Find the range of y=7x-34/x-5

Using a system of equations, it is found that Allison bought 2 bottles of juice and 9 bottles of soda.

A system of equations is when two or more variables are related, and equations are built to find the values of each variable.

For this problem, the variables are:

Allison purchased a total of 11 bottles of juice, hence:

x + y = 11 -> x = 11 - y.

These 11 bottles contain 445 grams of sugar, hence, considering the amounts of each bottle, we have that:

20x + 45y = 445

Since x = 11 - y:

20(11 - y) + 45y = 445

25y = 225

y = 225/25

y = 9.

x = 11 - y = 11 - 9 = 2.

She bought 2 bottles of juice and 9 bottles of soda.

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Answer: Logarithmic

Explanation:

This curve is a reflection of the exponential curve over the line y = x, to show that it is the inverse of exponentials. We use logs to help isolate the exponent among other useful properties.

[tex] \qquad \qquad \bf \huge\star \: \: \large{ \underline{Answer} } \huge \: \: \star[/tex]

[tex]\textsf{ \underline{\underline{Steps to solve the problem} }:}[/tex]

Take HJ = a, GH = b and GJ = c

put the value of a from equation 1 in equation 2

now, put the value of a and c in equation 3

Now, we need to find HJ (a)

[tex]{ \qquad \large \sf {Conclusion} :} [/tex]

Answer:

  y = -1/2n +40

Step-by-step explanation:

We are given two ordered pairs (stalks, ounces) and asked for the slope-intercept form equation of the line through them.

The slope of the desired line can be found from the formula ...

  m = (y2 -y1)/(x2 -x1)

For the given points (30, 25) and (32, 24), the slope is ...

  m = (24 -25)/(32 -30) = -1/2

The y-intercept of the desired line can be found from the formula ...

  b = y -mn

  b = 25 -(-1/2)(30) = 25 +15 = 40

The slope-intercept equation of a line is ...

  y = mn +b . . . . . line with slope m and y-intercept b

  y = -1/2n +40 . . . . . . line with slope -1/2 and y-intercept 40

The linear relationship between stalks (n) and yield (y) is ...

  y = -1/2n +40

It looks like the function is

[tex]f(x,y) = 3 + 4x^{3/2}[/tex]

We have

[tex]\dfrac{\partial f}{\partial x} = 6x^{1/2} \implies \left(\dfrac{\partial f}{\partial x}\right)^2 = 36x[/tex]

[tex]\dfrac{\partial f}{\partial y} = \left(\dfrac{\partial f}{\partial y}\right)^2 = 0[/tex]

Then the area of the surface over [tex]R[/tex] is

[tex]\displaystyle \iint_R f(x,y) \, dS = \iint_R \sqrt{1 + 36x + 0} \, dA \\\\ ~~~~~~~~ = \int_0^5 \int_0^2 \sqrt{1+36x} \, dx \, dy \\\\ ~~~~~~~~ = 5 \int_0^2 \sqrt{1+36x} \, dx \\\\ ~~~~~~~~ = \frac5{36} \int_1^{73} \sqrt u \, du \\\\ ~~~~~~~~ = \frac5{36}\cdot \frac23 \left(73^{3/2} - 1^{3/2}\right) = \boxed{\frac5{54} (73^{3/2} - 1)}[/tex]

Part (a)

[tex]z=\frac{85-75}{11.4} \approx \boxed{0.88}[/tex]

By similar logic, the answers to the other parts are

(b) 1.75

(c) -2.19

(d) 0

The equation of the parabola with the given vertex and directrix in vertex form is y = (1/16)( x + 5 )² - 9.

Hence, option C is the correct answer.

Given the data in the question;

To find the equation, we use the equation of the parabola that opens up or down since the directrix ( y = -13 ) is vertical.

The equation is expressed as;

( x - h )² = 4p( y - k )

First, we find the distance from the focus to the vertex.

|p| is the distance rom the focus to the vertex and from the vertex to the directrix.

p = -9 + 13

p = 4

We substitute the values into the equation;

( x - h )² = 4p( y - k )

( x - (-5) )² = 4(4)( y - (-9) )

( x + 5 )² = 16( y + 9 )

Multiply both side by 1/16

(1/16)( x + 5 )² =  y + 9

Make y the subject of the formula

(1/16)( x + 5 )² - 9 =  y

y = (1/16)( x + 5 )² - 9

The equation of the parabola with the given vertex and directrix in vertex form is y = (1/16)( x + 5 )² - 9.

Hence, option C is the correct answer.

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The number of strings that can be formed by ordering the letters SCHOOL using some or all of the letters is 1440 strings

Permutation can be defined as number of ways in which things can arranged.

