I need help again.lol

Mathematics

1 answer:

Monica [59]2 years ago

8 0

Answer:

$13\ units$

Step-by-step explanation:

One is asked to find the distance between two points on a coordinate plane. The easiest way to do so is to use the distance formula, this formula is the following:

$D=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2$

Where ( $(x_1,y_1)$ ) and ( $(x_2,y_2)$ ) are the points which one needs to find the distance between. In this case, these points are as follows:

$k\ (-3,5)$

$j\ (-8, -7)$

Substitute these points into the formula and solve for the distance between these points,

$D=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2$

$D=\sqrt{((-3)-(-8))^2+((5)-(-7))^2$

Simplify,

$D=\sqrt{((-3)-(-8))^2+((5)-(-7))^2$

$D=\sqrt{(-3+8)^2+(5+7)^2$

$D=\sqrt{(5)^2+(12)^2$

$D=\sqrt{25+144}$

$D=\sqrt{169}$

$D=13$

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Implicit differentiation of 1/x +1/y=5 y(4)= 4/19 y'(4)=?

Aneli [31]

$\frac{1}{x} +\frac{1}{y} = 5\\\\x^{-1}+y^{-1}=5\\$

Above, I changed the fraction form of x and y into exponential form so it is easier to see the differentiation. Now, we can differentiate:

$-1x^{-2}+-1y^{-2}\frac{dy}{dx}=5\\\\\frac{-1}{x^2}-\frac{1}{y^2}\frac{dy}{dx}=5\\\\-\frac{1}{y^2}\frac{dy}{dx}=5+\frac{1}{x^2}\\\\\frac{dy}{dx}=-5y^2-\frac{y^2}{x^2}$

Now that we have dy/dx, we can plug in the x, which is 4, and the y, which is 4/19. We know these values of x and y because your question stated y(4) = 4/19.

$\frac{dy}{dx}=-5(\frac{4}{19})^2-\frac{(\frac{4}{19})^2}{(4)^2}\\\\\frac{dy}{dx}=-5(\frac{16}{361})-\frac{(\frac{16}{361})}{16}\\\\\frac{dy}{dx}=\frac{-80}{361}-\frac{1}{361}\\\\\frac{dy}{dx}=\frac{-81}{361}$

5 0

3 years ago

Question 8: Find the equation of the straight line that:

Scorpion4ik [409]

Answer:

$y=4x+6$

Step-by-step explanation:

Hi there!

Linear equations are typically organized in slope-intercept form: $y=mx+b$ where m is the slope (also called the gradient) and b is the y-intercept (the value of y when x is 0)