BC=19
Explanation
Step 1
ABE
triangle ABE is rigth triangle, then let
![\begin{gathered} Angle=60 \\ adjacentside=BE \\ opposit\text{ side(the one in front of the angle)= AB=}\frac{19\sqrt[]{6}}{4} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20Angle%3D60%20%5C%5C%20adjacentside%3DBE%20%5C%5C%20opposit%5Ctext%7B%20side%28the%20one%20in%20front%20of%20the%20angle%29%3D%20AB%3D%7D%5Cfrac%7B19%5Csqrt%5B%5D%7B6%7D%7D%7B4%7D%20%5Cend%7Bgathered%7D)
so, we need a function that relates, angle, adjancent side and opposite side

replace
![\begin{gathered} \tan \theta=\frac{opposite\text{ side}}{\text{adjacent side}} \\ \tan 60=\frac{AB}{\text{BE}} \\ \text{cross multiply} \\ \text{BE}\cdot\tan \text{ 60=AB} \\ \text{divide both sides by tan 60} \\ \frac{\text{BE}\cdot\tan\text{ 60}}{\tan\text{ 60}}=\frac{\text{AB}}{\tan\text{ 60}} \\ BE=\frac{\text{AB}}{\tan\text{ 60}} \\ \text{if AB=}\frac{19\sqrt[]{6}}{4} \\ BE=\frac{\frac{19\sqrt[]{6}}{4}}{\sqrt[]{3}} \\ BE=\frac{19\sqrt[]{6}}{4\sqrt[]{3}} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20%5Ctan%20%5Ctheta%3D%5Cfrac%7Bopposite%5Ctext%7B%20side%7D%7D%7B%5Ctext%7Badjacent%20side%7D%7D%20%5C%5C%20%5Ctan%2060%3D%5Cfrac%7BAB%7D%7B%5Ctext%7BBE%7D%7D%20%5C%5C%20%5Ctext%7Bcross%20multiply%7D%20%5C%5C%20%5Ctext%7BBE%7D%5Ccdot%5Ctan%20%5Ctext%7B%2060%3DAB%7D%20%5C%5C%20%5Ctext%7Bdivide%20both%20sides%20by%20tan%2060%7D%20%5C%5C%20%5Cfrac%7B%5Ctext%7BBE%7D%5Ccdot%5Ctan%5Ctext%7B%2060%7D%7D%7B%5Ctan%5Ctext%7B%2060%7D%7D%3D%5Cfrac%7B%5Ctext%7BAB%7D%7D%7B%5Ctan%5Ctext%7B%2060%7D%7D%20%5C%5C%20BE%3D%5Cfrac%7B%5Ctext%7BAB%7D%7D%7B%5Ctan%5Ctext%7B%2060%7D%7D%20%5C%5C%20%5Ctext%7Bif%20AB%3D%7D%5Cfrac%7B19%5Csqrt%5B%5D%7B6%7D%7D%7B4%7D%20%5C%5C%20BE%3D%5Cfrac%7B%5Cfrac%7B19%5Csqrt%5B%5D%7B6%7D%7D%7B4%7D%7D%7B%5Csqrt%5B%5D%7B3%7D%7D%20%5C%5C%20BE%3D%5Cfrac%7B19%5Csqrt%5B%5D%7B6%7D%7D%7B4%5Csqrt%5B%5D%7B3%7D%7D%20%5Cend%7Bgathered%7D)
Step 2
BED
again, we have a rigth triangle,then let

so, we need a function that relates; angle, hypotenuse and adjacent side

replace.
![\begin{gathered} \cos \theta=\frac{adjacent\text{ side}}{\text{hypotenuse}} \\ \cos 45=\frac{6.71}{\text{BD}} \\ BD=\frac{6.71}{\cos \text{ 45}} \\ BD=\frac{\frac{19\sqrt[]{6}}{4\sqrt[]{3}}}{\frac{\sqrt[]{2}}{2}} \\ BD=\frac{38\sqrt[]{6}}{4\sqrt[]{6}} \\ BD=\frac{38}{4} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20%5Ccos%20%5Ctheta%3D%5Cfrac%7Badjacent%5Ctext%7B%20side%7D%7D%7B%5Ctext%7Bhypotenuse%7D%7D%20%5C%5C%20%5Ccos%2045%3D%5Cfrac%7B6.71%7D%7B%5Ctext%7BBD%7D%7D%20%5C%5C%20BD%3D%5Cfrac%7B6.71%7D%7B%5Ccos%20%5Ctext%7B%2045%7D%7D%20%5C%5C%20BD%3D%5Cfrac%7B%5Cfrac%7B19%5Csqrt%5B%5D%7B6%7D%7D%7B4%5Csqrt%5B%5D%7B3%7D%7D%7D%7B%5Cfrac%7B%5Csqrt%5B%5D%7B2%7D%7D%7B2%7D%7D%20%5C%5C%20BD%3D%5Cfrac%7B38%5Csqrt%5B%5D%7B6%7D%7D%7B4%5Csqrt%5B%5D%7B6%7D%7D%20%5C%5C%20BD%3D%5Cfrac%7B38%7D%7B4%7D%20%5Cend%7Bgathered%7D)
Step 3
finally BDE
let
angle=30
opposite side= BD
use sin function

so, the answer is 19
I hop
Answer:
Total number of puppies in the litter = 11
Step-by-step explanation:
Let the number of yellow puppies in the litter = y
And the number of black puppies in the litter = b
If there was one less black puppy, ratio of yellow to black puppies = 

2(b - 1) = 3y
3y - 2b = -2 -----(1)
If there was one more black puppy, ratio of yellow puppies to black puppies = 

2y = b + 1
2y - b = 1 -------(2)
Equation (1) - 2× equation (2)
(3y - 2b) - 2(2y - b) = -2 - 2
3y - 4y - 2b + 2b = -4
y = 4
From equation (1),
3×4 - 2b = -2
12 - 2b = -2
-2b = -14
b = 7
Total number of puppies in the litter = 7 + 4 = 11
Answer:
False
Step-by-step explanation:
V(cylinder)=πR²H
Radius of the cylinder R=x, height of the cylinder H=y.
We can write for the cylinder
V(cylinder)=πx²y
V(cone) =(1/3)πr²h
Radius of the cone r=2x.
We can write for the cone
V(cone)= (1/3)π(2x)²h=(1/3)π *4*x²h
V(cylinder) =V(cone)
πx²y=(1/3)π *4*x²h
y=(4/3)*h
h=(3/4)*y