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3 years ago

What’s the most common Multiple for 3 and 5

Mathematics

1 answer:

Ahat [919]3 years ago

8 0

Answer:

should be 15...

Step-by-step explanation:

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PLEASE HELP MEEE

STALIN [3.7K]

If there are 16 ounces in a pound, and 8 ounces costs $8, The best answer would be D. $16 per pound

4 0

4 years ago

Stan pays a 10% deposit to put a pool table on lay-by.

Lera25 [3.4K]

Answer:

$1431

Step-by-step explanation:

If Stan pays a 10% deposit, he pays $159. 10% of $1590 is simply 0.1 * 1590 = 159. Assuming this is the only amount he pays, he then simply needs to pay the full price minus $159. We can find this by simply subtracting 159 from 1590 to get 1431. Stan still needs to pay $1431.

6 0

2 years ago

What is the expanded form of 3×(x+3)^2

saul85 [17]

3x(x+3)^2
3x(x^2 + 6x + 9)
= 3x^3 + 18x^2 + 27x

answer
3x^3 + 18x^2 + 27x

7 0

3 years ago

you are the sales manager for solar panel company. you receive a commission based on the total sales of all the salespeople you

kondaur [170]

Answer:

The total commission is $8905

Step-by-step explanation:

The commission on first $80000 = 2%

The commission on next $80000 = 4%

The commission on the sales of more than $160000 = 5%

Now we have to calculate the total commission on the amount of $242100. Let divide the amount in three parts, $80000 + $80000 + $82100 = 242100

Now total commission = $80000×2% + $80000 ×4% + $82100×5%

Total commission = $8905

7 0

3 years ago

A large tank is filled to capacity with 300 gallons of pure water. brine containing 5 pounds of salt per gallon is pumped into t

Licemer1 [7]

If $A(t)$ is the amount of salt in the tank at time $t$ , then the rate at which this amount changes over time is given by the ODE

$A'(t)=\dfrac{3\text{ gal}}{1\text{ min}}\cdot\dfrac{5\text{ lb}}{1\text{ gal}}-\dfrac{3\text{ gal}}{1\text{ min}}\cdot\dfrac{A(t)\text{ lb}}{300+(3-3)t\text{ gal}}$

$A'(t)+\dfrac1{100}A(t)=15$

We're told that the tank initially starts with no salt in the water, so $A(0)=0$ .

Multiply both sides by an integrating factor, $e^{t/100}$ :

$e^{t/100}A'(t)+\dfrac1{100}e^{t/100}A(t)=15e^{t/100}$
$\left(e^{t/100}A(t)\right)'=15e^{t/100}$
$e^{t/100}A(t)=1500e^{t/100}+C$
$A(t)=1500+Ce^{-t/100}$

Since $A(0)=0$ , we have

$0=1500+C\implies C=-1500$

so that the amount of salt in the tank over time is given by

$A(t)=1500(1-e^{-t/100})$

After 10 minutes, the amount of salt in the tank is

$A(10)=1500(1-e^{-1/10})\approx142.74\text{ lb}$

8 0

4 years ago