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jeka94
3 years ago
15

Find the value of h(45) for the function below. h(x)=1/9(90 - x). A. 15 B. 5 C. -35 D. -315

Mathematics
1 answer:
kvasek [131]3 years ago
7 0

Answer:

Inverse Functions

One-to-one

Suppose f : A ⇥ B is a function. We call f one-to-one if every distinct

pair of objects in A is assigned to a distinct pair of objects in B. In other

words, each object of the target has at most one object from the domain

assigned to it.

There is a way of phrasing the previous definition in a more mathematical

language: f is one-to-one if whenever we have two objects a, c ⇤ A with

a ⌅= c, we are guaranteed that f(a) ⌅= f(c).

Example. f : R ⇥ R where f(x) = x2 is not one-to-one because 3 ⌅= 3

and yet f(3) = f(3) since f(3) and f(3) both equal 9.

Horizontal line test

If a horizontal line intersects the graph of f(x) in more than one point,

then f(x) is not one-to-one.

The reason f(x) would not be one-to-one is that the graph would contain

two points that have the same second coordinate – for example, (2, 3) and

(4, 3). That would mean that f(2) and f(4) both equal 3, and one-to-one

functions can’t assign two dierent objects in the domain to the same object

of the target.

If every horizontal line in R2 intersects the graph of a function at most

once, then the function is one-to-one.

Examples. Below is the graph of f : R ⇥ R where f(x) = x2. There is a

horizontal line that intersects this graph in more than one point, so f is not

one-to-one.

66

Inverse Functions

One-to-one

Suppose f : A —* B is a function. We call f one-to-one if every distinct

pair of objects in A is assigned to a distinct pair of objects in B. In other

words, each object of the target has at most one object from the domain

assigned to it.

There is a way of phrasing the previous definition in a more mathematical

language: f is one-to-one if whenever we have two objects a, c e A with

a ~ c, we are guaranteed that f(a) $ f(c).

Example. f : IR —* JR where f(x) = x2 is not one-to-one because 3 ~ —3

and yet f(3) = f(—3) since f(3) and f(—3) both equal 9.

Horizontal line test

If a horizontal line intersects the graph of f(.x) in more than one point,

then f(z) is not one-to-one.

The reason f(x) would not be one-to-one is that the graph would contain

two points that have the same second coordinate — for example, (2,3) and

(4,3). That would mean that f(2) and f(4) both equal 3, and one-to-one

functions can’t assign two different objects in the domain to the same object

of the target.

If every horizontal line in JR2 intersects the graph of a function at most

once, then the function is one-to-one.

Examples. Below is the graph of f : JR —, R where f(z) = z2. There is a

horizontal line that intersects this h in more than one point, so f is not

Step-by-step explanation:

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Answer:

\text{Length of AB is }\frac{ah}{a+h}

Step-by-step explanation:

Given △KMN, ABCD is a square where KN=a, MP⊥KN, MP=h.

we have to find the length of AB.

Let the side of square i.e AB is x units.

As ADCB is a square ⇒ ∠CDN=90°⇒∠CDP=90°

⇒ CP||MP||AB

In ΔMNP and ΔCND

∠NCD=∠NMP     (∵ corresponding angles)

∠NDC=∠NPM     (∵ corresponding angles)

By AA similarity rule,  ΔMNP~ΔCND

Also, ΔKAP~ΔKPM by similarity rule as above.

Hence, corresponding sides are in proportion

\frac{ND}{NP}=\frac{CD}{MP} \thinspace\thinspace and\thinspace\thinspace \frac{KA}{KP}=\frac{AB}{PM} \\\\\frac{ND}{NP}=\frac{x}{h} \thinspace\thinspace and\thinspace\thinspace \frac{KA}{KP}=\frac{x}{h}\\\\\frac{NP}{ND}=\frac{h}{x} \thinspace\thinspace and\thinspace\thinspace \frac{KP}{KA}=\frac{h}{x}\\\\\frac{PD}{ND}=\frac{h}{x}-1 \thinspace\thinspace and\thinspace\thinspace \frac{AP}{KA}=\frac{h}{x}-1\\

KA(\frac{h}{x}-1)=AP

ND(\frac{h}{x}-1)=PD

Adding above two, we get

(KA+ND)(\frac{h}{x}-1)=(AP+PD)

⇒ (KN-AD)=\frac{x}{(\frac{h}{x}-1)}

⇒ a-x=\frac{x}{(\frac{h}{x}-1)}

⇒ a-x=\frac{x^2}{h-x}

⇒ x^2=ah-ax-xh+x^2

⇒ x(h+a)=ah

⇒ x=\frac{ah}{a+h}

3 0
3 years ago
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