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3 years ago

1. What is the best estimate for the product of 37 and 19?

Mathematics

2 answers:

algol [13]3 years ago

7 0

Answer:

Step-by-step explanation:

40/20=2

Tomtit [17]3 years ago

6 0

Answer: 800.

Step-by-step explanation: Round them both to the nearest ten. That is 40x20.

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DaniilM [7]

Divide 5/8 by 150, to see how thick each is, that'll give you 150 even pieces that add up to 5/8

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3 years ago

Click an item in the list or group of pictures at the bottom of the problem and, holding the button down, drag it into the corre

fenix001 [56]

Answer:

x^6y^3

Step-by-step explanation:

Given polynomial:

8x^6y^5 - 3x^8y^3

Get prime factors:

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4 0

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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