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pav-90 [236]
2 years ago
7

solve the problem.a ball is kicked upward with an initial velocity of 68 feet per second. the ball's height, h (in feet), from t

he ground is modeled by h
Mathematics
1 answer:
joja [24]2 years ago
3 0

The practical domain of time is t ∈ [0, 4.25s].

  • The set of values that make feel in a particular context is referred to as the sensible area.
  • All feasible solutions are addressed by way of theoretical domains and stages.
  • domain names and stages restrict the solution sets such that they may be realistic in the parameters given
  • .The ball begins its motion at time t = 0s.
  • To locate the time while the ball hits the floor,
  • we solve the equation:
  • h(t) = -16t² + 68t
  • h(t) = 0
  • -16t² + 68t = 0
  • t(-16t + 68) = 0
  • t = 0 and -16t + 68 = 0.
  • 16t = 68
  • t = 68/16
  • t = 4.25
  • h(t) is non-negative within the interval [0, 4.25].
  • Hence, the practical domain is 0 ≤ t ≤ 4.25 s.

To learn more about domain, visit :

brainly.com/question/28135761

#SPJ4

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Consider parallelogram VWXY below.
olchik [2.2K]

note:Parallelogram WXYZ not VWXY & I will answer the questions in the image b'cos I don't get your question.

Answers:

a.(i.)Slope of line YZ = 4/5

(ii.)Line adjacent line YZ is line YX,slope of YX =5/4

b.(i.)Length of line YZ =√41 units

 (ii.)Length of line YX =√41 units

c.WXYZ is a rhombus.

Step-by-step explanation:

a.Slope=gradient(m)=(y₂-y₁)/(x₂-x₁)

(i.)m of line YZ=(y₂-y₁)/(x₂-x₁) where Y(-6,-1) & Z(-1,3)

(I believe you know what the y₂,y₁,x₂ & x₁ stand for.)

So m=(y₂-y₁)/(x₂-x₁),=(3-(-1))/(-1-(-6) =(4)/(5)=4/5

∴m of line YZ is 4/5

(ii.)m of line YX=(y₂-y₁)/(x₂-x₁) where Y(-6,-1) & X(-2,4)

So m=(y₂-y₁)/(x₂-x₁),=(4-(-1))/(-2-(-6) =(5)/(4)=5/4=1.25

∴m of line YX is 5/4

b.(i.)To find the length of line YZ,we have to express it as column vector YZ.

Vector YZ=OZ-OY,Z-Y=(-1,3)-(-6,-1)=(5,4)

Length=√x² plus y²=√5² plus 4²=√25 plus 16=√41 units

∴The length of line YZ is √41 units.

(ii.)To find the length of line YX,we have to express it as column vector YZ.

Vector YX=OX-OY,X-Y=(-2,4)-(-6,-1)=(4,5)

Length=√x² plus y²=√4² plus 5²=√16 plus 25=√41 units

∴The length of line YX is √41 units.

c.WXYZ is a rhombus.

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