A water rocket is projected vertically upward from the ground. Its a height(h) m above the ground after t seconds is given by h=20-5(t-2)^2. When will the rocket reach the ground?
MARKING THE BRAINLIEST

The rocket will reach the ground 4seconds after.

Given the height reached by the rocket expressed as shown;

The rocket will reach the ground when h(t) = 0

Substitute h(t) = 0 into the expression as shown:

Expand the expression:

This means that the rocket will reach the ground 4seconds after.

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after 4 seconds

Step-by-step explanation:

Given

h = 20 - 5(t - 2)²

When rocket is on ground then h = 0

Equate to zero and solve for t

20 - 5(t - 2)² = 0 ( subtract 20 from both sides )

- 5(t - 2)² = - 20 ( divide both sides by - 5 )

(t - 2)² = 4 ( take the square root of both sides )

t - 2 = ± = ± 2 ( add 2 to both sides )

t = 2 ± 2

Thus

t = 2 - 2 = 0 ← when rocket is launched

t = 2 + 2 = 4 ← when rocket hits the ground

The rocket reaches the ground after 4 seconds

22 dimes and 9 nickels

step-by-step explanation:

< 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)