Does every function have an inverse which is a function? Does every relation have an inverse which is a relation?

Answer: [tex]f\left(x\right)\le \:-4[/tex]

Step-by-step explanation:

[tex]\begin{bmatrix}\mathrm{Solution:}\:&\:f\left(x\right)\le \:-4\:\\ \:\mathrm{Interval\:Notation:}&\:(-\infty \:,\:-4]\end{bmatrix}[/tex]

Answer:

Step-by-step explanation:

A composite function exists as a function that depends on another function. The values for f(g(x)) = 4x-7 and g(f(x)) = 4x -28

In mathematics, function composition exists as an operation ∘ that takes two functions f and g, and has a function h = g ∘ f such that h(x) = g. In this operation, the function g exists used to the result of applying the function f to x.

Function composition is a mathematical procedure where two functions, f, and g, are taken and a function, h = g ∘ f, is produced such that h(x) = g(f(x)). The function g exists used in this operation on the outcome of applying the function f to x.

Let the two functions be f(x) = -x-7 and g(x) = 4 x

f(g(x)) = f(4x)

f(4x) = 4x-7

g(f(x)) = g(x-7)

simplifying the above equation, we get

g(x-7) = 4 (x - 7) = 4x -28

g(x-7) = 4x - 28

Therefore, the value of f(g(x)) = 4x-7 and g(f(x)) = 4x -28

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On day 1 they travel 40 miles , on day 2 they travel 120 miles, on day 3 they travelled 80 miles.

what is  linear equation ?

A linear equation is a kind of equation that has the highest power of the variable is always 1 is also called as a one-degree equation.

By the linear equation, lets assume family travelled x miles on day 1 then, According to the question, family travelled on day 2 is = 3x miles family travelled in day 3 is = 3x - 40 miles now the travel 240 miles in total then by addition x + 3x + 3x -40 = 240

7x -40 = 240,

7x = 280

x = 40 miles

day 1 =40 miles

day 2 = 120 miles

day 3 = 80 miles

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We line up the decimal point so that the place value of other number does not changes.

Adding and subtracting decimal numbers is the same as adding and subtracting whole numbers. The most important thing to remember is to align the decimal places directly on top of each other when stacked. This stacks the digit values ​​so that 1 is above 1, , the tens on the tens, the tenths on the tenths,, and so on. Note that if a number does not have the same number of decimal places as other numbers, you can pad all the digits to the right of the decimal point with zeros. Numerically, this looks like 3.15 = 3.150 = 3.150000000.

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The probability that the student will answer at least 10 questions correctly from a multiple-choice examination that has 15 questions, each with five possible answers and only one of which is correct is 0.00

Let P (C) = probability of a student picking a correct option is

P (C) = (Required outcome)/(Total outcome)

Required outcome = numbers of correct option

Total outcome = total numbers of option

P (C) = 1/5

Recall that

P (C) + P (C1) = 1

 Where P (C1) is the probability of picking wrong option

P(C)+P(C^1 )=1

1/5+P(C^1 )=1

P(C^1 )=1-1/5

P(C^1 )=4/5

The probability of answering at least 10 questions correctly can be solved by using Binomial probability which states that

P(x)=(nCx) (p)^x (q)^(n-x)

P(x) is the probability of an event

n is the number of trial (15)

x number of specific outcome

p is the probability that the event will happen (1/5)

q is the probability that the event will not happen (4/5)

      Probability of answering at least 10 questions correctly will be

P(10) + P(11) + P(12) + P(13) +P(14) + P(15)

P(x)=(nCx) (p)^x (q)^(n-x)

P(10)=(15C10) (1/5)^10 (4/5)^(15-10)

P(10)=(15C10) (1/5)^10 (4/5)^5

15C10=15!/((15-10)!(10!))

15C10=15!/((5)!(10!))

15C10=1307674368000/((120)(3628800))

15C10=1307674368000/43545600

15C10=3003

P(10)=(3003) (0.2)^10 (0.8)^5

P(10)=(3003)(1.024×〖10〗^(-7))(0.32768)

P(10)=0.0001

P(11)=(15C11) (1/5)^11 (4/5)^(15-11)

P(11)=(15C10) (1/5)^11 (4/5)^4

15C11=15!/((15-11)!(11!))

