Negative 1/3 in decimal form

Answer:

68.4210526316

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

Step-by-step explanation:

Answer:

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

Step-by-step explanation:

Answer:

x=1 1/3

Step-by-step explanation:

Answer:

Step-by-step explanation:

The two angles can be 48.75 which if you add all those together would be 180 degrees

Sorry if it might be wrong but if its right can you mark me branliest

Answer:

Step-by-step explanation:

The tree indicated angles form a straight angle:

The angle 2x:

The angle 3x:

Answer:

B: [tex]c(3c^2-4)[/tex]

Step-by-step explanation:

1. Substitute [tex]1[/tex] for c: [tex](24^6-32^4)/(8^3)[/tex]

2. Solve substitution: [tex]-1[/tex]

3. Compare [tex]-1[/tex] to calculated answers when [tex]c=1[/tex]

4. [tex]c(3c^2-4)[/tex] when [tex]c=1[/tex] is [tex]-1[/tex]

Answer:

Step-by-step explanation:

If one of the data points has the form (0,a)

, then a is the initial value. Using a, substitute the second point into the equatio f(x)=a(b)x, and solve for b.If neither of the data points have the form (0,a)

, substitute both points into two equations with the form f(x)=a(b)x Solve the resulting system of two equations in two unknowns to find a and b.Using the a and b found in the steps above, write the exponential function in the form f(x)=a(b)x.

Answer:

f(m-5) = 5m-34

Step-by-step explanation:

f(x)=5x-9

Replace x with m-5

f(m-5) = 5(m-5) -9

Distribute

         = 5m-25 -9

         = 5m-34

Answer:

answer is 5m - 34

Step-by-step explanation:

!!!!!!!!!!!!!!!!!!!!!!!!

Answer:

there are no like terms

Step-by-step explanation:

-x + 21 = -x + 21

Answer:

will u give the brainliest?

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).

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

Step-by-step explanation:

Answer:

8 millimeters 4x where x is 2 millimeters

Answer:

Step-by-step explanation:

Answer:

say what now-

Step-by-step explanation:

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

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:

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.

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:  60%

Step-by-step explanation:

81/135= 60/100

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

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

Answer:

x= 1.75

Step-by-step explanation:

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:

Can you put it in English pls..

Step-by-step explanation:

answer; you

Step-by-step explanation:

I dont know

sssooooooorrrrrrrrrryyyyyyyyyyyy

Answer:

w = 4

Step-by-step explanation:

6w+2=9w+14

9w-6w=14-2

3w=12

w=12/3

w=4

Answer:

23/90

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

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).

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