If one wanted to solve the equation below using the quadratic formula, what would be the b value used to substitute into the quadratic equation.

3x 2 − 2 = - 5x

well, with the assumption that a year has 365 days, that means one day is really just 1/365th of a year, so then 54 days will be 54/365 of a year.

[tex]~~~~~~ \textit{Simple Interest Earned} \\\\ I = Prt\qquad \begin{cases} I=\textit{interest earned}\\ P=\textit{original amount deposited}\dotfill & \$17100\\ r=rate\to 9.65\%\to \frac{9.65}{100}\dotfill &0.0965\\ t=years\dotfill &\frac{54}{365} \end{cases} \\\\\\ I = (17100)(0.0965)(\frac{54}{365})\implies \stackrel{\textit{interest owed}}{I\approx 244.13}~\hfill \underset{amount~owed}{\stackrel{17100~~ + ~~244.13}{\approx 17344.13}}[/tex]

a) The polynomial f(x) in expanded form is f(x) = x³ + 10 · x² - 20 · x - 24.

b) The rational function g(x) in factored form is g(x) = [(x - 6) · (x + 3)] / (x - 2). there is no slant asymptotes.

c) There is one evitable discontinuity at x = - 1, and one definitive discontinuity at x = 2, where there is a vertical asymptote.

a) In the first part of this question we need to determine the equation of a polynomial in expanded form, derived from its factor form defined below:

f(x) = Π (x - rₐ), for a ∈ {1, 2, 3, 4, ..., n}         (1)

Where rₐ is the a-th root of the polynomial.

If we know that r₁ = 6, r₂ = - 1 and r₃ = - 3, then the polynomial in factor form is:

f(x) = (x - 6) · (x + 1) · (x + 3)

f(x) = (x - 6) · (x² + 4 · x + 4)

f(x) = (x - 6) · x² + (x - 6) · (4 · x) + (x - 6) · 4

f(x) = x³ - 6 · x² + 4 · x² - 24 · x + 4 · x - 24

f(x) = x³ + 10 · x² - 20 · x - 24

The polynomial f(x) in expanded form is f(x) = x³ + 10 · x² - 20 · x - 24.

b) The rational function is introduced below:

g(x) = (x³ + 10 · x² - 20 · x - 24) / (x² - x - 2)

g(x) = [(x - 6) · (x + 1) · (x + 3)] / [(x - 2) · (x + 1)]

g(x) = [(x - 6) · (x + 3)] / (x - 2)

The slope of the slant asymptote is:

m = lim [g(x) / x] for x → ± ∞

m = [(x - 6) · (x + 3)] / [x · (x - 2)]

m = 1

And the intercept of the slant asymptote is:

n = lim [g(x) - m · x] for x → ± ∞

n = Non-existent

Hence, there is no slant asymptotes.

c) There is vertical asymptote at a x-point if the denominator is equal to zero. There is one evitable discontinuity at x = - 1, and one definitive discontinuity at x = 2, where there is a vertical asymptote.

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Answer:

So g(6) comes out to be 2.

Step-by-step explanation:

So f(6)=5 means when f acts on 6, the result is 5....so the inverse would take 5 back to 6...or g(5)=6

So f(2)=6 means that f reassigns 2 to 6....so the inverse would take 6 back to 2....or g(6)=2

hope this helps :)

Answer: 3

Step-by-step explanation:

If f(x) = y, then invf(y) = x
So, f(5) = invg(5) = 3

Answer:

the perpendicular bisector

Explanation:

The perpendicular bisector will intersect the segment at its midpoint.

Check the picture below.

notice, the dividend and divisor must be in descending order, and when one of the variables is "missing", is really not missing, it simply has a coefficient of 0.

Here is the solution process:

                     3 x² -  2 x  -   6                      

x² + 0x + 2 | 3 x⁴ -  2 x³ +  0  x² -     x -   4

                    3  x⁴ + 0 x³ +  6  x²                

                             - 2 x³  -  6  x² -      x -   4

                             - 2 x³  + 0  x² -   4 x        

                                        -  6  x² +  3 x  -  4

                                        -  6  x²  + 0 x  -  12

                                                        3 x  + 8

Long division:

Therefore, the correct option is "A".