15C11=15!/((4)!(11!))

15C11=1307674368000/((24)(39916800))

15C11=1307674368000/958003200

15C11=1365

P(11)=(1365) (0.2)^11 (0.8)^4

P(11)=(1365)(2.048×〖10〗^(-8))(0.4096)

P(11)=0.000011

P(12)=(15C12) (1/5)^12 (4/5)^(15-12)

P(12)=(15C12) (1/5)^12 (4/5)^3

15C12=15!/((15-12)!(12!))

15C12=15!/((3)!(12!))

15C12=1307674368000/((6)(479001600))

15C12=1307674368000/2874009600

15C12=455

P(12)=(455) (0.2)^12 (0.8)^3

P(12)=(455)(4.1×〖10〗^(-9))(0.512)

P(12)=0.0000009542

P(13)=(15C13) (1/5)^13 (4/5)^(15-13)

P(13)=(15C13) (1/5)^13 (4/5)^2

15C13=15!/((15-13)!(13!))

15C13=15!/((2)!(13!))

15C13=1307674368000/((2)(6227020800))

15C13=1307674368000/12454041600

15C13=105

P(13)=(105) (0.2)^13 (0.8)^2

P(13)=(105)(8.19×〖10〗^(-10))(0.64)

P(13)=0.00000005505

P(14)=(15C14) (1/5)^14 (4/5)^(15-14)

P(14)=(15C14) (1/5)^14 (4/5)^1

15C14=15!/((15-14)!(14!))

15C14=15!/((1)!(14!))

15C14=1307674368000/((1)(87178291200))

15C14=1307674368000/87178291200

15C14=15

P(14)=(15) (0.2)^14 (0.8)^1

P(14)=(15)(1.638×〖10〗^(-10))(0.8)

P(14)=0.00000000197

P(15)=(15C15) (1/5)^15 (4/5)^(15-15)

P(15)=(15C15) (1/5)^15 (4/5)^0

15C15=15!/((15-15)!(15!))

15C15=15!/((0)!(15!))

15C15=1307674368000/((1)(1307674368000))

15C15=1307674368000/1307674368000

15C15=1

P(15)=(1) (0.2)^15 (0.8)^0

P(15)=(1)(3.277×〖10〗^(-13))(1)

P(15)=3.277×〖10〗^(-13)

Probability of answering at least 10 questions correctly will be

P(10) + P(11) + P(12) + P(13) +P(14) + P(15)

P (at least 10)=0.0001+0.000011+ 0.0000009542+0.00000005505+0.00000000197+ 3.277×〖10〗^(-13)

P (at least 10)=0.00011250667

Approximately

P (at least 10)=0.0001

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Answer:

h(x) = 3x^2 - x^3 + 10x - 7 == 2

Step-by-step explanation:

x = 5 is not an extraneous solution of the equation.

✓(2x + 6) - ✓(x - 1) = 2

✓(2x + 6) = 2 + ✓(x - 1)

Squaring both the sides of equation, we get

(✓(2x + 6))² = (2 + ✓(x - 1))²

2x + 6 = (2)² + (✓(x - 1))² + 2(2)(✓(x - 1))

2x + 6 = 4 + x - 1 + 4(✓(x - 1))

x + 3 = 4✓(x - 1)

Squaring again on both the side of the equation, we get

(x + 3)² = (4✓(x - 1))²

x² + 3² + 2×3×x = 16(x - 1)

x² + 9 + 6x = 16x - 16

x² - 10x + 25 = 0

By applying middle term splitting on equation (1), we get

x² - 5x - 5x + 25 = 0

x(x - 5) -5(x - 5) = 0

(x - 5) (x - 5) = 0

x = 5 and 5.

Check for extraneous solutions by putting value of x in original equation -

✓(2×5 + 6) - ✓(5 - 1) = 2

✓16 - ✓4 = 2

4 - 2 = 2

2 = 2

LHS = RHS

x = 5 is not an extraneous solution of the equation.

