A military cannon is placed at the base of a hill. The cannon is fired at an angle toward the hill. The path of the cannon ball is parabolic and can be represented by y = -0.05x² + 5x + 1, where y represents
the height of the ball in meters and x represents the horizontal distance of the ball from the cannon.
The incline of the hill can be represented by the equation y = 0.725x. How far will the cannon ball have
moved horizontal from the cannon when it hits the hill?

9514 1404 393

  85.7 meters

Step-by-step explanation:

A graphing calculator finds the result easily. The ball moved 85.7 meters horizontally before it hit the hill.

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Equating y expressions gives ...

  0.725x = -0.05x^2 +5x +1

In standard form, this is ...

  0.05x^2 -4.275x -1 = 0

The quadratic formula tells us the positive solution is ...

  x = (4.275 +√(4.275^2 -4(0.05)(-1)))/(2(0.05)) = (4.275 +√18.475625)/.1

  x ≈ 85.733

The cannon ball hits the hill at a horizontal distance of 85.7 meters from the cannon.

< xcy ≅ < xby

step-by-step explanation:

in δcax and δbax,

so, δcax ≅ δbax (side-angle-side or sas)

since δcax ≅ δbax, we can conclude < acx ≅ < abx

in δacy and δaby,

so, δacy ≅ δaby (side-angle-side or sas)

since δacy ≅ δaby, we can conclude < acy ≅ < aby

now,

< acy ≅ < aby

=> < acx + < xcy ≅ < abx + < xby

=> < acx + < xcy ≅ < acx + < xby   (since < acx ≅ < abx already proved above)

subtracting < acx from both sides, we get

< acx + < xcy -< acx ≅ < acx + < xby -< acx

cancelling out < acx and -< acx from both sides, we get

< xcy ≅ < xby (proved)

hit or miss i guess they never miss huh

step-by-step explanation:

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