Ken's distance is: d(t)=1.7t
Ken's shadow is:
(14-6)/d(t)=14/s
8/d(t)=14/s
s=14d(t)/8
s=1.75d(t) and using the value for d(t) we have:
s=1.75(1.7t)
s=2.975t and if you want it in terms of d, d=1.7t, t=d/1.7 so
s=2.975(d/1.7)
s=1.75d
I know this is kinda late, but...
You can solve this easily as long as you remember how to find the volume of a cube. v=a^3, when a means length. The length is 1/2 cm, or .5 cm.
So, v+.5^3, which equals .125 cm. Each cube has a that volume. Now, multiply that by how many actual cubes there are, 96.
96 * .125 = 12
The given triangle has a right angle.
We use the mnemonics SOH-CAH-TOA.
1i) 
,
,
ii) 
,
,
,
,
2. We want to find the hypotenuse.
We know an angle to be 23 degrees.
We were also given the side opposite to this angle to be 1200km.
Therefore we use the sine ratio.
Answer:
Step-by-step explanation:
eq. of line parallel to AB is y=2x+a
if it passes through (3,-2) then
-2=2×3+a
a=-2-6=-8
y=2x-8