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Step-by-step explanation:

BQS=78

BQN=48

BQS+BQN=78+48=126

180-126=54°

NQS=54°

comfirm=BQN+BQS+NQS=180

48+78+54=180

Answer:

30°

Step-by-step explanation:

∠NQS + ∠BQN = ∠BQS

Knowing this information, we use the known measures to find ∠NQS

∠NQS + 48° = 78°

∠NQS = 78° - 48°

∠NQS = 30°

Thus, angle NQS is 30 degrees.

Hope this helps!

It should be noted that the graph which will correctly show the number of pages which Louise will have left to read after the number of days is D. A graph titled How much Louise has left to read has days on the x-axis and pages on the y-axis. A line goes through (0, 300) and (2, 270).

It should be noted that he a graph is a diagram that shows the relationships between two or more things.

In this case, it can be seen that Louise is reading a 300-page book and she reads 15 pages every day.

Therefore, the graph that correctly displays the number of pages which Louise has left to read after the number of days is a graph titled How Much Louise Has Left to Read has days on the x-axis and pages on the y-axis. A line goes through (0, 300) and (2, 270).

In conclusion, the correct option is D.

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The length of the third side of the right triangle is 12.65 feet.

It states in a right triangle the square of the length of the hypotenuse is equal to the sum of squares of the lengths of the other two sides of the right-angled triangle.

We have,

Right triangle:

Side = A = 6 feet

Hypotenuse = h = 14 feet

Let the 3rd side be = M

Applying Pythagoras theorem:

h² = A² + M²

14² = 6² + M²

196 = 36 + M²
M² = 196 - 36

M² = 160

M = √160

M = 2√10

M = 2 x 3.1622

M = 12.649

Rounding to the nearest hundredth we get,

M = 12.65

Thus the length of the third side of the right triangle is 12.65 feet.

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A function that gives the length of the hypotenuse of an isosceles right triangle with side length s is given by h(s)=s√2 and the length of the side of the triangle is approximately 4.24 inches.

The largest side of a right-angled triangle is referred to as the hypotenuse. Additionally, the hypotenuse is situated across from the right angle.

We know from the Pythagorean Theorem that the square of the hypotenuse is equal to the sum of the squares of the triangle's right-angled legs.

The right triangle's given legs are s. Let's think about the hypotenuse's function, h(s).

Therefore we can define the function h(s) as

h(s)=s√2

The inverse of the function can be written as

y=s√2

or, s=y²/2

so the inverse function is:

h⁻¹(s)=s²/2

Now for hypotenuse = 6 inches.

Each side can be calculated using the inverse function h⁻¹(s)=s²/2.

s=6÷√2

or, s=4.242...

Hence the length of the side of the triangle is approximately 4.24 inches.

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The expression (-1 ,-5) is not a solution to the inequality 4x + y < -3

The inequality is given as

4x + y < -3

The solution is given as

(-1, 5)

Substitiute (-1, 5) for (x, y) in the inequality expression 4x + y < -3

So, we have

4(-1) + 5 < -3

Open the bracket

-4 + 5 < -3

Evaluate the sum

1 < -3

The above inequality is false

Hence, (-1 ,-5) is not a solution to the inequality 4x + y < -3

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Using the properties of logarithm the value of the expression is   [tex]log_{2} 96 - log_{2} 3 = 5[/tex]

A logarithm is the power to which a number must be raised in order to get some other number.

According to the question we have been given the expression as

                        =  [tex]log_{2} 96 - log_{2} 3[/tex]

The properties of logarithm that we are going to use are :

         [tex]log(\frac{x}{y} ) = logx - logy[/tex]  and [tex]log_{a} a = 1[/tex]

Also we can see that the base are same so we can use the property of logarithm in the given equation.

