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Step2247 [10]
2 years ago
8

Find the slope of the line passing through the points (-6,7) and (-8,5)

Mathematics
1 answer:
kykrilka [37]2 years ago
4 0

Answer:

Slope = 1

Step-by-step explanation:

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Derivative of<br><img src="https://tex.z-dn.net/?f=%20%5Cfrac%7B%20%7B3x%7D%5E%7B2%7D%20-%202x%20-%201%20%7D%7B%20%7Bx%7D%5E%7B2
Anastaziya [24]

Answer:

\displaystyle  \frac{dy}{dx} =    \frac{2x + 2}{x^3}

Step-by-step explanation:

we would like to figure out the derivative of the following:

\displaystyle  \frac{ { 3x }^{2} - 2x - 1 }{ {x}^{2} }

to do so, let,

\displaystyle y =  \frac{ { 3x }^{2} - 2x - 1 }{ {x}^{2} }

By simplifying we acquire:

\displaystyle y =  3 -  \frac{2}{x}  -  \frac{1}{ {x}^{2} }

use law of exponent which yields:

\displaystyle y =  3 -  2 {x}^{ - 1}  -   { {x}^{  - 2} }

take derivative in both sides:

\displaystyle  \frac{dy}{dx} =  \frac{d}{dx}  (3 -  2 {x}^{ - 1}  -   { {x}^{  - 2} } )

use sum derivation rule which yields:

\rm\displaystyle  \frac{dy}{dx} =  \frac{d}{dx}  3 -   \frac{d}{dx} 2 {x}^{ - 1}  -     \frac{d}{dx} {x}^{  - 2}

By constant derivation we acquire:

\rm\displaystyle  \frac{dy}{dx} =  0 -   \frac{d}{dx} 2 {x}^{ - 1}  -     \frac{d}{dx} {x}^{  - 2}

use exponent rule of derivation which yields:

\rm\displaystyle  \frac{dy}{dx} =  0 -   ( - 2 {x}^{ - 1 -1} ) -     ( - 2 {x}^{  - 2 - 1} )

simplify exponent:

\rm\displaystyle  \frac{dy}{dx} =  0 -   ( - 2 {x}^{ -2} ) -     ( - 2 {x}^{  - 3} )

two negatives make positive so,

\displaystyle  \frac{dy}{dx} =   2 {x}^{ -2} +      2 {x}^{  - 3}

<h3>further simplification if needed:</h3>

by law of exponent we acquire:

\displaystyle  \frac{dy}{dx} =   \frac{2 }{x^2}+       \frac{2}{x^3}

simplify addition:

\displaystyle  \frac{dy}{dx} =    \frac{2x + 2}{x^3}

and we are done!

5 0
2 years ago
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Answer:

////

Step-by-step explanation:

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VARVARA [1.3K]

See attached graph: v is on the y-axis and t is on the x-axis

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What is the solution to the system of equations graphed below
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The point where the lines intersect is the point that satisfies both equations.

It is (1, 5), selection C.
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