<h3>Sin

</h3>
Solution given:
Cos

equating corresponding value
we get
adjacent=10
hypotenuse=17
perpendicular=x
now
by using Pythagoras law
Hypotenuse ²=perpendicular²+adjacent ²
substituting value
17²=x²+10²
17²-10²=x²
x²=17²-10²
x²=189
doing square root

x=
now
In I Quadrant sin angle is positive
Sin
<h3>Sin

</h3>
Answer:
12+35x3.14
Step-by-step explanation:
Answer:
-7
Step-by-step explanation:
In general, the volume

has total derivative

If the cylinder's height is kept constant, then

and we have

which is to say,

and

are directly proportional by a factor equivalent to the lateral surface area of the cylinder (

).
Meanwhile, if the cylinder's radius is kept fixed, then

since

. In other words,

and

are directly proportional by a factor of the surface area of the cylinder's circular face (

).
Finally, the general case (

and

not constant), you can see from the total derivative that

is affected by both

and

in combination.
Answer:
y = 19 (3)x
Explanation:
y = abx
Plug in the numbers from the 2 given points to find a and b.
19 = ab0 = a
a = 19
171 = 19b2
b = 3
y = 19 (3)x