The volume of a cone is V=(1/3)(area of the base)(height)= (1/3)(pi*r^2)(h)
r^2= (d/2)^2 = (d^2)/4. Given: V= 301.44cm^3 and h= 18cm.
r^2= V/[(1/3)(pi)(h)]
r^2= 3V/(pi)(h)
(d^2)/4= 3(301.44)/(3.14)(18)
d^2= 12(301.44)/(3.14)(18)
d^2= 63.967
d= 7.997
d~= 8cm.
The point-slope form of the equation for a line can be written as
... y = m(x -h) +k . . . . . . . for a line with slope m through point (h, k)
Your function gives
... f'(h) = m
... f(h) = k
a) The tangent line is then
... y = 5(x -2) +3
b) The normal line will have a slope that is the negative reciprocal of that of the tangent line.
... y = (-1/5)(x -2) +3
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You asked for "an equation." That's what is provided above. Each can be rearranged to whatever form you like.
In standard form, the tangent line's equation is 5x -y = 7. The normal line's equation is x +5y = 17.
First we use sin(a+b)= sinacosb+sinbcosa
and cos(a+b)=cosa cosb -sinasinb
tan(x+pi/2)= sin(x+pi/2) / cos(x+pi/2)
and sin(x+pi/2) = sinxcospi/2 + sinpi/2cosx =cosx,
<span>cos(x+pi/2) = cosxcospi/2- sinxsinpi/2= - sinx,
</span> because <span>cospi/2 =0, </span>and <span>sinpi/2=1
</span><span>=tan(x+pi/2)= sin(x+pi/2) / cos(x+pi/2)= cosx / -sinx = -1/tanx = -cotx
</span>from where <span>tan(x+pi/2)=-cotx</span>
3(9)=27
27+4Y=43
subtract 27 from both sides
4Y=16
devide both sides by 4
Y=4
Answer:
a = 6, b = 3, and c = 2
1. ab1 = 6 × 3 × 1 = 18
2. c + 42 = 2 + 42 = 44
3. 18 = 18
4. a − b4 = 6 - 3(4) = 6 - 12 = - 6
5. 2c3 = 2 × 2 × 3 = 12
6. b ÷ 35 = 3÷35 = 0.086
7. a − 1 6 = 6 - 16 = -10
8. 6 + c8 = 6 + 2 (8) = 6 + 16 = 22
Step-by-step explanation:
