PLS HELO THANK YOUUUUU

Answer:

Step-by-step explanation:

The cubes have sides of 2 cm, 3 cm and 4 cm respectively.

Their face area of the cubes is 4 cm², 9 cm² and 16 cm² respectively.

Total surface area of each cube is:

Total of three cubes:

When you glue the cubes together you lose an area equal to double face area of the smaller cube from the total surface area.

Possible decrease, provided full faces area covered:

The least possible surface area is:

3Cubes :

Soo :

Volume - 1 = 6 × 2² = 6 × 4 = 24cm²

Volume - 2 = 6 × 3² = 6 × 9 = 54cm²

Volume - 3 = 6 × 4² = 6 × 16 = 96cm²

[tex] \: [/tex]

surface area = 174 - 2(3² + 2²)

surface area = 174 - 2(9 + 4)

surface area = 174 - 2(13)

surface area = 174 - 26

serface area = 148cm²

[tex] \: [/tex]

-HayabusaBrainly

Answer:

can you add a picture or summerize this

Step-by-step explanation:

Answer:

3.0 E35

Step-by-step explanation:

Thanks.

We want to find the work done by lifting a book a given distance.

The answer is 200 N*m

We know that the book weighs 100 Newtons and that we lift it a distance of 2 meters.

Remember that the work needed to lift an object of mass M to a height H is:

W = M*g*H

where g is the gravitational acceleration.

Knowing that:

weight = M*g

We can replace the things we know to get the work:

W = 100N*2m = 200 N*m

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9514 1404 393

Answer:

  2.56

Step-by-step explanation:

Multiply by (1+6e^(-.7t))/1000:

  2000/1000 = 1 +6e^(-0.07t)

  1 = 6e^(-0.07t) . . . . . . . . . subtract 1

  e^(0.07t) = 6 . . . . . . . . . . multiply by e^(0.07t)

  0.07t = ln(6) . . . . . . . . . take the natural log

  t = ln(6)/0.07 . . . . . . . divide by the coefficient of t

  t ≈ 2.55966 ≈ 2.56

Answer:

terminating decimals

Step-by-step explanation:

they don't seem to repeat numbers

Answer:

repeating

Step-by-step explanation:

6÷7= 1.111111111111

Answer:

The answer is x≥9.

Step-by-step explanation:

Answer:

x≥9

Step-by-step explanation:

this is the answer all you have to do is solve it

The expression to represents the perimeter is,

54 = 2 (2k + k)

And, The width of the rectangle is, 9 units.

A rectangle is a two dimension figure with 4 sides, 4 corners and 4 right angles. The opposite sides of the rectangle are equal and parallel to each other.

Given that;

The perimeter of a rectangle is equal to 54 units.

And, The length of the rectangle is 2k units, and the width of the rectangle is k units.

Now, We know that;

Perimeter of rectangle = 2 (Length + Width)

Substitute the given values, we get;

Perimeter of rectangle = 2 (Length + Width)

54 = 2 (2k + k)

54 = 2 × 3k

54 = 6k

54/6 = k

k = 9

Hence, The width of the rectangle is, 9 units.

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Answer:

[tex]{ \rm{ {8}^{x} = \sqrt{2} }} \\ \\ { \rm{ {2}^{3x} = {2}^{ \frac{1}{2} } }} \\ \\ { \rm{3x = \frac{1}{2} }} \\ \\ { \rm{x = \frac{1}{6} }}[/tex]

Answer:

50.3 (rounded) I think? Could be wrong, however, it's what I got

9514 1404 393

Answer:

  $3

Step-by-step explanation:

The cost that hasn't been covered by bake sale proceeds is ...

  $750 -150 = $600

If that cost is divided among 200 students, the amount each student needs to pay is ...

  $600/(200 students) = $3/student

They need to charge $3 per ticket.

Answer:

II + III = 176

II = 44

III = 132

Step-by-step explanation:

Third round: 132

Second round: 3x=132

x=44 (second round)

2rounds: 132+44=176

9514 1404 393

Answer:

  y = 8

Step-by-step explanation:

If the variation is direct, any multiplier of the x-value will also multiply the y-value.

  old value × -1 = new value

  x: 4 × -1 = -4

  y: -8 × -1 = 8

The new value of y is 8.

Answer:

Step-by-step explanation:

Slope -  -3/2

y-intercept- (0,-3)

Answer:

37356

Step-by-step explanation:

Answer:

The first numbers in the sets are input and the second numbers are outputs.

1. {(1, 1), (2, 3), (3,5)}

2. {(6,2), (5, 3), (4,8)}​

Step-by-step explanation:

Given Question

Complete the following tables and plot the points on the graph paper yo represents the equations given below :-

[tex]\begin{gathered}\boxed{\begin{array}{c|c|c|c} \bf x & \bf 1 & \bf 2& \bf 3\\ \frac{\qquad}{} & \frac{\qquad}{}\frac{\qquad}{} &\frac{\qquad}{} & \frac{\qquad}{} &\\ \sf y = x + 1 & \sf & \sf & \sf \\ \\ \sf (x,y)& \sf & \sf & \sf \\ \end{array}} \\ \end{gathered} \\ [/tex]

and

[tex]\begin{gathered}\boxed{\begin{array}{c|c|c|c} \bf x & \bf 1 & \bf 2& \bf 3\\ \frac{\qquad}{} & \frac{\qquad}{}\frac{\qquad}{} &\frac{\qquad}{} & \frac{\qquad}{} &\\ \sf y = - 3x & \sf & \sf & \sf \\ \\ \sf (x,y)& \sf & \sf & \sf \\ \end{array}} \\ \end{gathered} \\ [/tex]

[tex]\large\underline{\sf{Solution-}}[/tex]

Given that,

[tex]\rm \longmapsto\:y = x + 1[/tex]

On substituting x = 1, we get

[tex]\rm \longmapsto\:y = 1 + 1[/tex]

[tex]\rm \longmapsto\:y = 2[/tex]

On substituting x = 2, we get

[tex]\rm \longmapsto\:y = 2 + 1[/tex]

[tex]\rm \longmapsto\:y = 3[/tex]

On substituting x = 3, we get

[tex]\rm \longmapsto\:y = 3 + 1[/tex]

[tex]\rm \longmapsto\:y = 4[/tex]

Hence,

[tex]\begin{gathered}\boxed{\begin{array}{c|c|c|c} \bf x & \bf 1 & \bf 2& \bf 3\\ \frac{\qquad}{} & \frac{\qquad}{}\frac{\qquad}{} &\frac{\qquad}{} & \frac{\qquad}{} &\\ \sf y = x + 1 & \sf 2 & \sf 3 & \sf 4\\ \\ \sf (x,y)& \sf (1,2) & \sf (2,3) & \sf (3,4)\\ \end{array}} \\ \end{gathered}[/tex]

Now, draw a graph using the points (1 , 2), (2 , 3) & (3 , 4)

---------------------------------------------

Given equation is

[tex]\rm \longmapsto\:y = - 3x[/tex]

On substituting x = 1, we get

[tex]\rm \longmapsto\:y = - 3 \times 1[/tex]

[tex]\rm \longmapsto\:y = - 3[/tex]

On substituting x = 2, we get

[tex]\rm \longmapsto\:y = - 3 \times 2[/tex]

[tex]\rm \longmapsto\:y = - 6[/tex]

On substituting x = 3, we get

[tex]\rm \longmapsto\:y = - 3 \times 3[/tex]

[tex]\rm \longmapsto\:y = - 9[/tex]

Hence,

[tex]\begin{gathered}\boxed{\begin{array}{c|c|c|c} \bf x & \bf 1 & \bf 2& \bf 3\\ \frac{\qquad}{} & \frac{\qquad}{}\frac{\qquad}{} &\frac{\qquad}{} & \frac{\qquad}{} &\\ \sf y = - 3x & \sf - 3 & \sf - 6 & \sf - 9\\ \\ \sf (x,y)& \sf (1, - 3) & \sf (2, - 6) & \sf (3, - 9)\\ \end{array}} \\ \end{gathered}[/tex]

Now, draw a graph using the points (1 , - 3), (2 , - 6) & (3 , - 9)

Answer:

(3a4 + 4)2

= (12a+4)2

= 24a+8

Answer:

Simple interest-?

Step-by-step explanation:

P= 12000

T=5yrs

R=5%

Now,

Simple interest - PTR/100

= 12000*5*5/100

= $3000

Again,

P=12000

T=3yrs

R=8%

Now,

Simple interest - PTR/100

= 12000*3*8/100

= $2880

Hope this helps.

Answer:

-2/3

Step-by-step explanation:

:)

Answer:

Step-by-step explanation:

Given equation:

Multiply both sides by - 3 and simplify;

Answer:8

Step-by-step explanation:

Answer:

73

Step-by-step explanation:

332÷2= 166

166 - 93 = 73

Answer:

42

Step-by-step explanation:

to find how many 0.25 there are in 10.5 we have to divide 10.5 by 0.25

10.5÷0.25

=42

Hope this helped you- have a good day bro cya)

Step-by-step explanation:

The prime numbers 5 and 3 are both factors that are common to 30 and 75. The product of these factors is 3×5 = 15. Thus, the “Greatest Common Factor” of 30 and 75 is 15.

Answer:

the Greatest common factor of 30 and 75 is 15

Step-by-step explanation:

The product of these factors is 3×5 = 15. Thus, the “Greatest Common Factor” of 30 and 75 is 15.

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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all of his sheep

Hope that helps :)