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3 years ago

H(j + i) - k5 + h when h = 3, i = 4, j = 2, and k = 1 pls put how you did it

Mathematics

1 answer:

VikaD [51]3 years ago

6 0

Answer:

Step-by-step explanation:

First you want to plug in all the numbers where the letters are

h ( j + i )- k(5) + h

3 ( 2 + 4 )- 1(5) + 3

After this step you will have to distribute the 3 outside the parenthesis. Which you do by multiplying everything in side the parenthesis by the outside number which is three.

6 + 12 - 1(5) + 3

Then you just follow PEMDAS to solve for the rest

6 + 12 - 5 + 3

18 - 5 + 3

13 + 3

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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

What is the slop-intercept of the line that passes through the points (7,-5) and (3,-9)?

bagirrra123 [75]

$\bf (\stackrel{x_1}{7}~,~\stackrel{y_1}{-5})\qquad (\stackrel{x_2}{3}~,~\stackrel{y_2}{-9}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{-9-(-5)}{3-7}\implies \cfrac{-9+5}{-4}\implies \cfrac{-4}{-4}\implies 1 \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-(-5)=1(x-7) \\\\\\ y+5=x-7\implies y=x-12$