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3 years ago

HELP! Which of the following pairs of functions are inverses of each other?

Mathematics

2 answers:

IrinaK [193]3 years ago

8 0

one way is to solve for the invers of the first function

remembe, to solve, replace f(x) or g(x) with y, switch x and y, solve for y and replace it with f⁻¹(x)

f(x)=x/2+8

y=x/2+8

x=y/2+8

x-8=y/2

2x-16=y

f⁻¹(x)=2x-16

nope, not A

f(x)=3x³+16

y=3x³+16

x=3y³+16

x-16=3y³

(x-16)/3=y³

∛((x-16)/3)=y

f⁻¹(x)=∛((x-16)/3)

nope, not the same

not B

f(x)=18/x-9

y=18/x-9

x=18/y-9

x+9=18/y

y(x+9)=18

y=18/(x+9)

f⁻¹(x)=18/(x+9)

correct, the answer is C

answer is C

qaws [65]3 years ago

4 0

Answer:

D.f(x)=3(x/2)-4 and g(x)=2(x+4)/3

Step-by-step explanation:

apexs

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Use two different methods to find an explain the formula for the area of a trapezoid that has parallel sides of length a and B a

evablogger [386]

Answer:

Formula of Trapezoid:

A = (a + b) × h / 2

The formula can be derived in different ways. for now, we have discussed two ways:

1. By using the formula of a triangle

2. By dividing into different sections

Step-by-step explanation:

1. By using the formula of a triangle

One of the ways to explain a formula for an area of a trapezoid using a formula for a triangle can be as follows.

Assume a trapezoid PQRS with lower base SR and upper base PQ (they are parallel) and sides PS and QR.

The image is attached below.

Connect vertices P and R with a diagonal.

Consider triangle ΔPQR as having a base PQ and an altitude from vertex R down to point M on base PQ (RM⊥PQ).

Its area is

S1= $\frac{1}{2} *PQ*RM$

Consider triangle ΔPRS as having a base SR and an altitude from vertex P up to point N on-base SR (PN⊥SR).

Its area is

S2= $\frac{1}{2} *SR*PN$

Altitudes RM and PN are equal and constitute the distance between two parallel bases PQ and SR.

They both are equal to the altitude of the trapezoid h.

Therefore, we can represent areas of our two triangles as

S1= $\frac{1}{2}*PQ*h$

S2= $\frac{1}{2}*SR*h$

Adding them together, we get the area of the whole trapezoid:

S=S1+S2= $\frac{1}{2} (PQ+SR)h,$

which is usually represented in words as "half-sum of the bases times the altitude".

2. By dividing into different sections

Trapezoid PQRS is shown below, with PQ parallel to RS.

Figure 1 - Trapezoid PQRS with PQ parallel to RS(image is attached below.)

We are going to derive the area of a trapezoid by dividing it into different sections.

If we drop another line from Q, then we will have two altitudes namely PT and QU.

Figure 2 - Trapezoid PQRS divided into two triangles and a rectangle. (image is attached below.)

From Figure 2, it is clear that Area of PQRS = Area of PST + Area of PQUT + Area of QRU. We have learned that the area of a triangle is the product of its base and altitude divided by 2, and the area of a rectangle is the product of its length and width. Hence, we can easily compute the area of PQRS. It is clear that

=> $A_{PQRS} = (\frac{ah}{2}) + b_{1}h + \frac{ch}{2}$