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Grace [21]
3 years ago
11

The 6:00 a.m. temperatures for four consecutive days in the town of Lincoln were -12.1 °C, -7.8°C, -14,3 C, and 7.2 C. What was

the average 6:00 a,m, temperature for the four days?
Mathematics
2 answers:
slamgirl [31]3 years ago
7 0
Well I would need a picture for this answer since there's a lot of explaining to do unless you can message me if you want
harkovskaia [24]3 years ago
3 0
To find average, you add up all the values then divide the sum by the amount of numbers you added up. I got -6.75 C.
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The perimeter of a rectangle is 38 units. The length<br>the dimensions of the rectangle?​
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6 0
2 years ago
You are given the following equation.
saul85 [17]

Answer:

Step-by-step explanation:

Given the equation  4x²+ 49y² = 196

a) Differentiating implicitly with respect to y, we have;

8x + 98y\frac{dy}{dx} = 0\\98y\frac{dy}{dx}  = -8x\\49y\frac{dy}{dx}  = -4x\\\frac{dy}{dx} = \frac{-4x}{49y}

b)  To solve the equation explicitly for y and differentiate to get dy/dx in terms of x,

First let is make y the subject of the formula from the equation;

If 4x²+ 49y² = 196

49y² = 196 - 4x²

y^{2} =  \frac{196}{49}  - \frac{4x^{2} }{49} \\y = \sqrt{\frac{196}{49}  - \frac{4x^{2} }{49} \\} \\

Differentiating y with respect to x using the chain rule;

Let u=  \frac{196}{49}  - \frac{4x^{2} }{49}

y =  \sqrt{u} \\y =u^{1/2} \\

\frac{dy}{dx}  = \frac{dy}{du} * \frac{du}{dx}

\frac{dy}{du} = \frac{1}{2}u^{-1/2} \\

\frac{du}{dx} =  0 - \frac{8x}{49} \\\frac{du}{dx} =\frac{-8x}{49} \\\frac{dy}{dx} = \frac{1}{2} ( \frac{196}{49}  - \frac{4x^{2} }{49})^{-1/2} *  \frac{-8x}{49}\\\frac{dy}{dx} = \frac{1}{2} (  \frac{196-4x^{2} }{49})^{-1/2} *  \frac{-8x}{49}\\\frac{dy}{dx} = \frac{1}{2} ( \sqrt{ \frac{49}{196-4x^{2} })} *  \frac{-8x}{49}\\\frac{dy}{dx} = \frac{1}{2} *{ \frac{7}\sqrt {196-4x^{2} }} *  \frac{-8x}{49}\\

\frac{dy}{dx} = \frac{-4x}{7\sqrt{196-4x^{2} } }

c) From the solution of the implicit differentiation in (a)

\frac{dy}{dx} = \frac{-4x}{49y}

Substituting y = \sqrt{\frac{196}{49}  - \frac{4x^{2} }{49} \\ into the equation to confirm the answer of (b) can be shown as follows

\frac{dy}{dx} = \frac{-4x}{49\sqrt{\frac{196-4x^{2} }{49} } }\\\frac{dy}{dx}  =  \frac{-4x}{49\sqrt{196-4x^{2}}/7} }\\\\\frac{dy}{dx}  = \frac{-4x}{7\sqrt{196-4x^{2}}}

This shows that the answer in a and b are consistent.

6 0
3 years ago
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