Suppose :
2^x = t ..**&**.. 3^y = u
_____________________________
t + u = 9 × 3 3t + 3u = 27
=====》
2t - 3u = 8 2t - 3u = 8
+ __________
5t = 35
5 × t = 5 × 7
t = 7
&
u = 2
________________________________
Thus :
2^x = 7 ===》 x = Log _ 2 { 7 }
3^y = 2 ===》 y = Log _ 3 { 2 }
Answer:
9.2, 11/12, 91.7, 91.6, .91
Step-by-step explanation:
The equation in standard form is 0.5x-y=-3.
Answer:
The equation that represents the new path is y=(1/2)x-2
Step-by-step explanation:
step 1
Find the slope of the give line
we have
y=-2x-7
so
the slope m is equal to
m=-2
step 2
Find the slope of the perpendicular line to the given line
Remember that
If two lines are perpendicular, then their slopes are opposite reciprocal of each other
so
we have
m=-2 -----> slope of the given line
therefore
The slope of the perpendicular line is equal to
m=1/2
step 3
With m=1/2 and the point (-2,-3) find the equation of the line
y-y1=m(x-x1)
substitute
y+3=(1/2)(x+2)
y=(1/2)x+1-3
y=(1/2)x-2 -----> equation that represent the new path
BE and CD are parallel, so the triangles ABE and ACD are similar. Then the following relation holds:
AE/AD = AB/AC
10/(10 + 8) = x/(x + 6)
Solve for x :
10 (x + 6) = x (10 + 8)
10x + 60 = 18x
8x = 60
x = 60/8 = 7/5