See below for the distance between the points and the lines

Question 2

The line and the points are given as:

x = y

P = (4, -2)

Rewrite the equation as:

y = x

The slope of the above equation is

m = 1

The slope of a line perpendicular to it is

m = -1

A linear equation is represented as:

y = mx + b

Substitute m = -1

y = -x + b

Substitute (4, -2) in y = -x + b

-2 = -4 + b

Solve for b

b = 2

Substitute b = 2 in y = -x + b

y = -x + 2

So, we have:

x = y and y = -x + 2

Substitute x for y

x = -x + 2

Solve for x

x = 1

Substitute x = 1 in y = x

y = 1

So, we have the following points

(1, 1) and (4, -2)

The distance between the above points is

d = √(x2 - x1)² + (y2 - y1)²

So, we have:

d = √(1 - 4)² + (1 + 2)²

Evaluate

d = 3√2

Hence, the distance between x = y and P = (4, -2) is 3√2 units

Question 3

The line and the points are given as:

y = 2x + 1

Q = (2, 10)

The slope of the above equation is

m = 2

The slope of a line perpendicular to it is

m = -1/2

A linear equation is represented as:

y = mx + b

Substitute m = -1/2

y = -1/2x + b

Substitute (2, 10) in y = -1/2x + b

10 = -1/2 * 2 + b

Solve for b

b = 11

Substitute b = 11 in y = -1/2x + b

y = -1/2x + 11

So, we have:

y = 2x + 1 and y = -1/2x + 11

Substitute 2x + 1 for y

2x + 1 = -1/2x + 11

Solve for x

x = 4

Substitute x = 4 in y = 2x + 1

y = 9

So, we have the following points

(4, 9) and (2, 10)

The distance between the above points is

d = √(x2 - x1)² + (y2 - y1)²

So, we have:

d = √(4 - 2)² + (9 - 10)²

Evaluate

d = √5

Hence, the distance between the line and the point is √5 units

Question 4

The line and the points are given as:

y = -x + 3

R = (-5, 0)

The slope of the above equation is

m = -1

The slope of a line perpendicular to it is

m = 1

A linear equation is represented as:

y = mx + b

Substitute m = 1

y = x + b

Substitute (-5, 0) in y = x + b

0 = 5 + b

Solve for b

b = -5

Substitute b = 5 in y = x + b

y = x + 5

So, we have:

y = x + 5 and y = -x + 3

Substitute x + 5 for y

x + 5 = -x + 3

Solve for x

x = -1

Substitute x = -1 in y = x + 3

y = 2

So, we have the following points

(-1, 2) and (-5, 0)

The distance between the above points is

d = √(x2 - x1)² + (y2 - y1)²

So, we have:

d = √(-1 + 5)² + (2 - 0)²

Evaluate

d = 2√5

Hence, the distance between the line and the point is 2√5 units

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Answer:

below

Step-by-step explanation:

1) slope = rise / run

2 coordinates are (-4, 0), (0, 2).

2 - 0 = 2

0 -- 4 = 4

2 / 4 = 1/2 so the slope is 0.5 or ½

2) it crosses the y axis at the average of the origin and 4.

4 + 0 = 4 / 2 = 2 so y intercept is 2.

3) in y= mx + b form

f(x) = ½x + 2, or, f(x) = 0.5x + 2

Answer:

Step-by-step explanation:

y - 4 = 5/4(x - 8)

y - 4 = 5/4x - 10

y = 5/4x - 6

Answer:

C

Step-by-step explanation:

the answer is c

The limit of j (x) as x approaches 3 is 4.

According to the question, A line begins at an open circle (2, 6) on a coordinate plane and descends through ( -2, 2). At, a complete circle is (3, 6). From a solid circle (2, 3), a curve leads to an open circle (3, 4). From the open circle to the closed circle, a line runs (6, 5).

From the graph, it can be seen that the limit of the function j(x) as the value of x approaches 3 is 4.

A diagram or pictorial representation that organizes the depiction of data or values is known as a graph.

The relationships between two or more items are frequently represented by the points on a graph.

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Answer:

4

Step-by-step explanation:

egg 2023

Answer: well one is

Bc its the smallest besides zero, but zero is neither odd or even

Step-by-step explanation:

The value of the probability is 0.30

Using the table of values, we have:

P(x <= -3) = P(x = -5) + P(x = -3)

From the table of values, we have:

P(x = -5) = 0.17

P(x = -3) = 0.13

Substitute the known values in the above equation

P(x <= -3) = 0.17 + 0.13

Evaluate the sum

P(x <= -3) = 0.30

Hence, the value of the probability is 0.30

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[tex]given \: that \: its \: linear \\ m = \frac{15 - 3}{3 - 0} = \frac{12}{3} = 4 [/tex]

[tex]b = y(0) = 3 \\ y = 4x + 3 \: [/tex]

Answer:

D) y = 4x + 3

Step-by-step explanation:

   [tex]\sf \boxed{\bf y = mx +b}[/tex]

Here, m is the slope and b is y intercept.

At y intercept, x = 0

         From the table, y intercept  = 3

Choose any two points from the table to find the slope.

(0 ,3)  & (3,15)

[tex]\sf \boxed{\bf Slope =\dfrac{y_2-y_1}{x_2-x_1}}[/tex]

           [tex]\sf =\dfrac{15-3}{3-0}\\\\=\dfrac{12}{3}\\\\=4[/tex]

m = 4 ; b = 3

 Equation of line:

                 y = 4x  + 3

Step-by-step explanation:

important of festival in nepali languages for class seven

Answer:

b

Step-by-step explanation:

The equation that can be used to find the cost of each binder n in dollars is 861=7n.

Given that the cost of 7 binders is $8.61.

We are required to form an equation that represents the total cost of each binder n in dollars.

Equation is like a relationship between all the variables that are expressed in equal to form.It may be linear equation or may be more types.

Suppose the cost of 1 binder is n dollar.

We know that the total cost is basically the product of price of 1 unit and number of quantities of units.

Total cost=Price of 1 unit* number of units

8.61=n*7

8.61=7n

Hence the equation that can be used to find the cost of each binder n in dollars is 861=7n.

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Answer:

-181 - 79x

Step-by-step explanation:

I am not sure your comment fits the given problem.

I get the following approach and solution :

T(1 + 4x) = 1×a + 4x×b = 2 + 2x

a = 2 + 2x - 4xb

T(4 + 15x) = 4×a + 15x×b = -3 + 3x

4×(2 + 2x - 4xb) + 15xb = -3 + 3x

8 + 8x - 16xb + 15xb = -3 + 3x

11 + 5x - xb = 0

11 + 5x = xb

b = (11 + 5x)/x

a = 2 + 2x - 4x(11 + 5x)/x = 2 + 2x - 44 - 20x = -42 - 18x

T(3 - 5x) = 3×a - 5xb = 3(-42 - 18x) - 5x(11 + 5x)/x =

= -126 - 54x - 55 - 25x = -181 - 79x

control :

T(1 + 4x) = a + 4xb = -42 - 18x + 44 + 20x = 2 + 2x

T(4 + 15x) = 4a + 15xb = -168 - 72x + 165 + 75x = -3 + 3x

Answer:

-25

Step-by-step explanation:

[tex] \dfrac{(20 - 5^2)(16 + 2^2)}{-2^3 + (3 \times 2^2)} = [/tex]

First, do all exponents.

[tex] = \dfrac{(20 - 25)(16 + 4)}{-8 + (3 \times 4)} [/tex]

Now do each operation in parentheses.

[tex] = \dfrac{(-5)(20)}{-8 + 12} [/tex]

Multiply in the numerator. Add in the denominator.

[tex] = \dfrac{-100}{4} [/tex]

Divide the numerator by the denominator.

[tex] = -25 [/tex]

Answer:

-25

Step-by-step explanation:

PEMDAS

The PEMDAS rule is an acronym representing the order of operations in math:

Given expression:

[tex]\sf \dfrac{(20-5^2)(16+2^2)}{-2^3+(3 \times 2^2)}[/tex]

As the given expression is a fraction, carry out the operations in the numerator and denominator first before finally dividing them.

