The measure of the angle formed by 2 chords that intersect inside the circle is 1/2<span> the sum of the chords' intercepted arcs.
</span>
(83 + x)/2 = 107
83 + x = 107 * 2
83 + x = 214
x = 214 - 83
x = 131°
Answer:
x = 0
Step-by-step explanation:
Subtract 25x^2 from both sides
24x^2 + bx^2 - 25x^2 - 25x^2
Simplify
bx^2 - x^2 = 0
Factor bx^2 - x^2: x^2(b - 1)
bx^2 - x^2
Factor out common term x^2
= x^2 (b - 1)
x^2(b - 1) = 0
Using the Zero Factor Principle: If ab = 0 then a = 0 or b = 0
x^2 = 0
Apply rule x^n = 0 x = 0
x = 0
a) A line can intersect a parabola at 2 points, 1 point, or no points respectively. The only possible way at which it intersects at 1 point, would be if it were tangent to the parabola. [Check first attachment]
b) Two different parabolas can intersect at 2 points, 1 point or no points respectively. [Check second attachment]
It is possible for 2 parabolas to intersect at 4 and even 3 points. But one of the parabolas would be on it's side. You can consider those cases.
Answer:
1+1=2
Fighting! You can do this
Answer:
![v=364.5\ m^3](https://tex.z-dn.net/?f=v%3D364.5%5C%20m%5E3)
Step-by-step explanation:
<u>Volume Of A Regular Solid</u>
When a solid has a constant cross-section, the volume can be found by multiplying the area of the base by the height. The area of a trapezium is
![\displaystyle A_t=\frac{b_1+b_2}{2}h](https://tex.z-dn.net/?f=%5Cdisplaystyle%20A_t%3D%5Cfrac%7Bb_1%2Bb_2%7D%7B2%7Dh)
where
and
are the lengths of the parallel sides and h the distance between them.
The figure shows a solid with a trapezoid as the constant cross-section and a height x. The volume of the solid is
![\displaystyle v=A_t\ x](https://tex.z-dn.net/?f=%5Cdisplaystyle%20v%3DA_t%5C%20x)
![\displaystyle v=\frac{b_1+b_2}{2}\ h\ x](https://tex.z-dn.net/?f=%5Cdisplaystyle%20v%3D%5Cfrac%7Bb_1%2Bb_2%7D%7B2%7D%5C%20h%5C%20x)
The image doesn't explicitly say if the length of 4.5 is the height of the trapezium or the length of that side. We'll assume the first, so our data is:
![\displaystyle b_1=7m,\ b_2=11m,\ h=4.5m,\ x=9m](https://tex.z-dn.net/?f=%5Cdisplaystyle%20b_1%3D7m%2C%5C%20b_2%3D11m%2C%5C%20h%3D4.5m%2C%5C%20x%3D9m)
We now compute the volume
![\displaystyle v=\frac{7+11}{2}.(4.5)(9)=364.5](https://tex.z-dn.net/?f=%5Cdisplaystyle%20v%3D%5Cfrac%7B7%2B11%7D%7B2%7D.%284.5%29%289%29%3D364.5)
![\boxed{\displaystyle v=364.5\ m^3}](https://tex.z-dn.net/?f=%5Cboxed%7B%5Cdisplaystyle%20v%3D364.5%5C%20m%5E3%7D)