representation of the function f(x)=-2x2+8x-6

Answer:

Step-by-step explanation:

This is a problem of ratios:

Set the problem up with matching similar side lengths equal:

50/10 = x/12

First simplify the left side:

50/10 = 5

Next, multiply both sides by 12.

We are left with:

x = 12(5)

x = 60

Answer:

there are no like terms

Step-by-step explanation:

-x + 21 = -x + 21

Answer:  60%

Step-by-step explanation:

81/135= 60/100

Step-by-step explanation:

√a = 1√a so we can solve them easily:

b) 3√7 -√7= 3√7 - 1√7 =( 3-1)√7= 2√7

d) 5√6 - 2√6+√6= (5-2+1)√6 = 4√6

g) √2+2√2= 3√2

j) √5+5√5 - 3√5 = 3√5

k) 2√3 + √3 - 5√3= -2√3

I) 5√11 + 7√11 - √11 = 11√11

10 quarters 6 nickels 18 dimes

Answer:

w = 4

Step-by-step explanation:

6w+2=9w+14

9w-6w=14-2

3w=12

w=12/3

w=4

Answer:

40

Step-by-step explanation:

0.28÷0.007

=280÷7

=40

_________

[tex] \: [/tex]

0,28 ÷ 0,007 = ...

[tex] = \frac{28}{100} \div \frac{7}{1.000} [/tex]

[tex] = \frac{28}{100} \times \frac{1.000}{7} [/tex]

[tex] = \frac{28.0 \cancel{00}}{7 \cancel{00}} [/tex]

[tex] = \frac{280}{7} [/tex]

[tex] \longmapsto \boxed{ \bold{ \red{40}}}[/tex]

Answer:

8 millimeters 4x where x is 2 millimeters

Answer:

0.333 < 1.03

Step-by-step explanation:

1.03 is greater than 0.333.

In fraction form, 1.03 is [tex]1\frac{3}{100}[/tex].

While 0.333 is [tex]\frac{33}{100}[/tex], which is a third of 1.

Therefore, 0.333 <  1.03.

Answer:

x=45 y=60

Step-by-step explanation:

the picture is kinda blurry but if im seeing it correctly its ^^^

Step-by-step explanation:

Content

Functions and their inverses

We begin with a simple example.

Example

Let f(x)=2x and g(x)=x2.

Apply the function g to the number 3, and then apply f to the result:

g(3)=32andf(32)=3.

A similar thing happens if we first apply f and then apply g:

f(3)=6andg(6)=3.

It is clear that this will happen with any starting number. This is expressed as

f(g(x))g(f(x))=x,for all x=x,for all x.

The function f reverses the effect of g, and the function g reverses the effect of f. We say that f and g are inverses of each other.

As another example, we have

(x−−√3)3=xandx3−−√3=x,

for all real x. So the functions f(x)=x3 and g(x)=x−−√3 are inverses of each other.

If x≥0, then (x−−√)2=x and x2−−√=x. If x<0, then x−−√ is not defined. So the functions f(x)=x2 and g(x)=x−−√ are inverses of each other, but we need to be careful about domains. We will look at this more carefully later in this section.

Basics

In an earlier section of this module, we defined the composite of two functions h and g by (g∘h)(x)=g(h(x)).

Definitions

The zero function 0–:R→R is defined by 0–(x)=0, for all x.

The identity function id:R→R is defined by id(x)=x, for all x.

Example

Consider a function f:R→R.

Prove that

0–∘f=0–

f∘id=f

id∘f=f.

Show that f∘0– does not necessarily equal 0–.

Solution

We have (0–∘f)(x)=0–(f(x))=0, for all x, and so 0–∘f=0–.

We have (f∘id)(x)=f(id(x))=f(x), for all x, and so f∘id=f.

We have (id∘f)(x)=id(f(x))=f(x), for all x, and so id∘f=f.

Consider the function given by f(x)=2, for all x. Then f∘0–(x)=f(0–(x))=f(0)=2, and so f∘0–≠0–.

Definition

Let f be a function with both domain and range all real numbers. Then the function g is the inverse of f if

f(g(x))g(f(x))=x,for all x,and=x,for all x.

That is, f∘g=id and g∘f=id.

Notes.

Clearly, if g is the inverse of f, then f is the inverse of g.

We denote the inverse of f by f−1. We read f−1 as 'f inverse'. Note that f inverse has nothing to do with the function 1f.

Example

Let f(x)=x+2 and let g(x)=x−2. Show that f and g are inverses of each other.

Solution

We have

f(g(x))=f(x−2)=x−2+2=x,for all x(f∘g=id)

and

g(f(x))=g(x+2)=x+2−2=x,for all x(g∘f=id).

Hence, the functions f and g are inverses of each other.

Exercise 5

Find the inverse of

f(x)=x+7

f(x)=4x+5.

Example

Let f(x)=ax+b with a≠0. Find the inverse of f.

Solution

We have x=f(x)−ba, for all x. So let g(x)=x−ba. Then

f(g(x))g(f(x))=f(x−ba)=a(x−ba)+b=x=g(ax+b)=(ax+b)−ba=x,

for all x. Hence, g is the inverse of f.

Exercise 6

Show that f(x)=x5 and g(x)=x15 are inverses of each other.

Find the inverse of f(x)=x3+2.

We do not yet have a general enough concept of inverses, since x2 and x−−√ do not fit into this framework, nor do ex and logex. We will give a definition that covers these functions later in this section.

The horizontal-line test

Consider the function f(x)=x2, which has domain the reals and range A={x:x≥0}. Does f have an inverse?

The following graph shows that it does not. We have f(−2)=f(2)=4, and so f−1(4) would have to take two values, −2 and 2! Hence, f does not have an inverse.

