Step-by-step explanation:
One gallon = 25 miles
8 gallon= 25×8 miles= 200 miles
Answer:
Step-by-step explanation:
2 cos x + √ 2 = 0
2 cos x = -√ 2
cos x = -√ 2 / 2
x = arcCos( -√ 2 / 2 )
so to solve we have to use "co-terminal " angles .. do you know what I'm saying? do you understand the words coming out of my mouth :DDDDD OKay back to math and not movie lines .. :P
x = arcCos( √ 2 / 2 )
x = 45 °
now find the "co terminal" angle that is on 45 ° but in the correct quadrant... since the -√ 2 is negative.. we now that we go down the y axis.. but also positive on the x axis.. soooo.. that put the angle in the 4th quadrant... so this is an angle of 315° if we go in the CCW ( counter clock wise ) direction but it's also -45° in the CW (clock wise ) direction
below is the table to remember the trig special angles
notice how it's 1,2,3,4 .. so it's super easy to remember.. the trig books don't show you this "trick" :P
copy and paste this to your computer some where handy
Sin(0) = 0/2 =0
Sin(30)= 
/2 = 1/2
Sing(45) =
/2 =/2
Sin(60)=
/2 = /2
Sin(90)=
/2 = 1
Cos is exactly the same but counts backwards from 90°
Cos(90) = 0/2 = 0
Cos(60) = /2 = 1/2
Cos(45) = /2 =/2
Cos(30) = /2 =/2
Cos(0) = /2 = 1
6/12 or 1/2 I hope this was the right answer and helped you
Answer:
<u><em>The area of the irregular figure is 140 sq. ft.</em></u>
Step-by-step explanation:
<em>Triangle:
</em>
<em>
</em>
<em>16-12=4
</em>
<em>
</em>
<em>6 x 4 = 24
</em>
<em>
</em>
<em>24 x 0.5 = 12
</em>
<em>
</em>
<em>Triangle = 12
</em>
<em>
</em>
<em>Rectangle:
</em>
<em>
</em>
<em>16 x 8
</em>
<em>
</em>
<em>= 128
</em>
<em>
</em>
<em>Rectangle = 128
</em>
<em>
</em>
<em>Total figure:
</em>
<em>
</em>
<em>128+12
</em>
<em>
</em>
<em>= 140
</em>
<u><em>Plzz give me brainlist!!!</em></u><u>
</u>
<span>12.3
Volume function: v(x) = ((18-x)(x-1)^2)/(4pi)
Since the perimeter of the piece of sheet metal is 36, the height of the tube created will be 36/2 - x = 18-x.
The volume of the tube will be the area of the cross section multiplied by the height. The area of the cross section will be pi r^2 and r will be (x-1)/(2pi). So the volume of the tube is
v(x) = (18-x)pi((x-1)/(2pi))^2
v(x) = (18-x)pi((x-1)^2/(4pi^2))
v(x) = ((18-x)(x-1)^2)/(4pi)
The maximum volume will happen when the value of the first derivative is zero. So calculate the first derivative:
v'(x) = (x-1)(3x - 37) / (4pi)
Convert to quadratic equation.
(3x^2 - 40x + 37)/(4pi) = 0
3/(4pi)x^2 - (10/pi)x + 37/(4pi) = 0
Now calculate the roots using the quadratic formula with a = 3/(4pi), b = -10/pi, and c = 37/(4pi)
The roots occur at x = 1 and x = 12 1/3. There are the points where the slope of the volume equation is zero. The root of 1 happens just as the volume of the tube is 0. So the root of 12 1/3 is the value you want where the volume of the tube is maximized. So the answer to the nearest tenth is 12.3</span>
