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Ludmilka [50]
3 years ago
10

Match the three dimensional shape with the HORIZONTAL cross section shown.

Mathematics
1 answer:
Maksim231197 [3]3 years ago
8 0
Square Shape - rectangular Pyramid, Rectangular Prism

Circle Shape - Cylinder, cone

Triangle Shape - Triangular Pyramid

Hope it helps! <span />
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5x - y =1 13htut yti rindu err err h is

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It takes Joey 1/16 of an hour to write one thank you card how many cards can he write in 3/4 of an hour ?
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to write one thank you  card it takes 1/16 of an hour


1 card  * 3/4 hour  

--------       ---------      =         the hours cancel and you are left with cards

1/16 hour       1



3/4

------                                  copy dot flip

1/16      



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A triangular prism was sliced perpendicular to its bases and through a vertex. What is the shape of the cross section shown in t
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Step-by-step explanation:

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4 years ago
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The head of the math department compared the scores of students in two classes on the Chapter 2 test. He found that the mean sco
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A population of bacteria is growing exponentially. At 7:00 a.m. the mass of the population is 12 mg. Five hours later it is 14 m
sergejj [24]

Answer with Step-by-step explanation:

The exponential growth function is given by

N(t)=N_oe^{mt}

where

N(t) is the population of the bacteria at any time 't'

N_o is the population of the bacteria at any time 't = 0'

'm' is a constant and 't' is time after 7.00 a.m in hours

Assuming we start our measurement at 7.00 a.m as reference time  t = 0

Thus we getN(0)=N_oe^{m\times 0}\\\\12=N_o

Now since it is given after 5 hours the population becomes 14 mg thus from the above relation we get

12\times e^{m\times 5}=14\\\\e^{5m}=\frac{14}{12}\\\\m=\frac{1}{5}\cdot ln(\frac{14}{12})\\\\m=0.031

Thus the population of bacteria at any time 't' is given by

N(t)=12e^{0.031t}

Part a)

Population of bacteria after another 5 hours equals the population after 10 hours from start

N(10)=12e^{0.031\times 10}=16.361mg

Part b)

Population of bacteria at 7:00 p.m is mass after 12 hours

N(1)=12e^{0.031\times 12}=17.41mg

Part c)

Population of bacteria at 8:00 p.m is mass after 1 hour

N(1)=12e^{0.031\times 1}=12.3378mg

Part d)

Differentiating the relation of population with respect to time we get

N'(t)=\frac{d(12\cdot e^{0.031t})}{dt}\\\\N'(t)=12\times 0.031=0.372e^{0.031t}

Thus we can see that the percentage increase varies with time initially the percentage increase is 37.2% but this percentage increase increases with increase in time

Part 4)

Since there are 24 hours in 1 day thus the percentage increase in the population is

\frac{N(24)-N_o}{N_o}\times 100\\\\=\frac{25.25-12}{12}\times 100=110.42

Thus there is an increase of 110.42% in the population each day.

7 0
4 years ago
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