All categories

3 years ago

Solve r = 4/3(p-q) for p

Mathematics

1 answer:

ddd [48]3 years ago

6 0

Answer:

q+3/4r=p

Step-by-step explanation:

r=4/3(p-q)

Distribute the 4/3

r=4/3p-4/3q

Add 4/3q to each side

4/3q+r=4/3p

Multiply ALL variables by 3/4 (undoes the 4/3)

q+3/4r=p

You might be interested in

(-2)2+3 × (-5) please helpe me answer this question

Tpy6a [65]

Answer:

(−2)(2)+(3)(−5)

=−19

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

Need help with Calculus 1 inverse trig functions

lidiya [134]

Answer:

$\displaystyle y'=3\frac{1+\frac{x}{\sqrt{1+x^2}}}{2+2x^2+2x\sqrt{1+x^2}}$

Step-by-step explanation:

The Derivative of a Function

The derivative of f, also known as the instantaneous rate of change, or the slope of the tangent line to the graph of f, can be computed by the definition formula

$\displaystyle f'(x)=\lim\limits_{\Delta x \rightarrow 0}\frac{f(x+\Delta x)-f(x)}{\Delta x}$

There are tables where the derivative of all known functions are provided for an easy calculation of specific functions.

The derivative of the inverse tangent is given as

$\displaystyle (tan^{-1}u)'=\frac{u'}{1+u^2}$

Where u is a function of x as provided:

$y=3tan^{-1}(x+\sqrt{1+x^2})$

If we set

$u=(x+\sqrt{1+x^2})$

Then

$\displaystyle u'=1+\frac{2x}{2\sqrt{1+x^2}}$

$\displaystyle u'=1+\frac{x}{\sqrt{1+x^2}}$

Taking the derivative of y

$y'=3[tan^{-1}(x+\sqrt{1+x^2})]'$

Using the change of variables

$\displaystyle y'=3[tan^{-1}u]'=3\frac{u'}{1+u^2}$

$\displaystyle y'=3\frac{u'}{1+u^2}=3\frac{1+\frac{x}{\sqrt{1+x^2}}}{1+(x+\sqrt{1+x^2})^2}$

Operating

$\displaystyle y'=3\frac{1+\frac{x}{\sqrt{1+x^2}}}{1+x^2+2x\sqrt{1+x^2}+1+x^2}$

$\boxed{\displaystyle y'=3\frac{1+\frac{x}{\sqrt{1+x^2}}}{2+2x^2+2x\sqrt{1+x^2}}}$

8 0

3 years ago

Prove that DJKL~ DJMN using SAS Similarity Theorem. Plot the points J (1,1), K(2,3), L(4,1) and J (1,1), M(3,5), N(7,1). Draw DJ

dolphi86 [110]

Answer and Step-by-step explanation: The triangles are plotted and shown in the attachment.

SAS Similarity Theorem is by definition: if two sides in one triangle are proportional to two sides of another triangle and the angles formed by those sides in each triangle is congruent, the triangles are similar.

For the triangles on the grid, we know that ΔJKL and ΔJMN have a congruent angle in J as shown in the image. To prove they are similar, we find the slope of sides KL and MN:

Slope of KL:

slope = $\frac{y_{K} - y_{L} }{x_{K} - x_{L} }$