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jenyasd209 [6]
3 years ago
14

Name a plane parallel to DBF.

Mathematics
1 answer:
cestrela7 [59]3 years ago
5 0

Answer:

C.CAE

because those two planes never touch

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The power generated by an electrical circuit (in watts) as a function of its current ccc (in amperes) is modeled by: P(c)=-20(c-
alekssr [168]

Answer:

The current that produces maximum power is 3A

Step-by-step explanation:

Given

P(c) = 20(c - 3)^2

Required [Missing from the question]

The current that produces maximum power

First, we represent the function in standard form

P(c) = 20(c - 3)^2

P(c) = 20(c - 3)(c - 3)

Open bracket

P(c) = 20(c^2 -6c+ 9)

P(c) = 20c^2 -120c+ 180

The maximum value of c is:

Max(c) = \frac{-b}{2a}

Where:

f(x) = ax^2 + b^2 + c

By comparison: P(c) = 20c^2 -120c+ 180

a = 20

b = -120

c = 180

So, we have:

Max(c) = \frac{-b}{2a}

Max(c) = \frac{-(-120)}{2 * 20}

Max(c) = \frac{120}{40}

Max(c) = 3

5 0
3 years ago
1/9(8 1/3(21x-51y)1x-108y+2x^2)
Ann [662]

Answer:

569/ 9  x^2−153xy−12y

Step-by-step explanation:

7 0
3 years ago
Find the dimensions of the open rectangular box of maximum volume that can be made from a sheet of cardboard 11 in. by 7 in. by
11111nata11111 [884]

Answer:

The dimension of the open rectangular box is 8.216\times 4.216\times 1.392.

The volume of the box is 8.217 cubic inches.

Step-by-step explanation:

Given : The open rectangular box of maximum volume that can be made from a sheet of cardboard 11 in. by 7 in. by cutting congruent squares from the corners and folding up the sides.

To find : The dimensions and the volume of the open rectangular box ?

Solution :

Let the height be 'x'.

The length of the box is '11-2x'.

The breadth of the box is '7-2x'.

The volume of the box is V=l\times b\times h

V=(11-2x)\times (7-2x)\times x

V=4x^3-36x^2+77x

Derivate w.r.t x,

V'(x)=4(3x^2)-2(36x)+77

V'(x)=12x^2-72x+77

The critical point when V'(x)=0

12x^2-72x+77=0

Solve by quadratic formula,

x=\frac{18+\sqrt{93}}{6},\frac{18-\sqrt{93}}{6}

x=4.607,1.392

Derivate again w.r.t x,

V''(x)=24x-72

Now, V''(4.607)=24(4.607)-72=38.568>0 (+ve)

V''(1.392)=24(1.392)-72=-38.592 (-ve)

So, there is maximum at x=1.392.

The length of the box is l=11-2x

l=11-2(1.392)=8.216

The breadth of the box is b=7-2x

b=7-2(1.392)=4.216

The height of the box is h=1.392.

The dimension of the open rectangular box is 8.216\times 4.216\times 1.392.

The volume of the box is V=l\times b\times h

V=8.216\times 4.216\times 1.392

V=48.217\ in.^3

The volume of the box is 8.217 cubic inches.

5 0
3 years ago
Suppose a football is kicked with an initial velocity of 82 ft/sec., at an angle of
satela [25.4K]

Answer:

The position P is:

P = 87\^x + 75\^y ft     <u><em> Remember that the position is a vector. Observe the attached image</em></u>

Step-by-step explanation:

The equation that describes the height as a function of time of an object that moves in a parabolic trajectory with an initial velocity s_0 is:

y(t) = y_0 + s_0t -16t ^ 2

Where y_0 is the initial height = 0 for this case

We know that the initial velocity is:

82 ft/sec at an angle of 58 ° with respect to the ground.

So:

s_0 = 82sin(58\°) ft/sec

s_0 = 69.54 ft/sec

Thus

y(t) = 69.54t -16t ^ 2

The height after 2 sec is:

y(2) = 69.54 (2) -16 (2) ^ 2

y(2) = 75\ ft

Then the equation that describes the horizontal position of the ball is

X(t) = X_0 + s_0t

Where

X_ 0 = 0 for this case

s_0 = 82cos(58\°) ft / sec

s_0 = 43.45 ft/sec

So

X(t) = 43.45t

After 2 seconds the horizontal distance reached by the ball is:

X (2) = 43.45(2)\\\\X (2) = 87\ ft

Finally the vector position P is:

P = 87\^x + 75\^y ft

6 0
3 years ago
Evaluate n/6 plus 2 when n=2
harina [27]

Step-by-step explanation:

n/6+2=0

n/6+2/1=0

find the LCM=6

n+12/6=0

cross multiply

n+12=0

n=-12

5 0
3 years ago
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