Answer:
ASA theorum
Step-by-step explanation:
we are given 1 side and 1 angle, other angle is vo of it corresponding equal angle
It's 28. here's proof 1/4 is 7, 1/2 is 14, 3/4 is 21, and 4/4 is 28.
General Idea:
The volume of cylinder is given by
, where r is the radius and h is the height of the cylinder.
Applying the concept:
Step 1: We need to find the volume of full cylinder with the given dimensions using the formula.
Volume of full cylinder ![=\pi r^{2} h=\pi* 8^{2} *9=576\pi cm^{3}](https://tex.z-dn.net/?f=%20%3D%5Cpi%20r%5E%7B2%7D%20h%3D%5Cpi%2A%208%5E%7B2%7D%20%2A9%3D576%5Cpi%20cm%5E%7B3%7D%20)
Volume of half cylinder ![=\frac{576\pi }{2} =288\pi cm^{3}](https://tex.z-dn.net/?f=%20%3D%5Cfrac%7B576%5Cpi%20%7D%7B2%7D%20%3D288%5Cpi%20cm%5E%7B3%7D%20%20)
Step 2: Let x be the number of minutes of filling the sand.
of sand filled every 15 seconds, there are four 15 seconds in a minute.
So volume of sand filled in 1 minute
.
of sand taken out of cylindrical vase every minute.
Net volume of sand filled in 1 minute = Volume of sand filled in the vase in one minute - Volume of sand taken out in 1 minute
Net volume of sand filled in 1 minute![=24 cm^{3} - 8cm^{3}=16cm^{3}](https://tex.z-dn.net/?f=%20%3D24%20cm%5E%7B3%7D%20-%208cm%5E%7B3%7D%3D16cm%5E%7B3%7D%20)
Volume of sand filled in x minutes
.
We need to set up an equation to find the number of minutes need to fill half the volume in cylindrical vase.
![16x = 288\pi \\ \\ x=\frac{288\pi}{16} \\ \\ x=18\pi \\ \\ x=57 minutes](https://tex.z-dn.net/?f=%2016x%20%3D%20288%5Cpi%20%5C%5C%20%5C%5C%20x%3D%5Cfrac%7B288%5Cpi%7D%7B16%7D%20%20%5C%5C%20%5C%5C%20x%3D18%5Cpi%20%5C%5C%20%5C%5C%20x%3D57%20minutes%20)
Conclusion:
The number of minutes required for the base be half filled with sand is 57
Standard form for the hyperbola with vertices ( – 12,0) and (12,0), and foci ( – 13,0) and (13,0) is ![\frac{x^{2} }{144}-\frac{y^{2} }{25}=1](https://tex.z-dn.net/?f=%5Cfrac%7Bx%5E%7B2%7D%20%7D%7B144%7D-%5Cfrac%7By%5E%7B2%7D%20%7D%7B25%7D%3D1)
Hyperbola is a plane curve generated by a point so moving that the difference of the distances from two fixed points is a constant.
Given,
The Vertices of the hyperbola = (-12,0) and (12,0)
Foci = (-13,0) and (13,0)
a=12
ac=13
c=![\frac{13}{12}](https://tex.z-dn.net/?f=%5Cfrac%7B13%7D%7B12%7D)
We know,
c=![\sqrt{1+\frac{b^{2} }{a^{2} } }](https://tex.z-dn.net/?f=%5Csqrt%7B1%2B%5Cfrac%7Bb%5E%7B2%7D%20%7D%7Ba%5E%7B2%7D%20%7D%20%7D)
![c^{2}=1+\frac{b^{2} }{a^{2} } \\(\frac{13}{12} )^{2}=1+\frac{b^{2} }{144} \\\frac{169}{144}=1+ \frac{b^{2} }{144} \\\frac{b^{2} }{144} =\frac{25}{144}\\ b^{2}=25](https://tex.z-dn.net/?f=c%5E%7B2%7D%3D1%2B%5Cfrac%7Bb%5E%7B2%7D%20%7D%7Ba%5E%7B2%7D%20%7D%20%20%5C%5C%28%5Cfrac%7B13%7D%7B12%7D%20%29%5E%7B2%7D%3D1%2B%5Cfrac%7Bb%5E%7B2%7D%20%7D%7B144%7D%20%20%5C%5C%5Cfrac%7B169%7D%7B144%7D%3D1%2B%20%5Cfrac%7Bb%5E%7B2%7D%20%7D%7B144%7D%20%5C%5C%5Cfrac%7Bb%5E%7B2%7D%20%7D%7B144%7D%20%3D%5Cfrac%7B25%7D%7B144%7D%5C%5C%20b%5E%7B2%7D%3D25)
The equation of the hyperbola is
![\frac{x^{2} }{a^{2} }-\frac{y^{2} }{b^{2} }=1\\ \frac{x^{2} }{144 }-\frac{y^{2} }{25 }=1](https://tex.z-dn.net/?f=%5Cfrac%7Bx%5E%7B2%7D%20%7D%7Ba%5E%7B2%7D%20%7D-%5Cfrac%7By%5E%7B2%7D%20%7D%7Bb%5E%7B2%7D%20%7D%3D1%5C%5C%20%20%5Cfrac%7Bx%5E%7B2%7D%20%7D%7B144%20%7D-%5Cfrac%7By%5E%7B2%7D%20%7D%7B25%20%7D%3D1)
Hence, the standard form for the hyperbola with vertices ( – 12,0) and (12,0), and foci ( – 13,0) and (13,0) is ![\frac{x^{2} }{144}-\frac{y^{2} }{25}=1](https://tex.z-dn.net/?f=%5Cfrac%7Bx%5E%7B2%7D%20%7D%7B144%7D-%5Cfrac%7By%5E%7B2%7D%20%7D%7B25%7D%3D1)
Learn more about hyperbola here
brainly.com/question/7098764
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