Now solving this step by step using the properties of logarithm we get

                       [tex]= log_{2} \frac{96}{3} \\= log_{2} 32\\= log_{2} 2^{5}\\ = 5* log_{2} 2\\= 5*1\\= 5[/tex]

Hence using the properties of logarithm the value of the expression is

[tex]log_{2} 96 - log_{2} 3 = 5[/tex]

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1.  The probability given as P(Y∩X) is; 2/25

2. The probability given as P(Y) * P(X) is; 4/15

We are given the following probabilities;

P(X) = 2/3 , P(Y) = 2/5 , and P(X|Y) = 1/5

1. We want to find P(Y ∩ X).

This means probability of the set intersection of Y and X

Since  P(X) = 2/3 , P(Y) =2/5 , and P(X|Y) =1/5, it means that we have a conditional probability.

Thus;  P(Y∩X) = P(Y) * P(X|Y)

P(Y∩X) = 2/5 × 1/5

P(Y∩X) = 2/25

2. We want to find P(Y) * P(X)

Plugging in the relevant values gives;

P(Y) * P(X) = 2/3 × 2/5

P(Y) * P(X) = 4/15

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Complete question is;

For two events X and Y, P(X) =2/3 , P(Y) =2/5 , and P(X|Y) =1/5 . Find the probabilities.

1.  P(Y∩X) = ____

2. P(Y)· P(X) =____

Answer: a. 1

b. 12

c. 2x+11

Step-by-step explanation:

f(x) = 2x+5

a. f(-2)=

2(-2)+5=

-4+5=1

b. f(1)+5=

2(1)+5+5=

2+5+5=12

c. f(x+3)=

2(x+3)+5=

2x+6+5=2x+11

Answer:

D= (-3,1]

R= [0,-4]

Step-by-step explanation:

[,] <-- used for when the point is included in the answer (closed circle or greater than/less than or equal to sign)

(,) <-- used for when the answer does not include point (open circle or greater than/less than sign   not equal to)

Plug in the points where shown on each line.

Hope that helps a bit!

Answer:

X= -4

Step-by-step explanation:

12x+2=5x−26

12+2−2=5−26−2

Answer:

x=−4

Step-by-step explanation:

Subtract  from both sides

Simplify then subtract 5x from both sides

Example:

The probability that 3 numbers chosen with replacement contain both even and odd is 0.75 while the probability that 3 numbers chosen without replacement contain both even and odd is 0.8.

The numbers include 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16

Case 1: with replacement

Let P (E) = probability of chosing even number is

P (E) = (Required outcome)/(Total outcome)

Required outcome = numbers of even number

Total outcome = total numbers

P (E) = 8/16

P (E) = 1/2

Let P (o) = probability of chosen odd number is

P (o) = (Required outcome)/(Total outcome)

Required outcome = numbers of odd number

Total outcome = total numbers

P (o) = 8/16

P (o) = 1/2

Since 3 numbers are chosen, there are 6 possible combination which are OEE, EOE, OEO, OOE EEO, and EOO

P (chosen 3 numbers) = P (OEE) + P (EOE) + P (OEO) + P (OOE) + P (EEO) + P (EOO)

P (OEE) =P(O)×P(E)×P(E)  

P (OEE) =1/2×1/2×1/2

P (OEE) =1/8

P (EOE) =P(E)×P(O)×P(E)  

P (EOE) =1/2×1/2×1/2

P (EOE) =1/8

P (OEO) =P(O)×P(E)×P(O)  

P (OEO) =1/2×1/2×1/2

P (OEO) =1/8

P (OOE) =P(O)×P(O)×P(E)  

P (OOE) =1/2×1/2×1/2

P (OOE) =1/8

P (EEO) =P(E)×P(E)×P(O)  

P (EEO) =1/2×1/2×1/2

P (EEO) =1/8

P (EOO) =P(E)×P(O)×P(O)  

P (EOO) =1/2×1/2×1/2

P (EOO) =1/8

P (chosen 3 numbers)=1/8+1/8+  1/8+1/8+1/8+1/8

P (chosen 3 numbers)=6(1/8)

P (chosen 3 numbers)=3/4

P (chosen 3 numbers)=0.75

Case 2: without replacement, when the first number is chosen there is a reduction in the amount of the remaining numbers.

