Answer:
3/4 pounds
Step-by-step explanation:
Given the weights :
weight of beans 1/4 pound, 1/2 pound, 3/4. pound and 1 pound.
Greatest weight = 1 pound
Least weight = 1/4 pound
Difference = greatest weight - Least weight
Difference = (1 - 1/4) pound
Difference in weight = 3/4 pounds
Answer:
which agrees with option"B" of the possible answers listed
Step-by-step explanation:
Notice that in order to solve this problem (find angle JLF) , we need to find the value of the angle defined by JLG and subtract it from 
, since they are supplementary angles. So we focus on such, and start by drawing the radii that connects the center of the circle (point "O") to points G and H, in order to observe the central angles that are given to us as 
and 
. (see attached image)
We put our efforts into solving the right angle triangle denoted with green borders.
Notice as well, that the triangle JOH that is formed with the two radii and the segment that joins point J to point G, is an isosceles triangle, and therefore the two angles opposite to these equal radius sides, must be equal. We see that angle JOH can be calculated by : 
Therefore, the two equal acute angles in the triangle JOH should add to:
resulting then in each small acute angle of measure 
.
Now referring to the green sided right angle triangle we can find find angle JLG, using: 
Finally, the requested measure of angle JLF is obtained via: 
The greatest six-digit number they could make is 998,877 I know this because I used the largest digits and put them in order from largest to smallest.
Hope this helps
Million has 6 zeoes
34,000,000+256,00*400
PEMDAS
multiply first
256,000*400=102,400,000
add to 34,000,000
136,400,000
now
must be in form
(x)(10^m)
such taht
1<x<10
and x times 10^m is the original number
1.364*10^8
<span>, y+2 = (x^2/2) - 2sin(y)
so we are taking the derivative y in respect to x so we have
dy/dx use chain rule on y
so y' = 2x/2 - 2cos(y)*y'
</span><span>Now rearrange it to solve for y'
y' = 2x/2 - 2cos(y)*y'
0 = x - 2cos(y)y' - y'
- x = 2cos(y)y' - y'
-x = y'(2cos(y) - 1)
-x/(2cos(y) - 1) = y'
</span><span>we know when f(2) = 0 so thus y = 0
so when
f'(2) = -2/(2cos(0)-1)
</span><span>2/2 = 1
</span><span>f'(2) = -2/(2cos(0)-1)
cos(0) = 1
thus
f'(2) = -2/(2(1)-1)
= -2/-1
= 2
f'(2) = 2
</span>
