Answer:
- At 1pm, the number of bacteria in the dish is 1,000,000
- At 12pm, the number of bacteria in the dish is 500,000
Step-by-step explanation:
Given the information on the table
![\left|\begin{array}{c|c}$Hour&$Number of Bacteria (in millions)\\---&--------------\\12 p.m.&\\1 p.m.&\\12 p.m.&\\2 p.m.&\\3 p.m.&\\4 p.m.&\\5 p.m.&32\\6 p.m.&64\end{array}\right|](https://tex.z-dn.net/?f=%5Cleft%7C%5Cbegin%7Barray%7D%7Bc%7Cc%7D%24Hour%26%24Number%20of%20Bacteria%20%28in%20millions%29%5C%5C---%26--------------%5C%5C12%20p.m.%26%5C%5C1%20p.m.%26%5C%5C12%20p.m.%26%5C%5C2%20p.m.%26%5C%5C3%20p.m.%26%5C%5C4%20p.m.%26%5C%5C5%20p.m.%2632%5C%5C6%20p.m.%2664%5Cend%7Barray%7D%5Cright%7C)
We are to follow the pattern to determine how many million bacteria were in the dish at 1 p.m. and 12 p.m.
Now:
![32=2^5\\64=2^6](https://tex.z-dn.net/?f=32%3D2%5E5%5C%5C64%3D2%5E6)
It therefore means that the bacteria population doubles every hour. Following this pattern and reducing the exponents by 1 backwards, we have:
![\left|\begin{array}{c|c}$Hour&$Number of Bacteria (in millions)\\---&--------------\\12 p.m.&2^{-1}=0.5\\1 p.m.&2^0=1\\12 p.m.&2^1=2\\2 p.m.&2^2=4\\3 p.m.&2^3=8\\4 p.m.&2^4=16\\5 p.m.&2^5=32\\6 p.m.&2^6=64\end{array}\right|](https://tex.z-dn.net/?f=%5Cleft%7C%5Cbegin%7Barray%7D%7Bc%7Cc%7D%24Hour%26%24Number%20of%20Bacteria%20%28in%20millions%29%5C%5C---%26--------------%5C%5C12%20p.m.%262%5E%7B-1%7D%3D0.5%5C%5C1%20p.m.%262%5E0%3D1%5C%5C12%20p.m.%262%5E1%3D2%5C%5C2%20p.m.%262%5E2%3D4%5C%5C3%20p.m.%262%5E3%3D8%5C%5C4%20p.m.%262%5E4%3D16%5C%5C5%20p.m.%262%5E5%3D32%5C%5C6%20p.m.%262%5E6%3D64%5Cend%7Barray%7D%5Cright%7C)
Therefore:
- At 1pm, the number of bacteria in the dish is 1,000,000
- At 12pm, the number of bacteria is 0.5 million=0.5 X 1,000,000=500,000
Answer:
The answer to your question is letter C
Step-by-step explanation:
Process
1.- Find two points of the line
A ( 0, -3) B ( 2, 2)
2.- Find the slope of the line
![m = \frac{y2 - y1}{x2 - x1} = \frac{2 + 3}{2 - 0}](https://tex.z-dn.net/?f=m%20%3D%20%5Cfrac%7By2%20-%20y1%7D%7Bx2%20-%20x1%7D%20%3D%20%5Cfrac%7B2%20%2B%203%7D%7B2%20-%200%7D)
![m = \frac{5}{2}](https://tex.z-dn.net/?f=m%20%3D%20%5Cfrac%7B5%7D%7B2%7D)
3.- Find the equation of the line
y - y1 = m(x - x1)
y + 3 = 5/2(x - 0)
y + 3 = 5/2 x
y = 5/2 x - 3
4.- Convert the line to an inequality, we want the upper area so use ">"
y > 5/2 x - 3
Hmmm to tell if they're orthogonal(perpendicular) to each other, we can simply get their dot-product, if their dot-product is 0, then they indeed are perpendicular, let's check
![\bf \ \textless \ 7,-4\ \textgreater \ \cdot \ \textless \ -28,16\ \textgreater \ \implies (7\cdot -28)+(-4\cdot 16)\implies -190](https://tex.z-dn.net/?f=%5Cbf%20%5C%20%5Ctextless%20%5C%207%2C-4%5C%20%5Ctextgreater%20%5C%20%5Ccdot%20%5C%20%5Ctextless%20%5C%20-28%2C16%5C%20%5Ctextgreater%20%5C%20%5Cimplies%20%287%5Ccdot%20-28%29%2B%28-4%5Ccdot%2016%29%5Cimplies%20-190)
well, no luck there... now, let's introspect the digits a little
![\bf \begin{array}{llll} \ \textless \ 7,-4\ \textgreater \ \qquad \ \textless \ &-28,&16\ \textgreater \ \\ &~~\uparrow &~\uparrow \\ &-4\cdot 7&-4\cdot -4 \end{array}](https://tex.z-dn.net/?f=%5Cbf%20%5Cbegin%7Barray%7D%7Bllll%7D%0A%5C%20%5Ctextless%20%5C%207%2C-4%5C%20%5Ctextgreater%20%5C%20%20%5Cqquad%20%5C%20%5Ctextless%20%5C%20%26-28%2C%2616%5C%20%5Ctextgreater%20%5C%20%5C%5C%0A%26~~%5Cuparrow%20%26~%5Cuparrow%20%5C%5C%0A%26-4%5Ccdot%207%26-4%5Ccdot%20-4%0A%5Cend%7Barray%7D)
notice, the second vector, -28, 16, is just a multiple of the first one, that simply means, they're
parallel.
you can always just do a b/a check and simplify both vectors, if they yield the same ratio, they're parallel, since they're just multiples of each other.
Figure 4 is the image of the square LMNP after the translation.
<u>Step-by-step explanation:</u>
Let us see the coordinates of the pre image LMNP as,
L (-3,1)
M(-1,1)
N(-1,-1)
P(-3,-1)
after translation of (x,y) → (x+5, y -3) the coordinates of the image obtained as,
L'(2,-2)
M'(4,-2)
N'(4,-4)
P'(2,-4) which matches the image 4.