Answer:
25
Step-by-step explanation:
We can use the Pythagorean theorem to determine the length of the hypotenuse
a^2 + b^2 = c^2
15^2 + 20 ^2 = c^2
225 + 400 = c^2
625 = c^2
Take the square root of each side
sqrt(625) = sqrt(c^2)
25 = c
Answer:
![\frac{r(1-\sqrt{2})}{-3}](https://tex.z-dn.net/?f=%5Cfrac%7Br%281-%5Csqrt%7B2%7D%29%7D%7B-3%7D)
Step-by-step explanation:
Volume of cone = ![\frac{1}{3} \pi r^{2} h](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B3%7D%20%5Cpi%20r%5E%7B2%7D%20h)
Since we are given that a circular cone has a base of radius r and a height of h that is the same length as the radius
=
=
Surface area of cone including 1 base = ![\pi r^{2} +\pi\times r \times \sqrt{r^2+h^2}](https://tex.z-dn.net/?f=%5Cpi%20r%5E%7B2%7D%20%2B%5Cpi%5Ctimes%20r%20%5Ctimes%20%5Csqrt%7Br%5E2%2Bh%5E2%7D)
Since r = h
So, area = ![\pi r^{2} +\pi\times r \times \sqrt{r^2+r^2}](https://tex.z-dn.net/?f=%5Cpi%20r%5E%7B2%7D%20%2B%5Cpi%5Ctimes%20r%20%5Ctimes%20%5Csqrt%7Br%5E2%2Br%5E2%7D)
=
= ![\pi r^{2} +\pi\times r^2 \times \sqrt{2}](https://tex.z-dn.net/?f=%5Cpi%20r%5E%7B2%7D%20%2B%5Cpi%5Ctimes%20r%5E2%20%5Ctimes%20%5Csqrt%7B2%7D)
Ratio of volume of cone to its surface area including base :
![\frac{\frac{1}{3} \pi r^{3}}{\pi r^{2} +\pi\times r^2 \times \sqrt{2}}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Cfrac%7B1%7D%7B3%7D%20%5Cpi%20r%5E%7B3%7D%7D%7B%5Cpi%20r%5E%7B2%7D%20%2B%5Cpi%5Ctimes%20r%5E2%20%5Ctimes%20%5Csqrt%7B2%7D%7D)
![\frac{\frac{1}{3}r}{1+\sqrt{2}}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Cfrac%7B1%7D%7B3%7Dr%7D%7B1%2B%5Csqrt%7B2%7D%7D)
![\frac{r}{3(1+\sqrt{2})}](https://tex.z-dn.net/?f=%5Cfrac%7Br%7D%7B3%281%2B%5Csqrt%7B2%7D%29%7D)
Rationalizing
![\frac{r}{3(1+\sqrt{2})} \times \frac{1-\sqrt{2}}{1-\sqrt{2}}](https://tex.z-dn.net/?f=%5Cfrac%7Br%7D%7B3%281%2B%5Csqrt%7B2%7D%29%7D%20%5Ctimes%20%5Cfrac%7B1-%5Csqrt%7B2%7D%7D%7B1-%5Csqrt%7B2%7D%7D)
![\frac{r(1-\sqrt{2})}{-3}](https://tex.z-dn.net/?f=%5Cfrac%7Br%281-%5Csqrt%7B2%7D%29%7D%7B-3%7D)
Hence the ratio the ratio of the volume of the candle to its surface area(including the base) is ![\frac{r(1-\sqrt{2})}{-3}](https://tex.z-dn.net/?f=%5Cfrac%7Br%281-%5Csqrt%7B2%7D%29%7D%7B-3%7D)
Let's find the discriminant of <span>x^2+9x+14=0. Here, a=1, b=9 and c=14.
The discriminant is b^2-4ac. Substituting the above numeric values,
9^2-4(1)(14) = 81-56 = 25
The sqrt of 25 is 5. Thus, your polynomial has two unequal, real roots.
Off the point example: If the discriminant were zero, your poly would have two real, equal roots.</span>