Which of the following is irrational?

The base and height of a rectangle are 134.4 and 98.46, find the sum of the lengths of the two diagonals of the rectangle?

Len [333]

Answer:

232.86

Step-by-step explanation:

134.4

+ 98.46

=232.86

232.86

+232.86

=465.72

Sorry that's all I know po

3 0

2 years ago

Find roots of the equation and state the multiplicity

olya-2409 [2.1K]

$({x + 2})^{3} = {x}^{3} + 6 {x}^{2} + 12x + 8$
so the answer is
$x = - 2$

7 0

3 years ago

Ashley earn 60 points every time she shops at a grocery store she needs a total of 2580 points to receive a prize so far she has

lutik1710 [3]

Answer: 35 more times
How: (if she gets 60 points for each grocery store visit then first do 2580-480=2100. she still needs 2100 points so then 2100/60 = 35 grocery visits!)

4 0

3 years ago

A swimming pool with a volume of 30,000 liters originally contains water that is 0.01% chlorine (i.e. it contains 0.1 mL of chlo

SpyIntel [72]

Answer:

$R_{in}=0.2\dfrac{mL}{min}$

$C(t)=\dfrac{A(t)}{30000}$

$R_{out}= \dfrac{A(t)}{1500} \dfrac{mL}{min}$

$A(t)=300+2700e^{-\dfrac{t}{1500}},$ A(0)=3000$

Step-by-step explanation:

The volume of the swimming pool = 30,000 liters

(a) Amount of chlorine initially in the tank.

It originally contains water that is 0.01% chlorine.

0.01% of 30000=3000 mL of chlorine per liter

A(0)= 3000 mL of chlorine per liter

(b) Rate at which the chlorine is entering the pool.

City water containing 0.001%(0.01 mL of chlorine per liter) chlorine is pumped into the pool at a rate of 20 liters/min.

$R_{in}=$ (concentration of chlorine in inflow)(input rate of the water)

$=(0.01\dfrac{mL}{liter}) (20\dfrac{liter}{min})\\R_{in}=0.2\dfrac{mL}{min}$

(c) Concentration of chlorine in the pool at time t

Volume of the pool =30,000 Liter

$Concentration, C(t)= \dfrac{Amount}{Volume}\\C(t)=\dfrac{A(t)}{30000}$

(d) Rate at which the chlorine is leaving the pool

$R_{out}=$ (concentration of chlorine in outflow)(output rate of the water)

$= (\dfrac{A(t)}{30000})(20\dfrac{liter}{min})\\R_{out}= \dfrac{A(t)}{1500} \dfrac{mL}{min}$

(e) Differential equation representing the rate at which the amount of sugar in the tank is changing at time t.

$\dfrac{dA}{dt}=R_{in}-R_{out}\\\dfrac{dA}{dt}=0.2- \dfrac{A(t)}{1500}$

We then solve the resulting differential equation by separation of variables.

$\dfrac{dA}{dt}+\dfrac{A}{1500}=0.2\\$The integrating factor: e^{\int \frac{1}{1500}dt} =e^{\frac{t}{1500}}\\$Multiplying by the integrating factor all through\\\dfrac{dA}{dt}e^{\frac{t}{1500}}+\dfrac{A}{1500}e^{\frac{t}{1500}}=0.2e^{\frac{t}{1500}}\\(Ae^{\frac{t}{1500}})'=0.2e^{\frac{t}{1500}}$

Taking the integral of both sides

$\int(Ae^{\frac{t}{1500}})'=\int 0.2e^{\frac{t}{1500}} dt\\Ae^{\frac{t}{1500}}=0.2*1500e^{\frac{t}{1500}}+C, $(C a constant of integration)\\Ae^{\frac{t}{1500}}=300e^{\frac{t}{1500}}+C\\$Divide all through by e^{\frac{t}{1500}}\\A(t)=300+Ce^{-\frac{t}{1500}}$

Recall that when t=0, A(t)=3000 (our initial condition)

$3000=300+Ce^{0}\\C=2700\\$Therefore:\\A(t)=300+2700e^{-\dfrac{t}{1500}}$

3 0

3 years ago

What is 6/100 + 2/10

Andreyy89

Answer:

$\frac{26}{100}$

Step-by-step explanation:

$\frac{6}{100}$ can stay untouched, but $\frac{2}{10}$ has to be changed to $\frac{20}{100}$ . Now that they have they have the same denominator, you can add them. $\frac{6}{100} + \frac{20}{100} = \frac{26}{100}$ , which can be simplified, which is $\frac{13}{50}$ .

3 0

3 years ago

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