Depending on how flexibly you interpret "about 20 times larger" to mean, the answers are B and D.
Check the ratios of the larger number to the smaller number:
A: (2.01 x 10^7)/(4.25 x 10^6) = 2.01/4.25 x 10^1 = 20.1/4.25 = 4.729
B: (8.21 x 10^-3)/(4.13 x 10^-4) = 8.21/4.13 x 10^1 = 82.1/4.13 = 19.879
C: (4.91 x 10^6)/(5.09 x 10^3) = 4.91/5.09 x 10^3 = 4910/5.09 = 964.637
D: (5.97 x 10^4)/(3.12 x 10^3) = 5.97/3.12 x 10^1 = 59.7/3.12 = 19.135
Answer:
choose "infinitely more solutions"
Step-by-step explanation:
Let's solve this system by substitution method
y+1=2x
- 1 = - 1
y = 2x -1 <-- sbstitute this in the next equation
---------------------------------------------------------------------------------
5y+5 = 10x
5(2x -1) + 5 = 10x
10x -5 + 5 = 10x
10x + 0= 10x
- 10x = -10x
0 = 0 <-- infinitely more solutions
Answer:
The venue has to host 47 events throughout the year
Step-by-step explanation:
Profit:
Profit is revenue subtracted by costs.
Per event:
$100,000 revenue
$50,000 costs
$100,000 - $50,000 = $50,000 profit.
If the venue wants to make a total profit margin of $2,350,000, how many events does the venue have to host throughout the year?
1 event - $50,000 profit.
x events - $2,350,000 profit.



The venue has to host 47 events throughout the year
<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>
Answer:
The full answer in the media.Good luck.