4q(-3q^3) simplified

Which is the inverse of the function a(d)=5d-3? And use the definition of inverse functions to prove a(d) and a-1(d) are inverse

Drupady [299]

Answer:

$a'(d) = \frac{d}{5} + \frac{3}{5}$

$a(a'(d)) = a'(a(d)) = d$

Step-by-step explanation:

Given

$a(d) = 5d - 3$

Solving (a): Write as inverse function

$a(d) = 5d - 3$

Represent a(d) as y

$y = 5d - 3$

Swap positions of d and y

$d = 5y - 3$

Make y the subject

$5y = d + 3$

$y = \frac{d}{5} + \frac{3}{5}$

Replace y with a'(d)

$a'(d) = \frac{d}{5} + \frac{3}{5}$

Prove that a(d) and a'(d) are inverse functions

$a'(d) = \frac{d}{5} + \frac{3}{5}$ and $a(d) = 5d - 3$

To do this, we prove that:

$a(a'(d)) = a'(a(d)) = d$

Solving for $a(a'(d))$

$a(a'(d)) = a(\frac{d}{5} + \frac{3}{5})$

Substitute $\frac{d}{5} + \frac{3}{5}$ for d in $a(d) = 5d - 3$

$a(a'(d)) = 5(\frac{d}{5} + \frac{3}{5}) - 3$

$a(a'(d)) = \frac{5d}{5} + \frac{15}{5} - 3$

$a(a'(d)) = d + 3 - 3$

$a(a'(d)) = d$

Solving for: $a'(a(d))$

$a'(a(d)) = a'(5d - 3)$

Substitute 5d - 3 for d in $a'(d) = \frac{d}{5} + \frac{3}{5}$

$a'(a(d)) = \frac{5d - 3}{5} + \frac{3}{5}$

Add fractions

$a'(a(d)) = \frac{5d - 3+3}{5}$

$a'(a(d)) = \frac{5d}{5}$

$a'(a(d)) = d$

Hence:

$a(a'(d)) = a'(a(d)) = d$

7 0

2 years ago

Please help worth 30 points it’s geometry.

Lyrx [107]

That would be a 45 degree angle

3 0

2 years ago

Read 2 more answers

Given gif and GHF, write three proportions involving geometric mean.

almond37 [142]

Answer:

$\frac{z}{h}=\frac{h}{y}\\\\\frac{a}{x}=\frac{x}{z}\\\\\frac{a}{w}=\frac{w}{y}$

Step-by-step explanation:

For this exercise it is important to remember that a Right triangle is a triangle that has an angle that measures 90 degrees.

According to the Altitude Rule, given a Right triangle, if you draw an altitude from the vertex of the angle that measures 90 degrees (The right angle) to the hypotenuse, the measure of that altitude is the geometric mean between the measures of the two segments of the hypotenuse.

In this case, you can identify that the altitude that goes from the vertex of the right angle ( $\angle G=90\°$ ) to the hypotenuse of the triangle, is:

$GF=h$

Then, based on the Altitude Rule, you can set up the following proportion:

$\frac{z}{h}=\frac{h}{y}$

According to the Leg Rule, each leg is the mean proportional between the hypotenuse and the portion of the hypotenuse that is located directly below that leg of the triangle.

Knowing this, you can set up the following proportions:

$\frac{a}{x}=\frac{x}{z}\\\\\frac{a}{w}=\frac{w}{y}$