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arlik [135]
3 years ago
14

A ball is thrown into the air with an upward velocity of 20 feet per second. Its height, h, in feet after t seconds is given by

the function h of t equals negative 16t squared plus 20t plus 2 . How long does it take the ball to reach its maximum height? What is the ball’s maximum height? Round to the nearest hundredth, if necessary.
Mathematics
1 answer:
musickatia [10]3 years ago
5 0
<span>height of the ball = h(t) = -16t2 + 20t + 2 The maximum value for h(t) will occur when t = -20/2(-16) = 5/8 It will take 5/8 of a second, or 0.625 seconds, for the ball to reach its maximum height. To find the ball's maximum height, plug in 5/8 for t . h( 5/8 ) = -16( 5/8 )2 + 20( 5/8 ) + 2 h( 5/8 ) = -16( 25/64 ) + 50/4 + 2 h( 5/8 ) = -25/4 + 50/4 + 8/4 h( 5/8 ) = 33/4 The maximum height is 33/4 feet, or 8.25 feet.</span>
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Answer:

C. 3 gallons

Step-by-step explanation:

30 ft × 18 ft = 540 ft²

540 ft² ÷ 250 = 2.16

2.16 is not an option so you would have to buy 3 gallons and only use 2.16 gallons.

3 0
3 years ago
A garden is in the shape of a rectangle.if the garden has a length of 4 feet and a width of 6 feet, what is the area of the gard
nlexa [21]
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3 0
3 years ago
A pilot flies on a bearing of 160° for 30 miles. The pilot then executes a quick turn and flies another 15 miles at a bearing of
vivado [14]

Answer: 34.65 miles at an angle of 325.39°

Step-by-step explanation:

Ok, the initial position is (0,0)

Then she flies 30 miles at an angle of 160°, if we count the angle counterclokwise from the x-axis, the new position will be:

p = (30*cos(120°), 30*sin(120°))

Then she travels another 15 miles at an angle of 205°, the new position is:

p = (30*cos(120°) + 15*cos(205°), 30*sin(120°) + 15*sin(205°))

p = (-28.59 , 19.64)

If she now travles X miles at an angle Y, we must have that the final position is the point (0,0)

this means that:

X*cos(Y) = -(-28.59) = 28.59

X*sin(Y) = -19.64

Now, we can find the quotient between those two equations and use that tan(x) = sin(x)/cos(x)

X*(sin(Y))/(X*cos(Y)) = -19.64/28.59

Tg(Y) = -0.69

Y = ATg(-0.69) = -34.61°

If we use only positive angles, this angle is equivalent to:

360° - 34.61° = 325.39°

now lets find the distance:

Xcos(325.39°) = 28.59

X = 28.59/cos(325.39°) = 34.65 miles.

4 0
3 years ago
Solve: 0 <img src="https://tex.z-dn.net/?f=%5Cleq" id="TexFormula1" title="\leq" alt="\leq" align="absmiddle" class="latex-formu
Salsk061 [2.6K]

\huge{ \mathrm{  \underline{ Answer  }}}࿐

  • 6 \sin(x)  - 3 \sqrt{2}  = 0

  • 6 \sin(x)  = 3 \sqrt{2}

  • \sin(x)  =  \dfrac{3 \sqrt{2} }{6}

  • \sin(x)  =  \dfrac{ \sqrt{2} }{2}

  • \sin(x)  =  \dfrac{1}{ \sqrt{2} }

\large\boxed{x = 45° \:  \: or \:  \: 135°}

\#TeeNForeveR

6 0
3 years ago
Exercise 7.3.5 The following is a Markov (migration) matrix for three locations        1 5 1 5 2 5 2 5 2 5 1 5 2 5 2 5 2
fredd [130]

Answer:

Both get the same results that is,

\left[\begin{array}{ccc}140\\160\\200\end{array}\right]

Step-by-step explanation:

Given :

\bf M=\left[\begin{array}{ccc}\frac{1}{5}&\frac{1}{5}&\frac{2}{5}\\\frac{2}{5}&\frac{2}{5}&\frac{1}{5}\\\frac{2}{5}&\frac{2}{5}&\frac{2}{5}\end{array}\right]

and initial population,

\bf P=\left[\begin{array}{ccc}130\\300\\70\end{array}\right]

a) - After two times, we will find in each position.

P_2=[P].[M]^2=[P].[M].[M]

M^2=\left[\begin{array}{ccc}\frac{1}{5}&\frac{1}{5}&\frac{2}{5}\\\frac{2}{5}&\frac{2}{5}&\frac{1}{5}\\\frac{2}{5}&\frac{2}{5}&\frac{2}{5}\end{array}\right]\times \left[\begin{array}{ccc}\frac{1}{5}&\frac{1}{5}&\frac{2}{5}\\\frac{2}{5}&\frac{2}{5}&\frac{1}{5}\\\frac{2}{5}&\frac{2}{5}&\frac{2}{5}\end{array}\right]

     =\frac{1}{25} \left[\begin{array}{ccc}7&7&7\\8&8&8\\10&10&10\end{array}\right]

\therefore\;\;\;\;\;\;\;\;\;\;\;P_2=\left[\begin{array}{ccc}7&7&7\\8&8&8\\10&10&10\end{array}\right] \times\left[\begin{array}{ccc}130\\300\\70\end{array}\right] = \left[\begin{array}{ccc}140\\160\\200\end{array}\right]

b) - With in migration process, 500 people are numbered. There will be after a long time,

After\;inifinite\;period=[M]^n.[P]

Then,\;we\;get\;the\;same\;result\;if\;we\;measure [M]^n=\frac{1}{25} \left[\begin{array}{ccc}7&7&7\\8&8&8\\10&10&10\end{array}\right]

                                   =\left[\begin{array}{ccc}140\\160\\200\end{array}\right]

4 0
3 years ago
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