Let P (E) = probability of chosen even number is

P (E) = (Required outcome)/(Total outcome)

Required outcome = numbers of even number

Total outcome = total numbers

P (E) = 8/16

P (E) = 1/2

Let P (o) = probability of chosing odd number is

P (o) = (Required outcome)/(Total outcome)

Required outcome = numbers of odd number

Total outcome = total numbers

P (o) = 8/16

P (o) = 1/2

Since 3 numbers are chosen, there are 6 possible combination which are OEE, EOE, OEO, OOE EEO, and EOO

P (chosen 3 numbers) = P (OEE) + P (EOE) + P (OEO) + P (OOE) + P (EEO) + P (EOO)

P (OEE) =P(O)×P(E)×P(E)  

P (OEE) =8/16×8/15×7/14

P (OEE) =1/2×8/15×1/2

P (OEE) =8/60

P (OEE) =2/15

P (EOE) =P(E)×P(O)×P(E)  

P (EOE) =8/16×8/15×7/14

P (EOE) =1/2×8/15×1/2

P (EOE) =8/60

P (EOE) =2/15

P (OEO) =P(O)×P(E)×P(O)  

P (OEO) =8/16×8/15×7/14

P (OEO) =1/2×8/15×1/2

P (OEO) =8/60

P (OEO) =2/15

P (OOE) =P(O)×P(O)×P(E)  

P (OOE) =8/16×7/15×8/14

P (OOE) =1/2×7/15×4/7

P (OOE) =28/210

P (OOE) =2/15

P (EEO) =P(E)×P(E)×P(O)  

P (EEO) =8/16×7/15×8/14

P (EEO) =1/2×7/15×4/7

P (EEO) =28/210

P (EEO) =2/15

P (EOO) =P(E)×P(O)×P(O)  

P (EOO) =8/16×8/15×7/14

P (EOO) =1/2×8/15×1/2

P (EOO) =8/60

P (EOO) =2/15

P (chosen 3 numbers)=2/15+2/15+  2/15+2/15+2/15+  2/15

P (chosen 3 numbers)=6(2/15)

P (chosen 3 numbers)=4/5

P (chosen 3 numbers)=0.8

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The solution for the compound inequality is 4 < b ≤ 5.

It is an expression mathematical that represents a non-equal relationship between a number or another algebraic expression. Therefore, it is common the use following symbols: ≤ (less than or equal to), ≥ (greater than or equal to), < (less than), and > (greater than).

The solutions for inequalities can be given by: a graph in a number line or numbers.

The question gives:

3b + 2 < 5b - 6 ≤ 2b + 9

You can write the previous compound inequality as:

3b + 2 < 5b - 6 (1)

5b - 6 ≤ 2b + 9 (2)

         3b + 2 < 5b - 6

         3b-5b < -6 -2

          -2b < -8

            2b >8

            b >4

         5b - 6 ≤ 2b + 9

         5b-2b  ≤ 9+6

          3b ≤ 15

          b ≤ 5

Combining the intervals, you have 4 < b ≤ 5.

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Speed in still water= [tex]\sqrt{43}[/tex]miles

Let take the speed in still water be x and  the speed of current in the river be  y.

The speed of the current in a river= 5 miles

Let the distance be s which is 1.2 miles.

The time(t) difference is 40 minutes.

We have an equation,

[tex]\frac{s}{x-y}[/tex] - [tex]\frac{s}{x+y}[/tex] = t

[tex]\frac{1.2}{x - 5}[/tex] - [tex]\frac{1.2}{x + 5}[/tex] = [tex]\frac{40}{60}[/tex]

1.2 × 60(2 × 5) = 40([tex]x^{2}[/tex] - 25)

x =[tex]\sqrt{43}[/tex]

Therefore the speed in still water is [tex]\sqrt{43}[/tex].

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Step-by-step explanation:

6 = 5/2x + k

6 = ( 5/2 . 2/1) + k

6 = 10/2 +k

6-5 = k

k = 1

The calorie serving in simplest form is 56.5

The rate in mentioned question will be number of calories present per serving.  