Following PEMDAS, carry out the calculations inside the parentheses first, then carry out the rest of the calculations following the order of operations:

Parentheses

Calculate the exponents inside the parentheses:

[tex]\implies \sf \dfrac{(20-25)(16+4)}{-2^3+(3 \times 4)}[/tex]

Multiply:

[tex]\implies \sf \dfrac{(20-25)(16+4)}{-2^3+(12)}[/tex]

Add and subtract:

[tex]\implies \sf \dfrac{(-5)(20)}{-2^3+(12)}[/tex]

Exponents

Calculate the exponent:

[tex]\implies \sf \dfrac{(-5)(20)}{-8+(12)}[/tex]

Multiply and Divide

Multiply:

[tex]\implies \sf \dfrac{-100}{-8+(12)}[/tex]

Add and Subtract

Add:

[tex]\implies \sf \dfrac{-100}{4}[/tex]

Finally, divide the numerator by the denominator:

[tex]\implies \sf -25[/tex]

The length of the diagonal that he cut to the nearest whole inch is 36 inches

A red dotted line crosses the rectangle from the upper left corner to the lower right corner to form a triangle.

Length of the diagonal of a rectangle;

Hypotenuse² = adjacent² + opposite²

= 18² + 36²

= 324 + 1296

hyp² = 1620

hyp = √1620

hyp = 35.4964786985976

Approximately,

hypotenuse = 36 inches

Therefore, the length of the diagonal that he cut to the nearest whole inch is 36 inches.

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Step-by-step explanation:

sin(90-A) = sin90cosA-sinAcos90

= cosA*1-0*sinA

= cos90

hence proved

Step-by-step explanation:

sin(90-A)=cosA

sin(90-A)=sin(90-A)

90-A=90-A

-A+A=90-90

=0

Step-by-step explanation:

21) f(x)=1/x-6. g(x)=7/x+6

f(g(x))=f(7/x+6)=1÷7/x+6 - 6=x+6/7 - 6

g(f(x))=g(1/x-6)=7÷1/x-6 - 6 =7(x-6) - 6

simplify forward

25)f(x)=|x| g(x)=5x+1

f(g(x))=f(5x+1)=|5x+1|=5x+1=g(x)

g(f(x))=g(|x|)=5|x|+1=5x+1=g(x)

Answer: 3

Step-by-step explanation:

Substituting into vertex form, the equation is

[tex]y=a(x+2)^2 +2[/tex]

Substituting in the coordinates (-3, 5),

[tex]5=a(-3+2)^2 +2\\\\5=a+2\\\\a=3[/tex]

[tex]~~~~~~ \textit{Simple Interest Earned} \\\\ I = Prt\qquad \begin{cases} I=\textit{interest earned}\dotfill & \$132.90\\ P=\textit{original amount deposited}\dotfill & \$6000\\ r=rate\to 5.5\%\to \frac{5.5}{100}\dotfill &0.055\\ t=years \end{cases} \\\\\\ 132.90 = (6000)(0.055)(t)\implies \cfrac{132.90}{(6000)(0.055)}=t\implies \cfrac{443}{1100}=t \\\\\\ \stackrel{\textit{converting that to days}}{\cfrac{443}{1100}\cdot 365} ~~ \approx ~~ 147~days[/tex]

now, if we move back from August 14th by 147 days backwards, that'd put us on March 20th.

Answer:

g(0)⁻¹ = 6

Step-by-step explanation:

First, you must find the inverse of the function. Remember, another way of representing g(x) is with "y".  To find the inverse, you must swap the positions of the "x" and "y" variables in the equation. Then, you must rearrange the equation and isolate "y".

g(x) = 18 - 3x                                    <----- Original function

y = 18 - 3x                                        <----- Plug "y" in for g(x)

x = 18 - 3y                                        <----- Swap the positions of "x" and "y"

x + 3y = 18                                        <----- Add 3y to both sides

3y = 18 - x                                         <----- Subtract "x" from both sides

y = (18 - x) / 3                                    <----- Divide both sides by 3

y = 6 - (1/3)x                                     <----- Divide both terms by 3

Now that we have the inverse function, we need to plug x = 0 into the equation and solve for the output. In the inverse function, "y" is represented by the symbol g(x)⁻¹.

g(x)⁻¹ = 6 - (1/3)x                                  <----- Inverse function

g(0)⁻¹ = 6 - (1/3)(0)                               <----- Plug 0 in for "x"

g(0)⁻¹ = 6 - 0                                       <----- Multiply 1/3 and 0

g(0)⁻¹ = 6                                             <----- Subtract

The point estimate of difference of the sample his will be -20.

