How to explain what comes next 1,2,2,3,3,3,4,4,4,4 for third graders?

Mathematics

1 answer:

lesya692 [45]3 years ago

6 0

Next it would be 5,5,5,5,5. Because, when you go up one in numbers you also go up one the amount.

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HELP ME PLEASE!! ASAP!!

Levart [38]

The total area is given by:
A1 = (300) * (500)
A1 = 150000 feet ^ 2
The area occupied by 5 people is:
A2 = (5) * (5)
A2 = 25 feet ^ 2
Then, we can make the following rule of three:
5 --------> 25
x --------> 150000
Clearing x we have:
x = (150000/25) * (5)
x = 30000
Answer:
D. 30,000 people

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ABC is an isosceles triangle with

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A certain breed of mouse was introduced onto a small island with an initial population of 320 mice, and scientists estimate that

Viefleur [7K]

Y=2x+320
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3 years ago

What is the coefficient of y in the expression 3 ⋅ 4 + 5y?

ZanzabumX [31]

<h3>Answer: 5</h3>

The coefficient is the term just to the left of the variable. So for the term 5y, the number to the left of y is 5.

extra info: 5y is the same as 5*y or "5 times y", where y is a placeholder for some unknown number.

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3 years ago

Read 2 more answers

A tank contains 30 lb of salt dissolved in 300 gallons of water. a brine solution is pumped into the tank at a rate of 3 gal/min

sesenic [268]

$A'(t)=(\text{flow rate in})(\text{inflow concentration})-(\text{flow rate out})(\text{outflow concentration})$
$\implies A'(t)=\dfrac{3\text{ gal}}{1\text{ min}}\cdot\left(2+\sin\dfrac t4\right)\dfrac{\text{lb}}{\text{gal}}-\dfrac{3\text{ gal}}{1\text{ min}}\cdot\dfrac{A(t)\text{ lb}}{300+(3-3)t\text{ gal}}$
$A'(t)+\dfrac1{100}A(t)=6+3\sin\dfrac t4$

We're given that $A(0)=30$ . Multiply both sides by the integrating factor $e^{t/100}$ , then

$e^{t/100}A'(t)+\dfrac1{100}e^{t/100}A(t)=6e^{t/100}+3e^{t/100}\sin\dfrac t4$
$\left(e^{t/100}A(t)\right)'=6e^{t/100}+3e^{t/100}\sin\dfrac t4$
$e^{t/100}A(t)=600e^{t/100}-\dfrac{150}{313}e^{t/100}\left(25\cos\dfrac t4-\sin\dfrac t4\right)+C$
$A(t)=600-\dfrac{150}{313}\left(25\cos\dfrac t4-\sin\dfrac t4\right)+Ce^{-t/100}$

Given that $A(0)=30$ , we have

$30=600-\dfrac{150}{313}\cdot25+C\implies C=-\dfrac{174660}{313}\approx-558.02$

so the amount of salt in the tank at time $t$ is

$A(t)\approx600-\dfrac{150}{313}\left(25\cos\dfrac t4-\sin\dfrac t4\right)-558.02e^{-t/100}$

3 0

3 years ago