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Lena [83]
3 years ago
13

Which statement is true about the extreme value of the given quadratic equation? y = -3x2 + 12x − 33

Mathematics
1 answer:
Zina [86]3 years ago
8 0

Answer:

The equation has a maximum value with a y-coordinate of -21.

Step-by-step explanation:

Given

y =-3x^2 + 12x - 33

Required

The true statement about the extreme value

First, write out the leading coefficient

Leading = -3

-3 < 0 means that the function would be a downward parabola;

Downward parabola always have their vertex on top of the parabola and as such, the function has a maximum value.

The maximum value is:

x = -\frac{b}{2a}

Where:

a= -3; b =12; c =-33

So, we have:

x = -\frac{12}{2 * -3}

x = -\frac{12}{-6}

x =2

Substitute x =2 in y =-3x^2 + 12x - 33

y = -3*2^2 + 12 * 2 - 33

y = -21

<em>Hence, the maximum is -21.</em>

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An airplane flying at 600 miles per hour has a bearing of 53°. After flying for 2.5 hours, how far north and how far east will t
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Answer:

Eastward distance of airplane = 903 miles

Northward distance of airplane = 1198 miles

Step-by-step explanation:

Consider the sketch attached below.

Let the Eastward distance be represented by x

let the Northward distance be represerted by y

The distance traveled = speed of plane x time

= 600 x 2.5 = 1500 miles  

from basic trigonometry of a right angled triangle we have that

sin(53)=\frac{x}{1500}

x=1500\timessin(53) \approx  903 miles

cos (53) = \frac{y}{1500}

y= 1500cos(53)\approx1198 miles

∴ Northward distance = 1198 miles

   Eastward distance =903 miles

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Find the missing right triangle side lengths in the models of the Pythagorean thereon below
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Step-by-step explanation:

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In △ABC , GE=27 in.   What is the length of BE¯¯¯¯¯ ?
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I hope it's not to late but the answer would be 81
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Volleyball A volleyball is hit when it is 4 ft above the ground and 12 ft from a 6-ft-high net. It leaves the point of impact wi
Natali [406]

Answer:

a) the vector equation is r(x,y) = (35cos27°t, 16t^2 -35sin27° - 4)

b) Maximum height = 7.945ft

c) Time of flight = 1.201 secs and range = 37.45ft

d) when t = 0.74, the distance above the ground = 14.37ft

When t = 0.254, the distance above the ground = 26.53ft

e) if the ball is raised to 8ft, the ball won't be able to reach it. This is because the maximum height is 7.954ft

Step-by-step explanation:

a) From the diagram

x0 = 0

y0= 4

V0(initial velocity) = 35

α= 27°

y = 4 + 35sin27°t - 16t^2

y = -16t^2 + 35sin27°t+ 4

y = 16t^2 - 35sin27°t - 4

x = 35cos27°t

Vector equation for the path of the volleyball = r(x,y)

r(x,y) = (35cos27°t, 16t^2 -35sin27° - 4)

b) the ball reaches maximum when dy/dt = 0

y = 16t^2 - 35sin27°t - 4

dy/dt = 32 - 35sin27°

0 = 32 - 35sin27°t

-32 = -35sin27°t

t = -32 / -35sin27°

t = 0.4966 seconds

Put t = 0.4966 into y = 16t^2 - 35sin27°t - 4

y = 16(0.4966)^2 - 35sin27°(0.4966) - 4

y = 7.945 ft

The maximum height is 7.945 ft

c) To find the range and flight time, put y= 0

0 = 16t^2 - 35sin27°t - 4

0 = 16t^2 - 15.89t - 4

Using quadratic equation general formula,

[-b +/- √b^2 -4ac] / 2a

a = 16, b = -15.89, c = -4

= [-(-15.89) +/- √ (-15.89)^2 - 4(16)(-4)] /2(16)

= [15.89 +/- √(15.89)^2 + 256] / 32

= (15.59 +/- 22.55) / 32

= (15.89 + 22.55) / 32 or (15.89 - 22.55) / 32

= 1.201 or -0.208

Time of flight = 1.201 secs

Range = x = 35cos27°t

Range = 32cos27°(1.201)

= 37.45 ft

d) when the volley is 7ft above ground, y = 7

Recall that y = 16t^2 - 35sin27°t - 4

7 = 16t^2 - 35sin27°t - 4

0 = 16t^2 - 35sin27°t - 4 +7

0 = 16t^2 - 35sin27°t + 3

0 = 16t^2 - 15.89t + 3

Using quadratic equation general formula,

[-b +/- √b^2 -4ac] / 2a

a = 16, b = -15.89, c = 3

= [-(-15.89) +/- √ (-15.89)^2 - 4(16)(3)] /2(16)

= [15.89 +/- √(15.89)^2 - 192] / 32

= (15.59 +/- 7.7778) / 32

= (15.89 + 7.7778) / 32 or (15.89 - 7.7778) / 32

= 0.74 or 0.254

When t = 0.74,

x = 35cos27°t

x = 35cos27°(0.74)

x = 23.08 ft

Therefore , R - x(0.74)

= 37.45 - 23.08

= 14.37ft

When t = 0.254,

x = 35cos27°t

x = 35cos27°(0.254)

x = 7.92 ft

Therefore R - x(0.254) =

37.45 - 7.92 = 29.53ft

e) if the ball is raised to 8ft, the ball won't be able to reach it. This is because the maximum height is 7.954ft

7 0
3 years ago
What is 2+2-2X2 divided by 2??
Free_Kalibri [48]

Answer:

0

Step-by-step explanation:

cause 2+2 is 4 and 2x2 is four and since u put - it take it away a then divide 0 by 2 u get 0

4 0
3 years ago
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