Help with following questions please.

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

A book is in the shape of a right rectangular prism. It measures 8 1/2 inches tall, 7 inches wide, and 2 1/2 inches thick. What

Alenkasestr [34]

Hello!

The formula for volume of a rectangular prism is:

V = lwh

Let's convert the dimensions to decimals, and then use the formula to solve.

8 1/2 = 8.5 as a decimal.

7 = 7 as a decimal.

2 1/2 = 2.5 as a decimal.

Multiply:

V = 8.5 × 7 × 2.5

V = 59.5 × 2.5

V = 148.75

148.75 = 148 3/4 as a fraction because if you convert 3/4 to a decimal by dividing the numerator by the denominator, it equals 0.75.

ANSWER:

The volume of the book is 148 3/4 inches cubed.

6 0

3 years ago

Read 2 more answers

Write the equation in spherical coordinates. (a) 4x2 − 3x + 4y2 + 4z2 = 0

spayn [35]

4x^2-3x+4y^2+4z^2=0
here we shall proceed as follows:
x=ρcosθsinφ
y=ρsinθsinφ
z=ρcosφ
thus
4x^2-3x+4y^2+4z^2=
4(ρcosθsinφ)^2-3(ρcosθsinφ)+4(ρsinθsinφ)^2+4(ρcosφ)
but
ρ=1/4cosθsinφ

hence we shall have:
4x^2-3x+4y^2+4z^2
=1/4cosθsinθ(cosθ(4-3sinφ))+4sin^2(φ)

6 0

3 years ago

Is the centroid of a triangle equidistant from the three vertices?

Xelga [282]

The point that is equidistant<span> to all sides of a </span>triangle is <span>called the incenter.

</span>

4 0

3 years ago

Issuing a note payable for cash automatically results in a

Dmitry_Shevchenko [17]

Sry not sure this one is very confusing

5 0

3 years ago