H(t) = -5t^2+20t+1 what time does the ball reach the same height it was kicked at again? When does the ball reach its max height? What is the max height?

Answers

Answer 1

Answer: h(t) = -5t² + 20t + 1
-5t² + 20t + 1 = 0
t = -(20) +/- √((20)² - 4(-5)(1))
      2(-5)
t = -20 +/- √(400 + 20)
      -10
t = -20 +/- √(420)
-10
t = -20 +/- 2√(105)
-10
t = -20 + 2√(105) t = -20 - 2√(105)
   -10    -10
t = 2 - 0.2√(105) t = 2 + 0.2√(105)
h(t) = -5t² + 20t + 1
h(2 - 0.2√(105)) = -5(2 - 0.2√(105))² + 20(2 - 0.2√(105)) + 1
h(2 - 0.2√(105)) = -5(2 - 0.2√(105))(2 - 0.2√(105)) + 20(2) - 20(0.2√(105)) + 1
h(2 - 0.2√(105)) = -5(4 - 0.4√(105) - 0.4√(105) + 0.04√(11025)) + 40 - 4√(105) + 1
h(2 - 0.2√(105)) = -5(4 - 0.8√(105) + 0.04(105)) + 40 + 1 - 4√(105)
h(2 - 0.2√(105)) = -5(4 - 0.8√(105) + 4.2) + 41 - 4√(105)
h(2 - 0.2√(105)) = -5(4 + 4.2 - 0.8√(105)) + 41 - 4√(105)
h(2 - 0.2√(105)) = -5(8.2 - 0.8√(105)) + 41 - 4√(105)
h(2 - 0.2√(105)) = -5(8.2) - 5(-0.8√(105)) + 41 - 4√(105)
h(2 - 0.2√(105)) = -41 + 4√(105) + 41 - 4√(105)
h(2 - 0.2√(105)) = -41 + 41 + 4√(105) - 4√(105)
h(2 - 0.2√(105)) = 0 + 0
h(2 - 0.2√(105)) = 0
(t, h(t)) = (2 - 0.2√(105), 0)
or
h(t) = -5t² + 20t + 1
h(2 + 0.2√(105)) = -5(2 + 0.2√(105))² + 20(2 + 0.2√(105)) + 1
h(2 + 0.2√(105)) = -5(2 + 0.2√(105))(2 + 0.2√(105)) + 20(2) + 20(0.2√(105)) + 1
h(2 + 0.2√(105)) = -5(4 + 0.4√(105) + 0.4√(105) + 0.04√(11025)) + 40 + 4√(105) + 1
h(2 + 0.2√(105)) = -5(4 + 0.8√(105) + 0.04(105)) + 40 + 4√(105) + 1
h(2 + 0.2√(105)) = -5(4 + 0.8√(105) + 4.2) + 40 + 1 + 4√(105)
h(2 + 0.2√(105)) = -5(4 + 4.2 + 0.8√(105)) + 41 + 4√(105)
h(2 + 0.2√(105)) = -5(8.2 + 0.8√(105)) + 41 + 4√(105)
h(2 + 0.2√(105)) = -5(8.2) - 5(0.8√(105)) + 41 + 4√(105)
h(2 + 0.2√(105)) = -41 - 4√(105) + 41 + 4√(105)
h(2 + 0.2√(105)) = -41 + 41 - 4√(105) + 4√(105)
h(2 + 0.2√(105)) = 0 + 0
h(2 + 0.2√(105)) = 0
(t, h(t)) = (2 + 0.2√(105), 0)

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Answers

Answer:

Step-by-step explanation:

if you remove the low value outlier, the variation from the mean will decrease therefore decreasing the varianc.

Answer:

D. Removing the outlier will decrease the spread of the graph of the data set.

Step-by-step explanation:

Outliers increase the spread of the data set.

So removing an outlier will generally decrease the spread or variance or standard deviation of a data set.

The answer is

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If A, B,C are the angles of a triangle then prove: (the following in picture)Please help me to prove this.

Answers

Answer: see proof below

Step-by-step explanation:

Given: A + B + C = π → A + B = π - C

→ C = π - (A + B)

Use Sum to Product Identity: cos A + cos B = 2 cos [(A + B)/2] · cos [(A - B)/2]

Use Product to Sum Identity: 2 sin A · sin B = cos [(A + B)/2] - cos [(A - B)/2]

Use the Double Angle Identity: cos 2A = 1 - 2 sin² A

Use the Cofunction Identity: cos (π/2 - A) = sin A

Proof LHS → RHS:

LHS: cos A + cos B + cos C

= (cos A + cos B) + cos C

$\text{Sum to Product:}\qquad 2\cos \bigg((A+B)/(2)\bigg)\cdot \cos \bigg((A-B)/(2)\bigg)+\cos C$

$\text{Given:}\qquad 2\cos \bigg((\pi -C)/(2)\bigg)\cdot \cos \bigg((A-B)/(2)\bigg)+\cos C\n\n\n.\qquad \qquad =2\cos \bigg((\pi)/(2) -(C)/(2)\bigg)\cdot \cos \bigg((A-B)/(2)\bigg)+\cos C$

