Y=kx
35=7k
35/7=k
k=5
y=5x
y=5×11
y=55
Answer:
Step-by-step explanation:
Δy =-4-1 = -5
Δx =-8-(-6) = -2
distance = √((Δx)^2+(Δy)^2) = √(4+25) = √29
Answer:
3x - 13 +58 = 90, so 3x +45 = 90
Then 3x = 90 - 45 = 45 and x= 45/3 = 15.
m∠MNQ = 3x-13 = 3(15) - 13
= 45-13
= 32
Answer: answer is b.58
Step-by-step explanation:
The number of microorganisms that I will have after 10 days would be: 245.41 microoganisms.
<h3>How to calculate the number of microorganisms that I will have after 10 days?</h3>
To calculate the number of microorganisms that I will have after 10 days, I must perform the following operations:
We need to find 37% of 55 as shown below:
- 55 ÷ 100 = 0.55
- 0.55 x 37% = 20.35
- 55 + 20.35 = 75.35
So after 2 days I will have 75.35 microorganisms.
- 75.35 ÷ 100 = 0.75
- 0.7535 x 37% = 27.87
- 75.35 + 27.87 = 103.22
On the fourth day I will have: 103.22 microorganisms.
- 103.22 ÷ 100 = 1.0322
- 1.0322 x 37% = 38.19
- 103.22 + 38.19 = 141.41
On the sixth day I will have 141.41 microorganisms.
- 141.41 ÷ 100 = 1.4141
- 1.4141 x 37% = 52.32
- 141.41 + 52.32 = 193,73
On the eighth day I will have 193.73 microorganisms.
- 193.73 ÷ 100 = 1.9373
- 1.9373 x 37% = 71.68
- 193.73 + 71.68 = 265.41
On the tenth day I will have 245.41 microoganisms.
Learn more about percentages in: brainly.com/question/13450942
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