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3 years ago

2.16 or 2.061 larger

1 answer:

4 0

The larger number is 2.16

That is because 2.061 = 2.06 when rounded to 2 decimals, which is smaller than 2.16. If you add a zero to 2.16 you'll get a 2.160. which is still larger than 2.061.

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Find the perimeter of ABC with vertices A(0,-4), B(4,-4), and C(0,-1).

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3 years ago

A fisherman uses a spring scale to weigh a tilapia fish. He records the fish weight as a kilograms and notices that the spring s

belka [17]

Answer:

k = $\frac{980a}{b}$

Step-by-step explanation:

Fisherman noticed a stretch in the spring = 'b' centimetres

Weight of the fish = a kilograms

If force applied on a spring scale makes a stretch in the spring then Hook's law for the force applied is,

F = kΔx

Where k = spring constant

Δx = stretch in the spring

F = weight applied

F = mg

Here 'm' = mass of the fish

g = gravitational constant

F = a(9.8)

= 9.8a

Δx = b centimetres = 0.01b meters

Therefore, 9.8a = k(0.01b)

k = $\frac{9.8a}{0.01b}$

k = $\frac{980a}{b}$

Therefore, spring constant of the spring will be determined by the expression, k = $\frac{980a}{b}$

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3 years ago

During an experiment, Juan rolled a six-sided number cube 18 times. The number two occurred four times. Juan claimed the experim

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3 years ago

Jenifer had 30$ to spend on her self she spent 1/5 Of the money in a sandwich 1/6 for a ticket to a museum and 1/2 Of It On a Bo

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3 years ago

If N is the population of the colony and t is the time in days, express N as a function of t. Consider Upper N 0 is the origina

lesya [120]

Answer:

(a) $N(t)=Noe^{kt}$

(b)5,832 Mosquitoes

Step-by-step explanation:

(a)Given an original amount $N_o$ at t=0. The population of the colony with a growth rate $k \neq 0$ , where k is a constant is given as:

$N(t)=Noe^{kt}$

(b)If $N_o=1000$ and the population after 1 day, N(1)=1800

Then, from our model:

N(1)=1800

$1800=1000e^{k}\\$Divide both sides by 1000\\e^{k}=1.8\\$Take the natural logarithm of both sides\\k=ln(1.8)$

Therefore, our model is:

$N(t)=1000e^{t*ln(1.8)}\\N(t)=1000\cdot1.8^t$

In 3 days time

$N(3)=1000\cdot1.8^3=5832$

The population of mosquitoes in 3 days time will be approximately 5832.

(c)If the population N(t)=20,000,we want to determine how many days it takes to attain that value.

From our model

$N(t)=1000\cdot1.8^t\\20000=1000\cdot1.8^t\\$Divide both sides by 1000\\20=1.8^t\\$Convert to logarithm form\\Log_{1.8}20=t\\\frac{Log 20}{Log 1.8}=t\\ t=5.097\approx 5\; days$

In approximately 5 days, the population of mosquitoes will be 20,000.

7 0

3 years ago

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