$\text{Cofunction:}\qquad 2\sin \bigg((C)/(2)\bigg)\cdot \cos \bigg((A-B)/(2)\bigg)+\cos C$

$\text{Double Angle:}\qquad 2\sin \bigg((C)/(2)\bigg)\cdot \cos \bigg((A-B)/(2)\bigg)+\cos\bigg(2\cdot (C)/(2)\bigg)\n\n\n.\qquad \qquad \qquad =2\sin \bigg((C)/(2)\bigg)\cdot \cos \bigg((A-B)/(2)\bigg)+1-2\sin^2 \bigg((C)/(2)\bigg)\n\n\n.\qquad \qquad \qquad =1+2\sin \bigg((C)/(2)\bigg)\cdot \cos \bigg((A-B)/(2)\bigg)-2\sin^2\bigg((C)/(2)\bigg)$

$\text{Factor:}\qquad 1+2\sin \bigg((C)/(2)\bigg)\bigg[\cos \bigg((A-B)/(2)\bigg)-\sin\bigg((C)/(2)\bigg)\bigg]$

$\text{Given:}\qquad 1+2\sin \bigg((C)/(2)\bigg)\bigg[\cos \bigg((A-B)/(2)\bigg)-\sin\bigg((\pi-(A+B))/(2)\bigg)\bigg]\n\n\n.\qquad \qquad 1+2\sin \bigg((C)/(2)\bigg)\bigg[\cos \bigg((A-B)/(2)\bigg)-\sin\bigg((\pi)/(2)-(A+B)/(2)\bigg)\bigg]$

$\text{Cofunction:}\qquad 1+2\sin \bigg((C)/(2)\bigg)\bigg[\cos \bigg((A-B)/(2)\bigg)-\cos\bigg((A+B)/(2)\bigg)\bigg]$

$\text{Product to Sum:}\qquad 1+2\sin \bigg((C)/(2)\bigg)\bigg[2\sin \bigg((A)/(2)\bigg)\cdot \sin\bigg((B)/(2)\bigg)\bigg]\n\n\n.\qquad \qquad \qquad \qquad =1+4\sin \bigg((C)/(2)\bigg)\bigg[\sin \bigg((A)/(2)\bigg)\cdot \sin\bigg((B)/(2)\bigg)\bigg]\n\n\n.\qquad \qquad \qquad \qquad =1+4\sin \bigg((A)/(2)\bigg)\sin \bigg((B)/(2)\bigg) \sin\bigg((C)/(2)\bigg)$

$\text{LHS = RHS:}\ 1+4\sin \bigg((A)/(2)\bigg)\sin \bigg((B)/(2)\bigg) \sin\bigg((C)/(2)\bigg)=1+4\sin \bigg((A)/(2)\bigg)\sin \bigg((B)/(2)\bigg) \sin\bigg((C)/(2)\bigg)\quad \checkmark$

The proof for this is simple. Let's say that A + B + C = π. From here on we require several trigonometric identities that must be applied.

$\cos \left(A\right)+\cos \left(B\right)+\cos \left(C\right) \n= 2 * cos((A + B) / 2) * cos((A - B) / 2) + \cos C \n= 2 * cos((\pi /2) - (C/2)) * cos((A - B) / 2) +\cos C \n= 2 * sin(C/2) * cos((A - B) / 2) + (1 - 2 * sin^2 (C/2)) \n= 1 + 2 sin (C/2) * cos((A - B) / 2) - sin (C/2) \n= 1 + 2 sin (C/2) * cos((A - B) / 2) - sin((\pi /2) - (A + B)/2 ))\n= 1 + 2 sin (C/2) * cos((A - B) / 2) - cos((A + B)/ 2)\n= 1 + 2 sin (C/2) * 2 sin (A/2) * sin(B/2) \n= 1 + 4 sin(A/2) sin(B/2) sin(C/2)$

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Answers

Answer:

1. 300%

2. 108

3. 60

Step-by-step explanation:

Hope this helps!

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What divides a design so that every point on one side of the line coincides with a point on the other side of it

Answers

Answer:

line of symmetry

Step-by-step explanation:

Final answer:

The term that represents a design divided so that every point on one side coincides with a point on the other side is known as symmetry. This involves an axis of symmetry dividing the design into two contrasting fields.

Explanation:

The concept in question is commonly known as symmetry, and it is often a key factor in design and visual arts. In a symmetrical design, every point on one side of the line, often referred to as the axis of symmetry, coincides with a corresponding point on the other side. This alignment creates a mirroring effect, dividing the entire design into two contrasting fields, like images divided on a poster. For instance, if you have a vertical line dividing a rectangle directly down the middle, and you have a circle on the left side of the line, a symmetrical design would have a corresponding circle on the right side.

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brainly.com/question/34365664

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What is the final result when either algebraic or numerical fractions are added together?A single fraction
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Answers

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Im a year late so this for the next person who might need it like I did Lol

Answer:

That is incorrect the answer is SINGLE FRACTION

I just did this problem.

Step-by-step explanation:

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