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3 years ago

Evaluate √x^2 − y^2 when x^2 =33 and y^2 =15

Mathematics

1 answer:

Andre45 [30]3 years ago

3 0

Answer:

3√2

Step-by-step explanation:

In this question, we are asked to get the value of √ ( x²-y² ), while the value of x² and y² are given to be 33 and 15 respectively.

Hence, x² =33 and y² = 15.

Putting the values of x² and y² in the above expression we get,

√ ( x²-y² ) = √ ( 33-15) = √18 =3√2. (Answer)

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Tresset [83]

Answer:

D is the correct answer

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3 years ago

Read 2 more answers

Please help thank you

Elodia [21]

Answer:

a1 = 6

a2 = 15

a3 = 24

a4 = 33

Step-by-step explanation:

Let k =1

a1 =-3 +9(1) = -3 +9 = 6

Let k = 2

a2 = -3 +9(2) =-3+18 = 15

Let k =3

a3 =-3 +9(3) = -3 +27 = 24

Let k = 4

a4 = -3 +9(4) =-3+36 = 33

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3 years ago

Problem 10: A tank initially contains a solution of 10 pounds of salt in 60 gallons of water. Water with 1/2 pound of salt per g

AysviL [449]

Answer:

The quantity of salt at time t is $m_{salt} = (60)\cdot (30 - 29.833\cdot e^{-\frac{t}{10} })$ , where t is measured in minutes.

Step-by-step explanation:

The law of mass conservation for control volume indicates that:

$\dot m_{in} - \dot m_{out} = \left(\frac{dm}{dt} \right)_{CV}$

Where mass flow is the product of salt concentration and water volume flow.

The model of the tank according to the statement is:

$(0.5\,\frac{pd}{gal} )\cdot \left(6\,\frac{gal}{min} \right) - c\cdot \left(6\,\frac{gal}{min} \right) = V\cdot \frac{dc}{dt}$

Where:

$c$ - The salt concentration in the tank, as well at the exit of the tank, measured in $\frac{pd}{gal}$ .

$\frac{dc}{dt}$ - Concentration rate of change in the tank, measured in $\frac{pd}{min}$ .

$V$ - Volume of the tank, measured in gallons.

The following first-order linear non-homogeneous differential equation is found:

$V \cdot \frac{dc}{dt} + 6\cdot c = 3$

$60\cdot \frac{dc}{dt} + 6\cdot c = 3$

$\frac{dc}{dt} + \frac{1}{10}\cdot c = 3$

This equation is solved as follows:

$e^{\frac{t}{10} }\cdot \left(\frac{dc}{dt} +\frac{1}{10} \cdot c \right) = 3 \cdot e^{\frac{t}{10} }$

$\frac{d}{dt}\left(e^{\frac{t}{10}}\cdot c\right) = 3\cdot e^{\frac{t}{10} }$

$e^{\frac{t}{10} }\cdot c = 3 \cdot \int {e^{\frac{t}{10} }} \, dt$

$e^{\frac{t}{10} }\cdot c = 30\cdot e^{\frac{t}{10} } + C$

$c = 30 + C\cdot e^{-\frac{t}{10} }$

The initial concentration in the tank is:

$c_{o} = \frac{10\,pd}{60\,gal}$

$c_{o} = 0.167\,\frac{pd}{gal}$

Now, the integration constant is:

$0.167 = 30 + C$

$C = -29.833$

The solution of the differential equation is:

$c(t) = 30 - 29.833\cdot e^{-\frac{t}{10} }$

Now, the quantity of salt at time t is:

$m_{salt} = V_{tank}\cdot c(t)$

$m_{salt} = (60)\cdot (30 - 29.833\cdot e^{-\frac{t}{10} })$

Where t is measured in minutes.

7 0

3 years ago

Round to the nearest half 6.385

iragen [17]

Answer:

6.5

Step-by-step explanation:

I hope you have a great day!!

3 0

3 years ago

The front and back covers of a textbook are each 0.3 cm thick. Between the covers are 200 sheets of paper, each 0.008 cm thick.

zysi [14]

Answer:

22 cm

Step-by-step explanation:

Start off by multiplying the number of pages by the thickness.

200*0.008=1.6

Add the thickness of the front and back cover.

1.6+0.6=2.2

Multiply by 10.

2.2*10=22 cm

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3 years ago