What is square footage of 83"x 68"?
1 answer:
You might be interested in
Get into ax²+bx+c=0 form
subsitute u=x⁴
x⁸-3x⁴+2=0
(x⁴)²-3(x⁴)+2=0
u²-3u+2=0
factor
(u-2)(u-1)=0
set to zero
u-2=0
u=2
u-1=0
u=1
wait, u=x⁴
2=x⁴
![\sqrt[4]{2}=x](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7B2%7D%3Dx)
1=x⁴
+/-1=x
the subsitution is u=x⁴
and the possible values for x are -1,, and 1
subsitute u=x⁴
x⁸-3x⁴+2=0
(x⁴)²-3(x⁴)+2=0
u²-3u+2=0
factor
(u-2)(u-1)=0
set to zero
u-2=0
u=2
u-1=0
u=1
wait, u=x⁴
2=x⁴
1=x⁴
+/-1=x
the subsitution is u=x⁴
and the possible values for x are -1,
Answer:
The point C is 12.68 km away from the point A on a bearing of S23.23°W.
Step-by-step explanation:
Given that AB is 50 km and BC is 40 km as shown in the figure.
From the figure, the length of x-component of AC = |AB sin 80° - BC cos 20°|
=|50 sin 80° - 40 cos 20°|=11.65 km
The length of y-component of AC = |AB cos 80° - BC sin 20°|
=|50 cos 80° - 40 sin 20°|= 5 km
tan= 5/11.65
=23.23°
AC= km
Hence, the point C is 12.68 km away from the point A on a bearing of S23.23°W.
