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3 years ago

PLZ help I need to turn this in by today also!!

Mathematics

2 answers:

Vlad [161]3 years ago

8 0

Answer:

you cant view any of the pictures in this question. (is it just my computer?)

Romashka [77]3 years ago

8 0

I would answer but I can’t view the pictures

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What is the value of the expression below when y = 5?

muminat

Answer:

Step-by-step explanation:

10y+6

10(5)+6

10*5=50

50+6=56

3 0

3 years ago

Read 2 more answers

Determine an expression for dy<br> dx given that x = sin3<br> (t)andy = cos3<br> (t)

shtirl [24]

Answer:

$\frac{dy}{dx} =-\sqrt[3]{\frac{y}{x} }$

Step-by-step explanation:

Recall that using the chain rule we can state:

$\frac{dy}{dt} =\frac{dy}{dx}*\frac{dx}{dt}$

and therefore solve for dy/dx as long as dx/dt is different from zero.

Then we find dy/dt and dx/dt,

Given that

$x=sin^3(t)\\dx/dt = 3 sin^2(t)* cos(t)$

And similarly:

$y=cos^3(t)\\dy/dt=-3\,cos^2(t)*sin(t)$

Therefore, dy/dx can be determined by the quotient of the expressions we just found:

$\frac{dy}{dx} =\frac{dy/dt}{dx/dt} =\frac{-3\,cos^2(t)*sin(t)}{3\,sin^2(t)*cos(t)} =-\frac{cos(t)}{sin(t)}$

now notice that we can find $cos(t) = \sqrt[3]{y}$ from the expression for y,

and $sin(t) = \sqrt[3]{x}$ from its expression for x.

Therefore dy/dx can be written in terms of x and y as:

$\frac{dy}{dx} =-\frac{cos(t)}{sin(t)}=-\sqrt[3]{\frac{y}{x} }$

4 0

3 years ago

A certain country has $10 billion in paper currency in circulation, and each day $50 million comes into the country's banks. The

mash [69]

Answer:

$\bf x(t)=10(1-e^{-0.005*t})$

Step-by-step explanation:

The differential equation

$\bf \displaystyle\frac{dx}{dt}=0.005(10-x)$

can be solved by separation of variables. Write the equation as

$\bf \displaystyle\frac{dx}{10-x}=0.005dt$

Integrate on both sides

$\bf \int\displaystyle\frac{dx}{10-x}=\int0.005dt\Rightarrow -ln(10-x)=0.005t+C\Rightarrow\\\\\Rightarrow ln(10-x)^{-1}=0.005t+C\Rightarrow (10-x)^{-1}=e^{0.005t}e^C$

where C is a constant.

$\bf e^C$ is also a constant and we will keep calling it C, (there is no reason to change the letter). We have then

$\bf (10-x)^{-1}=Ce^{0.005t}\Rightarrow \displaystyle\frac{1}{10-x}=Ce^{0.005t}\Rightarrow 10-x=\displaystyle\frac{1}{Ce^{0.005t}}\Rightarrow\\\\\Rightarrow x(t)=10-(1/C)e{-0.005t}$