Based on the information given, the. following can be depicted:

n1 = 13

x1 = 141

s1 = 13

n2 = 9

x2 = 161

s2 = 12

The point estimate of difference will be:

= 141 - 161

= -20

The margin of error to be used in constructing the confidence interval will be calculated by multiplying the standard error which is 5.467 and the critical value. This will be:

= 5.467 × 2.528

= 13.822

The margin of error is 13.822.

The confidence interval will now be:

= (-20 + 13.822) and (-20 - 13.822)

= -6.178 and -33.822

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Answer:

Length = 30ft, Width = 5ft

Step-by-step explanation:

Let x be the width.

Area of Rectangle = Length * Width

Given from the information in the question,

Length = 6x ft

Width = x ft

Substitute the values into the formula:

6x * x = 150

[tex]6x^{2}[/tex] = 150

[tex]x^{2} =\frac{150}{6}[/tex]

[tex]x^{2} =25[/tex]

[tex]x=\sqrt{25}[/tex]

x = 5 ft.

Therefore,

Length = 6 * 5 = 30ft

Width = 5 ft.

Answer:

Simplified: 4X^3

Step-by-step explanation:

Simplify the expression.

9. The curve passes through the point (-1, -3), which means

[tex]-3 = a(-1) + \dfrac b{-1} \implies a + b = 3[/tex]

Compute the derivative.

[tex]y = ax + \dfrac bx \implies \dfrac{dy}{dx} = a - \dfrac b{x^2}[/tex]

At the given point, the gradient is -7 so that

[tex]-7 = a - \dfrac b{(-1)^2} \implies a-b = -7[/tex]

Eliminating [tex]b[/tex], we find

[tex](a+b) + (a-b) = 3+(-7) \implies 2a = -4 \implies \boxed{a=-2}[/tex]

Solve for [tex]b[/tex].

[tex]a+b=3 \implies b=3-a \implies \boxed{b = 5}[/tex]

10. Compute the derivative.

[tex]y = \dfrac{x^3}3 - \dfrac{5x^2}2 + 6x - 1 \implies \dfrac{dy}{dx} = x^2 - 5x + 6[/tex]

Solve for [tex]x[/tex] when the gradient is 2.

[tex]x^2 - 5x + 6 = 2[/tex]

[tex]x^2 - 5x + 4 = 0[/tex]

[tex](x - 1) (x - 4) = 0[/tex]

[tex]\implies x=1 \text{ or } x=4[/tex]

Evaluate [tex]y[/tex] at each of these.

[tex]\boxed{x=1} \implies y = \dfrac{1^3}3 - \dfrac{5\cdot1^2}2 + 6\cdot1 - 1 = \boxed{y = \dfrac{17}6}[/tex]

[tex]\boxed{x = 4} \implies y = \dfrac{4^3}3 - \dfrac{5\cdot4^2}2 + 6\cdot4 - 1 \implies \boxed{y = \dfrac{13}3}[/tex]

11. a. Solve for [tex]x[/tex] where both curves meet.

[tex]\dfrac{x^3}3 - 2x^2 - 8x + 5 = x + 5[/tex]

[tex]\dfrac{x^3}3 - 2x^2 - 9x = 0[/tex]

[tex]\dfrac x3 (x^2 - 6x - 27) = 0[/tex]

[tex]\dfrac x3 (x - 9) (x + 3) = 0[/tex]

[tex]\implies x = 0 \text{ or }x = 9 \text{ or } x = -3[/tex]

Evaluate [tex]y[/tex] at each of these.

[tex]A:~~~~ \boxed{x=0} \implies y=0+5 \implies \boxed{y=5}[/tex]

[tex]B:~~~~ \boxed{x=9} \implies y=9+5 \implies \boxed{y=14}[/tex]

[tex]C:~~~~ \boxed{x=-3} \implies y=-3+5 \implies \boxed{y=2}[/tex]

11. b. Compute the derivative for the curve.

[tex]y = \dfrac{x^3}3 - 2x^2 - 8x + 5 \implies \dfrac{dy}{dx} = x^2 - 4x - 8[/tex]

Evaluate the derivative at the [tex]x[/tex]-coordinates of A, B, and C.

[tex]A: ~~~~ x=0 \implies \dfrac{dy}{dx} = 0^2-4\cdot0-8 \implies \boxed{\dfrac{dy}{dx} = -8}[/tex]

[tex]B:~~~~ x=9 \implies \dfrac{dy}{dx} = 9^2-4\cdot9-8 \implies \boxed{\dfrac{dy}{dx} = 37}[/tex]

[tex]C:~~~~ x=-3 \implies \dfrac{dy}{dx} = (-3)^2-4\cdot(-3)-8 \implies \boxed{\dfrac{dy}{dx} = 13}[/tex]

12. a. Compute the derivative.

