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2 years ago

Find the equation of the surface. a standard cone with vertex at (6, 0 , 0) opening up on the positive x-axis.

Mathematics

1 answer:

DerKrebs [107]2 years ago

7 0

Solution: Since, it’s a standard cone opening up on the positive x- axis

Therefore, x=√y2+z2

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Allison is 5 ft tall. One afternoon her shadow measured 2.5 ft. At the same time, the shadow of her favorite tree was 14.5 ft. W

luda_lava [24]

At the same time, the sun casts a shadow on the girl and tree creating two similar right triangles. You can use a ratio of girl's height to her shadow will be the same as the tree's height to it's shadow.

5 / 2.5 = T / 14.5
2 = T / 14.5
2 * 14.5 = T
29 ft = Tree

8 0

2 years ago

Drew creates a table of ordered pairs representing the width and area of a dog pen. Which situation could describe Drew’s plans

Feliz [49]

Answer:

the answer is A :)

Step-by-step explanation:

8 0

2 years ago

Read 2 more answers

A 400 gallon tank initially contains 100 gal of brine containing 50 pounds of salt. Brine containing 1 pound of salt per gallon

posledela

Answer:

The amount of salt in the tank when it is full of brine is 393.75 pounds.

Step-by-step explanation:

This is a mixing problem. In these problems we will start with a substance that is dissolved in a liquid. Liquid will be entering and leaving a holding tank. The liquid entering the tank may or may not contain more of the substance dissolved in it. Liquid leaving the tank will of course contain the substance dissolved in it. If Q(t) gives the amount of the substance dissolved in the liquid in the tank at any time t we want to develop a differential equation that, when solved, will give us an expression for Q(t).

The main equation that we’ll be using to model this situation is:

Rate of change of Q(t) = Rate at which Q(t) enters the tank – Rate at which Q(t) exits the tank

where,

Rate at which Q(t) enters the tank = (flow rate of liquid entering) x

(concentration of substance in liquid entering)

Rate at which Q(t) exits the tank = (flow rate of liquid exiting) x

(concentration of substance in liquid exiting)

Let y(t) be the amount of salt (in pounds) in the tank at time t (in seconds). Then we can represent the situation with the below picture.

Then the differential equation we’re after is

$\frac{dy}{dt} = (Rate \:in)- (Rate \:out)\\\\\frac{dy}{dt} = 5 \:\frac{gal}{s} \cdot 1 \:\frac{pound}{gal}-3 \:\frac{gal}{s}\cdot \frac{y(t)}{V(t)} \:\frac{pound}{gal}\\\\\frac{dy}{dt} =5\:\frac{pound}{s}-3 \frac{y(t)}{V(t)} \:\frac{pound}{s}$

$V(t)$ is the volume of brine in the tank at time t. To find it we know that at time 0 there were 100 gallons, 5 gallons are added and 3 are drained, and the net increase is 2 gallons per second. So,

$V(t)=100 + 2t$

We can then write the initial value problem:

$\frac{dy}{dt} =5-\frac{3y}{100+2t} , \quad y(0)=50$

We have a linear differential equation. A first-order linear differential equation is one that can be put into the form

$\frac{dy}{dx}+P(x)y =Q(x)$

where P and Q are continuous functions on a given interval.

In our case, we have that

$\frac{dy}{dt}+\frac{3y}{100+2t} =5 , \quad y(0)=50$

The solution process for a first order linear differential equation is as follows.

Step 1: Find the integrating factor, $\mu \left( x \right)$ , using $\mu \left( x \right) = \,{{\bf{e}}^{\int{{P\left( x \right)\,dx}}}$

$\mu \left( t \right) = \,{{e}}^{\int{{\frac{3}{100+2t}\,dt}}}\\\int \frac{3}{100+2t}dt=\frac{3}{2}\ln \left|100+2t\right|\\\\\mu \left( t \right) =e^{\frac{3}{2}\ln \left|100+2t\right|}\\\\\mu \left( t \right) =(100+2t)^{\frac{3}{2}$

Step 2: Multiply everything in the differential equation by $\mu \left( x \right)$ and verify that the left side becomes the product rule $\left( {\mu \left( t \right)y\left( t \right)} \right)'$ and write it as such.

$\frac{dy}{dt}\cdot \left(100+2t\right)^{\frac{3}{2}}+\frac{3y}{100+2t}\cdot \left(100+2t\right)^{\frac{3}{2}}=5 \left(100+2t\right)^{\frac{3}{2}}\\\\\frac{dy}{dt}\cdot \left(100+2t\right)^{\frac{3}{2}}+3y\cdot \left(100+2t\right)^{\frac{1}{2}}=5 \left(100+2t\right)^{\frac{3}{2}}\\\\\frac{dy}{dt}(y \left(100+2t\right)^{\frac{3}{2}})=5\left(100+2t\right)^{\frac{3}{2}}$

Step 3: Integrate both sides.

$\int \frac{dy}{dt}(y \left(100+2t\right)^{\frac{3}{2}})dt=\int 5\left(100+2t\right)^{\frac{3}{2}}dt\\\\y \left(100+2t\right)^{\frac{3}{2}}=(100+2t)^{\frac{5}{2} }+ C$

Step 4: Find the value of the constant and solve for the solution y(t).

$50 \left(100+2(0)\right)^{\frac{3}{2}}=(100+2(0))^{\frac{5}{2} }+ C\\\\100000+C=50000\\\\C=-50000$

$y \left(100+2t\right)^{\frac{3}{2}}=(100+2t)^{\frac{5}{2} }-50000\\\\y(t)=100+2t-\frac{50000}{\left(100+2t\right)^{\frac{3}{2}}}$

Now, the tank is full of brine when:

$V(t) = 400\\100+2t=400\\t=150$

The amount of salt in the tank when it is full of brine is

$y(150)=100+2(150)-\frac{50000}{\left(100+2(150)\right)^{\frac{3}{2}}}\\\\y(150)=393.75$

6 0

2 years ago

What do the symbols of mirrors and glass represent in this

luda_lava [24]

Question:

Read the excerpt from Julius Caesar, act 1, scene 2.

CASSIUS: ‘Tis just; And it is very much lamented, Brutus, That you have no such mirrors as will turn Your hidden worthiness into your eye, That you might see your shadow. I have heard 5 Where many of the best respect in Rome– Except immortal Caesar‐speaking of Brutus, And groaning underneath this age’s yoke, Have wished that noble Brutus had his eyes.

BRUTUS: Into what dangers would you lead me, Cassius, 10 That you would have me seek into myself For that which is not in me?

CASSIUS: Therefore, good Brutus, be prepared to hear. And since you know you cannot see yourself So well as by reflection, I, your glass, 15 Will modestly discover to yourself That of yourself which you yet know not of.

Answer:

The correct choice is D)

Explanation:

Cassius speaks of Brutus as one who is unable to see or know his own value and presumes to help him therewith. He does this by pointing out that many of the well respected people in Rome wish that he were in Caesars shoes as King.

Cheers!

7 0

2 years ago

Use technology to approximate the solution(s) to the system of equations to the nearest tenth of a unit.

kherson [118]

Answer:

A.) (3.6, 0.6)
D.) (2.6, 0.4)

Step-by-step explanation:

See below for a graph.

___

Choices B, C, E can be eliminated on the basis that neither x nor g(x) can be negative. The domain of f(x) is x>0; the range of g(x) is x≥0.

7 0

2 years ago