Graph of y = x squared and the line y = 4 on the one set of axes.

This idea can be formulated as a test.

Horizontal-line test

Let f be a function. If there is a horizontal line y=c that meets the graph y=f(x) at more than one point, then f does not have an inverse.

Notes. Remember that the vertical-line test determines whether a relation is a function.

Example

Consider the function

f(x)=x3−x=(x+1)x(x−1).

Its graph is shown in the following diagram.

Graph of y = x cubed minus x.

Does f have an inverse?

Solution

The line y=0 meets the graph at three points. By the horizontal-line test, the function f does not have an inverse.

The function whose inverse does not exist is f(x) = x(x - 1)

The correct option is (D)  f(x) = x(x - 1)

An inverse is a function that serves to “undo” another function. That is, if f(x) produces y, then putting y into the inverse of f produces the output x.

First,  f(x)= [tex]\frac{1}{2} x -\frac{1}{2}[/tex]

let y= [tex]\frac{1}{2} x -\frac{1}{2}[/tex]

On solving for x we get a unique value

Then replace x and y.

It shows that the function have a unique value, which satisfies the condition of inverse.

Now, f(x) =[tex](x - 1)^3 + 2[/tex]

Again, solving for y we can get a cube root function which is a inverse of cube.

Hence, the inverse of [tex](x - 1)^3 + 2[/tex] exists.

Next, f(x) =[tex]2^x[/tex]

Solving for above we get logarithmic value. Log function are inverse of exponential function.

Hence, the inverse of [tex]2^x[/tex] exists.

Last, f(x)= x(x-1)

Solving for above create a square value.

The inverse of square never exist because having square root gives two value one is positive and other is negative.

Hence, the inverse of x(x-1) not exists.

Hence the function whose inverse does not exist is x(x-1).

Learn more about inverse of function here:

https://brainly.com/question/2541698

#SPJ2

Answer:

978

explanation is a calculator

Answer:

68.4210526316

you first divide 19/100 and then you divide the out put 13/ 0.19

Step-by-step explanation:

[tex] \frac{1}{12} \times \frac{1}{3} \\ = \frac{1 \times 1}{12 \times 3} \\ = \frac{1}{36} [/tex]

Answer:

23/90

Step-by-step explanation:Sorry if it's wrong!!!

Answer:

a. The real GDP increases by $200,000.

a. The real GDP increases by $150,000.

Step-by-step explanation:

a. What is the eventual effect on real GDP if the government increases its purchases of goods and services by $50,000?

Eventual effect on real GDP = Amount of increase in government spending * (1 /(1 - MPC)) = $50,000 * (1 / (1 – 0.75)) = $200,000

Therefore, the real GDP increases by $200,000.

a. What is the eventual effect on real GDP if the government, instead of changing its spending, increases transfers by $50,000?

Eventual effect on real GDP = (Amount of increase in government transfers * (1 /(1 - MPC))) - Amount of increase in government transfers = ($50,000 * (1 / (1 – 0.75))) - $50,000 = $150,000

Therefore, the real GDP increases by $150,000.

answer; you

Step-by-step explanation:

I dont know

sssooooooorrrrrrrrrryyyyyyyyyyyy

The sine law of triangles can be proved by dividing the triangle into right-triangles.

Option (c) proves the first equality in the law of sines​

From the question, we have:

h equals to the distance from point C to segment AB

The sine of angle is:

[tex]\mathbf{sin(\theta) = \frac{Opposite}{Hypotenuse}}[/tex]

For angle at A, we have:

[tex]\mathbe{\theta = A}[/tex]

[tex]\mathbf{Hypotenuse = b}\\\mathbf{Opposite = h}[/tex]

So, we have:

[tex]\mathbf{sin(\theta) = \frac{Opposite}{Hypotenuse}}[/tex]

Substitute values for Opposite, Hypotenuse and theta

[tex]\mathbf{sin A = \frac{h}{b}}[/tex]

Similarly

[tex]\mathbf{sin B = \frac{h}{a}}[/tex]

Make h the subject

[tex]\mathbf{h = a\ sinB}[/tex]

Substitute [tex]\mathbf{h = a\ sinB}[/tex] in [tex]\mathbf{sin A = \frac{h}{b}}[/tex]

[tex]\mathbf{sin A = \frac{a\ sinB}{b}}[/tex]

Divide both sides by a

[tex]\mathbf{\frac{sin A}{a} = \frac{sinB}{b}}[/tex]

Hence, the correct option is (c).

Read more about proof of sine law at:

https://brainly.com/question/4637169

Answer:

Step-by-step explanation:

Answer:

say what now-

Step-by-step explanation:

Answer:

-7/4 < -2/3 < 0 < 0.6 < 3/4 < 3

Step-by-step explanation:

Answer:

250

Step-by-step explanation:

As a decimal 253 and 19/23 can be written as 253.73

If we think about 253.73 being closer to 250 or 260, we can see that it is closer to 250, so that is the nearest ten we will round to.

Even if it is too difficult to convert to a decimal, if we round 19/23 to 23/23, our number becomes 254. The nearest ten is still 250.

Answer:

will u give the brainliest?

Answer:

x=1 1/3

Step-by-step explanation:

Answer:Each side of x would be 7 and each side of y would be 4.

Step-by-step explanation:

Answer:

(2/12 & 9/12)

Step-by-step explanation:

12 is the least common denominator for the fractions, so they must be rewritten to have the same denominator, but hold the same value.

( not written by me )

Answer:

24,000

Step-by-step explanation:

615 ≈ 600

38 ≈ 40

600 × 40 = 24,000

Therefore, 615 × 38 ≈ 24,000

Answer:

AC=5

in a basic right triangle the small side is always one shorter than the big side which is always one smaller than the hypotenuse.