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3 years ago

3.47472 rounded to the nearest 10th

Mathematics

2 answers:

Neporo4naja [7]3 years ago

8 0

Answer:

It would be 0 its nowhere near a ten

Step-by-step explanation:

hope i helped

ser-zykov [4K]3 years ago

5 0

Good evening ,

______

Answer:

3.5

___________________

Step-by-step explanation:

the number that comes after the comma is called the 10th

In our case is 4

We look at the 100th which is 7 and because 7 ≥ 5

then our number rounded to nearest 10th is : 3.5

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If 3x x 10=22, what is the value of 5x +8?

scZoUnD [109]

Step-by-step explanation:

3x x 10 = 22

3x x 10 - 10 = 22 - 10

3x = 12

3x ÷ 3 = 12 ÷ 3

x = 4

5x + 8

5(4) + 8

20 + 8

7 0

3 years ago

Find the area of a trapezoid with bases 14 cm and 18 cm and height 10 cm.

4vir4ik [10]

You should add the two bases and muliply them with the height and then divide with 2
(14+18)×10÷2=160

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4 years ago

Susie is visiting China from the U.S. One day she stopped by a market and found a stall selling tomatoes for 15.65 yuan per kilo

liq [111]

Answer:

1st blank is 0,14 dollars/1yuan, and 2nd blank is 1kg/2.2lb

Step-by-step explanation:

5 0

4 years ago

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 get $N(0)=N_oe^{m\times 0}\\\\12=N_o$

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

Claire wants to buy enough sand to fill a sandbox that is shaped like a rectangular prism. The sandbox is 4 feet long, 4 feet wi

nataly862011 [7]

First, find the volume of the sand box
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8 divided by 0.5 is 16, so you need 16 bags
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Hope this helps :D

7 0

3 years ago