have;

Time, t = Horizontal distance/(Horizontal velocity)

t = 20/10 = 2 seconds

The vertical velocity is given as follows;

s = u·t - 1/2·g·t²

Where;

u = The vertical velocity

s = The height above the initial height from which the ball is released

s = y - y₀

Where;

y = The final height of the ball in the calculation = The height of the basket = 10 foot

y₀ = The initial height from which the ball is released = 8 feet

∴ s = 10 - 8 = 2 feet

g = The acceleration due to gravity = 9.81 m/s²

Substituting gives;

2 = u × 2 - 1/2 × 9.81× 2²

u = (2 + 1/2 × 9.81× 2²)/2 = 10.81 m/s

The vertical velocity of the ball when released = 10.81 m/s.

The vertical velocity of the ball in the basket is given by the following relation;

v² = u² - 2·g·s

v² = 10.81² - 2×9.81×2 = 77.6161

v = √(77.6161) = 8.81 m/s

The vertical velocity of the ball in the basket = 8.81 m/s

7 0

3 years ago

The area of the triangle is eqaula to the sqare show that x²-3x-2=0

agasfer [191]

Answer:

Step-by-step explanation:

Area of square = area of triangle

$side*side = \dfrac{1}{2}b*h\\\\\\x*x=\dfrac{1}{2}*(x+1)(x+2)\\x^{2}=\dfrac{1}{2}[x*x +x*2+1*x+1*2]\\\\\\x^{2}=\dfrac{1}{2}[x^{2}+2x+x+2]\\\\\\x^{2}=\dfrac{1}{2}[x^{2}+3x+2]\\\\\\Multiply \ both \ sides \ by \ 2\\\\2x^{2}=x^{2}+3x+2\\\\2x^{2}-x^{2}-3x-2 = 0$