9514 1404 393
Answer:
see attached
Step-by-step explanation:
Perhaps you want a graph of ...
x^2/49 +(y +1)^2/4 = 1
This is an ellipse centered at (x, y) = (0, -1) with a major axis in the x-direction of 14, and a minor axis in the y-direction of 4.
Lol
trouble varies directly as distance
lets say t=trouble and d=distance
t=kd
k is constant
given
when t=20, and d=400
find k
20=400k
divide by 400 both sides
1/20=k
t=(1/20)d
given, d=60
find t
t=(1/20)60
t=60/20
t=3
3 troubles
Answer:
tan(2u)=[4sqrt(21)]/[17]
Step-by-step explanation:
Let u=arcsin(0.4)
tan(2u)=sin(2u)/cos(2u)
tan(2u)=[2sin(u)cos(u)]/[cos^2(u)-sin^2(u)]
If u=arcsin(0.4), then sin(u)=0.4
By the Pythagorean Identity, cos^2(u)+sin^2(u)=1, we have cos^2(u)=1-sin^2(u)=1-(0.4)^2=1-0.16=0.84.
This also implies cos(u)=sqrt(0.84) since cosine is positive.
Plug in values:
tan(2u)=[2(0.4)(sqrt(0.84)]/[0.84-0.16]
tan(2u)=[2(0.4)(sqrt(0.84)]/[0.68]
tan(2u)=[(0.4)(sqrt(0.84)]/[0.34]
tan(2u)=[(40)(sqrt(0.84)]/[34]
tan(2u)=[(20)(sqrt(0.84)]/[17]
Note:
0.84=0.04(21)
So the principal square root of 0.04 is 0.2
Sqrt(0.84)=0.2sqrt(21).
tan(2u)=[(20)(0.2)(sqrt(21)]/[17]
tan(2u)=[(20)(2)sqrt(21)]/[170]
tan(2u)=[(2)(2)sqrt(21)]/[17]
tan(2u)=[4sqrt(21)]/[17]
14^(7x) = 17^(-8-x)
Take the log of both sides:
7x*log 14 = -(8+x)*log 17, or 7x^log 14 = -8*log 17 - x*log 17.
Grouping the x terms, we get x(7*log 14 + log 17)= -8*log 17
Then:
-8*log 17
x= -------------------------- (answer)
7*log 14 + log 17