So writing the given information in the form of rate -

Rate of calorie per serving = 452/8

Now, to convert the rate of calories in simplest form, we will perform division of both numerator and denominator of fraction by a common number. We can directly divide the numerator and denominator of fraction with 8. Else, we can perform division by simpler numbers too.

Indirect method -

Lets begin the division by 4

Rate of calorie per serving = 113/2

Further dividing by 2

Rate of calorie per serving = 56.5

Direct method will also give the same result, which is rate of calorie per serving = 56.5.

Hence, in simplest form (decimal) the rate of calorie per serving is 56.5.

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The new mean 101.5 is greater than the previous mean 94.

By adding up all the numbers in a data collection and dividing by the total number of values in the set, the mean (average) of the data set can be determined. The median is the middle when a data set is sorted from least to greatest.

The distribution of a basketball team's score with a mean of 94 points per game and a high of 109 for the first 20 games.

Mean = 94,

Points per game = 94, 110, 102, 93, 98, 99, 121, 96, 94, and 108 points.

Therefore,

The new mean = ( sum of observations) / Number of observations

New mean = ( 94 + 110 + 102 + 93 + 98 + 99 + 121 + 96 + 94 + 108) / 10

New mean = 1015 / 10

New mean = 101.5

101.5 > 94

Now, the new mean is greater than the previous mean.

Hence, the correct option is (A).

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x = 512 is not an extraneous solution of the equation.

[tex] \sqrt[3]{x} [/tex] = 8

Cubing both the side of the equation, we get

[tex] { \sqrt[3]{x} }^{3} [/tex]

= 8³

x = 512

Check for extraneous solution by putting the value x = 512 in the original equation

[tex] \sqrt[3]{x} [/tex] = 8

[tex] \sqrt[3]{512} [/tex] = 8

8 = 8

⇒ LHS = RHS

∴ x = 512 is not an extraneous solution of the equation.

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Step-by-step explanation:

so, MN + NO = MO

MO = 50 miles

MN = 45 miles

therefore,

45 + NO = 50

NO = 50 - 45 = 5 miles

Answer:

23 and 26? which numbers would you like to know?

Step-by-step explanation:

The values of x and y of the parallelogram PQRS are x = 2 and y = 11

We're given the dimensions of parallelogram PQRS as;

PT = y

TR = 5x + 1

QT = 2y

TS = 6x + 10

T is the intersection of the two diagonals PR and QS and so the diagonals bisect each other.

The diagonal PR is cut into PT and TR, both of which are congruent or equal. Thus, PT = TR.

Similarly, the other diagonal QS is split in half as well. The two equal pieces are QT and TS. So QT = TS.

PT = TR

y = 5x + 1

and

QT = TS

2y = 6x + 10

Plugging the values gives;

2y = 6x + 10

2(y) = 6x + 10

2(5x + 1) = 6x + 10

2*5x + 2*1 = 6x + 10

10x + 2 = 6x + 10

10x + 2 - 6x = 6x + 10 - 6x

4x + 2 = 10

4x + 2 - 2 = 10 - 2

4x = 8

4x/4 = 8/4

x = 2

If x = 2, then y is...

y = 5x+1

y = 5*x+1

y = 5*2+1

y = 10+1

y = 11

Thus, the values of x and y of the parallelogram PQRS are x = 2 and y = 11

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Answer:

Probably puberty

After solving the expression 3-√(x-3)=x algebraically, we get the result as  3.

Isolate the square root from the left hand side √x-3 = 3-x

Eliminate the radical on the left hand side by squaring on both sides-

Both sides must be raised to the second power.    

(√x-3)² = (3-x)²  

After squaring we get       

x-3 = x²-6x+9

Solving the quadratic equation-

We get rearranged equation as -

x²  - 7x  + 12 = 0

(x-4)(x-3)=0

x=4,3

Check if the first solution is correct by putting it in original equation-

putting x=3 in 3-√(x-3)=x  we get 3-0=3

Hence 3 is correct.

Check if the second solution is correct by putting it in original equation-

putting x=4 in  3-√(x-3)=x  we get 3-1=4 which is not true

hence 4 is rejected.

Therefore, after solving the expression 3-√(x-3)=x algebraically, we get the result as  3.

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