[tex]y = 4x^3 + 3x^2 - 6x - 1 \implies \boxed{\dfrac{dy}{dx} = 12x^2 + 6x - 6}[/tex]

12. b. By completing the square, we have

[tex]12x^2 + 6x - 6 = 12 \left(x^2 + \dfrac x2\right) - 6 \\\\ ~~~~~~~~ = 12 \left(x^2 + \dfrac x2 + \dfrac1{4^2}\right) - 6 - \dfrac{12}{4^2} \\\\ ~~~~~~~~ = 12 \left(x + \dfrac14\right)^2 - \dfrac{27}4[/tex]

so that

[tex]\dfrac{dy}{dx} = 12 \left(x + \dfrac14\right)^2 - \dfrac{27}4 \ge 0 \\\\ ~~~~ \implies 12 \left(x + \dfrac14\right)^2 \ge \dfrac{27}4 \\\\ ~~~~ \implies \left(x + \dfrac14\right)^2 \ge \dfrac{27}{48} = \dfrac9{16} \\\\ ~~~~ \implies \left|x + \dfrac14\right| \ge \sqrt{\dfrac9{16}} = \dfrac34 \\\\ ~~~~ \implies x+\dfrac14 \ge \dfrac34 \text{ or } -\left(x+\dfrac14\right) \ge \dfrac34 \\\\ ~~~~ \implies \boxed{x \ge \dfrac12 \text{ or } x \le -1}[/tex]

13. a. Compute the derivative.

[tex]y = x^3 + x^2 - 16x - 16 \implies \boxed{\dfrac{dy}{dx} = 3x^2 - 2x - 16}[/tex]

13. b. Complete the square.

[tex]3x^2 - 2x - 16 = 3 \left(x^2 - \dfrac{2x}3\right) - 16 \\\\ ~~~~~~~~ = 3 \left(x^2 - \dfrac{2x}3 + \dfrac1{3^2}\right) - 16 - \dfrac13 \\\\ ~~~~~~~~ = 3 \left(x - \dfrac13\right)^2 - \dfrac{49}3[/tex]

Then

[tex]\dfrac{dy}{dx} = 3 \left(x - \dfrac13\right)^2 - \dfrac{49}3 \le 0 \\\\ ~~~~ \implies 3 \left(x - \dfrac13\right)^2 \le \dfrac{49}3 \\\\ ~~~~ \implies \left(x - \dfrac13\right)^2 \le \dfrac{49}9 \\\\ ~~~~ \implies \left|x - \dfrac13\right| \le \sqrt{\dfrac{49}9} = \dfrac73 \\\\ ~~~~ \implies x - \dfrac13 \le \dfrac73 \text{ or } -\left(x-\dfrac13\right) \le \dfrac73 \\\\ ~~~~ \implies \boxed{x \le 2 \text{ or } x \ge \dfrac83}[/tex]

[tex]f(x) = \sqrt[3]{11x} \\ y = \sqrt[3]{11x} \\ x = \sqrt[3]{11y} \\ x {}^{3} = 11y \\ y = \frac{x {}^{3} }{11} [/tex]

[tex]f(x) = \frac{x}{7} + 10 \\ y = \frac{x}{7} + 10 \\ x = \frac{y}{7} + 10 \\ x - 10 = \frac{y}{7} \\ y = \frac{x - 10}{7} [/tex]

[tex]f(x) = \frac{7}{x} - 2 \\ y = \frac{7}{x} - 2 \\ x = \frac{7}{y} + 2 \\ x - 2 = \frac{7}{y} \\ y = \frac{7}{x - 2} [/tex]

[tex]f(x) = 9x - 6 \\ y = 9x - 6 \\ x = 9y - 6 \\ x + 6 = 9y \\ y = \frac{x + 6}{9} = g(x)[/tex]

Answer: 19/40

Step-by-step explanation:

First Write 0.475 as 0.4751Multiply both numerator and denominator by 10 for every number after the decimal point0.475 × 10001 × 1000 = 4751000. Reducing the fraction gives The answer