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3 years ago

Can anyone help me understand this

Mathematics

1 answer:

bagirrra123 [75]3 years ago

5 0

Answer:

(x, y) = (2, 9)

Step-by-step explanation:

For the triangles to be congruent, the hypotenuses must be the same length:

y = x + 7

and the marked leg must be the same length in each triangle:

y -3 = 4x -2

These are two equations in two unknowns (a "system" of equations) that can be solved in any of the usual ways. Since the first equation gives an expression for y, it is convenient to substitute that into the second equation:

(x +7) -3 = 4x -2

x +4 = 4x -2 . . . . . . collect terms

x +6 = 4x . . . . . . . . .add 2

6 = 3x . . . . . . . . . . . subtract x

2 = x . . . . . . . . . . . . divide by 3

y = 2 + 7 = 9 . . . . . .substitute for x in the first equation

The values you're looking for are x = 2, y = 9.

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Write an algebraic expression for<br><br> "10 less than h"<br><br> Answer (no spaces in answer)

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Answer:

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Suppose f(x) = 4x – 2 and g(x) = -2x + 1. Find the value of f(g(3)).

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4 years ago

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3 0

3 years ago

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

3 years ago