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3 years ago

The function N(X) = 90(0.86)* + 69 can be used to predict the temperature of a

Mathematics

1 answer:

dezoksy [38]3 years ago

4 0

Answer: -8.93

Step-by-step explanation:

Given the function:

N(X) = 90(0.86)^x+ 69

X in the interval [0, 6]

For X = 0

N(0) = 90(0.86)^0 + 69

90 + 69 = 159

For X = 6

N(6) = 90(0.86)^6 + 69

90(0.404567235136) + 69

= 105.41105116224

Therefore, average change of change in temperature ;

(temp 2 - temp 1) / ( time 2 - time)

(105.41105116224 - 159) / (6 - 0)

= - 53.58894883776 / 6

= - 8.93149147296

= - 8.93

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If 2y^2+2=x^2, then find d^2y/dx^2 at the point (-2, -1) in simplest form.

horrorfan [7]

Answer:

$\frac{d^2y}{dx^2}_{(-2, -1)}=\frac{1}{2}$

Step-by-step explanation:

We have the equation:

$2y^2+2=x^2$

And we want to find d²y/dx² at the point (-2, -1).

So, let's take the derivative of both sides with respect to x:

$\frac{d}{dx}[2y^2+2]=\frac{d}{dx}[x^2]$

On the left, let's implicitly differentiate:

$4y\frac{dy}{dx}=\frac{d}{dx}[x^2]$

Differentiate normally on the left:

$4y\frac{dy}{dx}=2x$

Solve for the first derivative. Divide both sides by 4y:

$\frac{dy}{dx}=\frac{x}{2y}$

Now, let's take the derivative of both sides again:

$\frac{d}{dx}[\frac{dy}{dx}]=\frac{d}{dx}[\frac{x}{2y}]$

We will need to use the quotient rule:

$\frac{d}{dx}[f/g]=\frac{f'g-fg'}{g^2}$

So:

$\frac{d^2y}{dx^2}=\frac{\frac{d}{dx}[(x)](2y)-x\frac{d}{dx}[(2y)]}{(2y)^2}$

Differentiate:

$\frac{d^2y}{dx^2}=\frac{(1)(2y)-x(2\frac{dy}{dx})}{4y^2}$

Simplify:

$\frac{d^2y}{dx^2}=\frac{2y-2x\frac{dy}{dx}}{4y^2}$

Substitute x/2y for dy/dx. This yields:

$\frac{d^2y}{dx^2}=\frac{2y-2x\frac{x}{2y}}{4y^2}$

Simplify:

$\frac{d^2y}{dx^2}=\frac{2y-\frac{2x^2}{2y}}{4y^2}$

Simplify. Multiply both the numerator and denominator by 2y. So:

$\frac{d^2y}{dx^2}=\frac{4y^2-2x^2}{8y^3}$

Reduce. Therefore, our second derivative is:

$\frac{d^2y}{dx^2}=\frac{2y^2-x^2}{4y^3}$

We want to find the second derivative at the point (-2, -1).

So, let's substitute -2 for x and -1 for y. This yields:

$\frac{d^2y}{dx^2}_{(-2, -1)}=\frac{2(-1)^2-(-2)^2}{4(-1)^3}$

Evaluate:

$\frac{d^2y}{dx^2}_{(-2, -1)}=\frac{2(1)-(4)}{4(-1)}$

Multiply:

$\frac{d^2y}{dx^2}_{(-2, -1)}=\frac{2-4}{-4}$

Subtract:

$\frac{d^2y}{dx^2}_{(-2, -1)}=\frac{-2}{-4}$

Reduce. So, our answer is:

$\frac{d^2y}{dx^2}_{(-2, -1)}=\frac{1}{2}$

And we're done!

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