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Snezhnost [94]
2 years ago
5

What rate of growth is modeled by this bacteria colony per hour of observation??

Mathematics
1 answer:
romanna [79]2 years ago
3 0
Answer is C because 95+52% is 144.4 so jus yeah it’s C
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Solve x-5/y = z for y​
kogti [31]

Answer:

y=5x−z

Step-by-step explanation:

Step 1: Multiply both sides by y.

xy−5=yz

Step 2: Add -yz to both sides.

xy−5+−yz=yz+−yz

xy−yz−5=0

Step 3: Add 5 to both sides.

xy−yz−5+5=0+5

xy−yz=5

Step 4: Factor out variable y.

y(x−z)=5

Step 5: Divide both sides by x-z.

y(x−z)x−z=5x−z

y=5x−z

5 0
2 years ago
Find the equation of the line containing the points (20,-7) and (10,-2). Type your answer in this order and don't use spaces or
aliina [53]

(\stackrel{x_1}{20}~,~\stackrel{y_1}{-7})\qquad (\stackrel{x_2}{10}~,~\stackrel{y_2}{-2}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{-2}-\stackrel{y1}{(-7)}}}{\underset{run} {\underset{x_2}{10}-\underset{x_1}{20}}}\implies \cfrac{-2+7}{-10}\implies -\cfrac{1}{2}

\begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{(-7)}=\stackrel{m}{-\cfrac{1}{2}}(x-\stackrel{x_1}{20}) \\\\\\ y+7=-\cfrac{1}{2}x+10\implies y=-\cfrac{1}{2}x+3

4 0
2 years ago
Use the fact that the bacteria is doubling every five
victus00 [196]

Answer:

I think it's 1/16

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
I need help with #3 plz
Nadusha1986 [10]
The correct answer is A. 10 bracelets

5 0
3 years ago
Read 2 more answers
A midwestern music competition awarded 38 ribbons. The number of blue ribbons awarded was 1 less than the number of white ribbon
Artist 52 [7]
Starting from what I know (the difference between ribbons), I decided to go from the bottom and work my way up. I then noticed a pattern (each sum was three more of the previous one), and decided to keep my pattern of the three numbers but not have to do any major mental work and instead add three to the previous sum until I got to 38.

4 0
3 years ago
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