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3 years ago

What is the solution to the system of equations below?

Mathematics

1 answer:

kifflom [539]3 years ago

4 0

Answer:

(-5,-14)

Step-by-step explanation:

To find the solution of the system of two equations

$\left\{\begin{array}{l}y=4x+6\\y=2x-4\end{array}\right.$

equate right sides of these two equations:

$4x+6=2x-4$

Rewrite terms with x into the left side and without x into the right side:

$4x-2x=-4-6\\ \\2x=-10\ [\text{divide by 2}]\\ \\x=-5$

Substitute x=-5 into the first equation:

$y=4\cdot (-5)+6=-20+6=-14$

The solution is (-5,-14)

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This seems to be calculus 1.

Question a

We have $-2x^3 + 5x^2 + 9$

m = slope = derivative

Find the derivative / slope of $-2x^3 + 5x^2 + 9$

We do this by differentiating the polynomials. There are a few methods to do this but I am going to use the power rule, which we multiply the constant by the exponent on the variable and subtract one from the exponent.

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Now that we have this information, we can answer question b

Question b

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$m = 4$

Now that we have our slope, use point slope form to find our tangent line

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Now lets do the same for the Point (2, 13)

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show that the sum of the x- and y- intercepts of any tangent line to the curve sqrt(x) sqrt(y) = sqrt(c) is equal to c.

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The answer is √x + √y = √c 
=> 1/(2√x) + 1/(2√y) dy/dx = 0 
=> dy/dx = - √y/√x 

Let (x', y') be any point on the curve 
=> equation of the tangent at that point is 
y - y' = - (√y'/√x') (x - x') 

x-intercept of this tangent is obtained by plugging y = 0 
=> 0 - y' = - (√y'/√x') (x - x') 
=> x = √(x'y') + x' 

y-intercept of the tangent is obtained by plugging x = 0 
=> y - y' = - (√y'/√x') (0 - x') 
=> y = y' + √(x'y') 

Sum of the x and y intercepts 
= √(x'y') + x' + y' + √(x'y') 
= (√x' + √y')^2 
= (√c)^2 (because (x', y') is on the curve => √x' + √y' = √c) 
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