We were given to find how many strings that can be formed by ordering the letters SCHOOL using some or all of the letters.

First of all, How many distinct letters are in the word SCHOOL ?

The distinct letters are 5 in numbers.

What is the total number of letters in the word SCHOOL ?

The total number of letters is 6.

Then

6! + 5[tex]P_{5}[/tex]

That is, 6 factorial + 6 permutation 5

( 6 x 5 x 4 x 3 x 2 x 1 ) + 6!/( 6 - 5)!

720 + 720

1440 strings

Therefore, the number of strings that can be formed by ordering the letters SCHOOL using some or all of the letters is 1440 strings

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Answer:

A:The Reqd. No. of Permutations=3360

Step-by-step explanation:

I couldn’t find the answer to B..sorry

The solutions to the quadratic equation of the sprinkler in the exact form are x = 7 + 2√110 and x = 7 - 2√110

Quadratic equations are second-order polynomial equations and they have the form y = ax^2 + bx + c or y = a(x - h)^2 + k

A quadratic equations can be split to several equations and it can be solved as a whole

In this case, the quadratic equation is given as

-x^2 + 14x + 61 = 0

Using the form of the quadratic equation y = ax^2 + bx + c, we have

a = -1, b = 14 and c = 61

The quadratic equation can be solved using the following formula

x = (-b ± √(b^2 - 4ac))/2a

Substitute the known values of a, b and c in the above equation

x = (-14 ± √(14^2 - 4*-1*61))/2*-1

Evaluate the exponent

x = (-14 ± √(196 - 4*-1*61))/2*-1

Evaluate the products

x = (-14 ± √(196 + 244))/-2

Evaluate the sum

x = (-14 ± √(440))/-2

Express 440 as 4 * 110

x = (-14 ± √(4 * 110))/-2

Take the square root of 4

x = (-14 ± 2√110)/-2

Evaluate the quotient

x = 7 ± 2√110

Split the equation

x = 7 + 2√110 and x = 7 - 2√110

Hence, the solutions to the quadratic equation of the sprinkler in the exact form are x = 7 + 2√110 and x = 7 - 2√110

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The total amount in the transaction account is

= $625.

The amount in the transaction account is

=75

Your salary and any other payments that you get may be deposited into an account known as a transaction account, which is an account that you utilize on a day-to-day basis. When connected to a Visa card, your transaction account may also be used to pay bills, go shopping, and make other types of payments that are commonplace. Balances in transaction accounts do not accrue any interest for the account holder.

b) Now if the reserve requirement changes to 5%, we need to reconstruct the balance sheet of the total banking system.

Calculate the initial total reserve using the new reserve requirement as follows:

= 4% x500

= 0.04×500

=20

Calculate the excess reserve as follows:

=30-20

= $10

Calculate the money multiplier as follows:

=1/4%

=1/0.04

=25

Calculate the increase in transaction account as follows:

=25x $5

= $125

Hence, the total amount in the transaction account is

$500+ $125

= $625.

c)

Therefore, the amount in the transaction account is

=600-525

=75

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The nth term of the sequence is 2n - 8

The nth term of an arithmetic progression is expressed as;

Tn = a + (n - 1)d

where

a is the first term

d is the common difference

n is the number of terms

Given the following parameters

a = f(1)=−6

f(2) = −4

Determine the common difference

d = f(2) - f(1)

d = -4 - (-6)
d = -4 + 6

d = 2

Determine the nth term of the sequence

Tn = -6 + (n -1)(2)

Tn = -6+2n-2
Tn = 2n - 8

Hence the nth term of the sequence is 2n - 8

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By definition, we have

[tex]f(n) = f(n - 1) + f(n - 2)[/tex]

so that by substitution,

[tex]f(n-1) = f(n-2) + f(n-3) \implies f(n) = 2f(n-2) + f(n-3)[/tex]

[tex]f(n-2) = f(n-3) + f(n-4) \implies f(n) = 3f(n-3) + 2f(n-4)[/tex]

[tex]f(n-3) = f(n-4) + f(n-5) \implies f(n) = 5f(n-4) + 3f(n-5)[/tex]

[tex]f(n-4) = f(n-5) + f(n-6) \implies f(n) = 8f(n-5) + 5f(n-6)[/tex]

and so on.

Recall the Fibonacci sequence [tex]F(n)[/tex], whose first several terms for [tex]n\ge1[/tex] are

[tex]\{1, 1, 2, 3, 5, 8, 13, 21, 34, 55, \ldots\}[/tex]

Let [tex]F_n[/tex] denote the [tex]n[/tex]-th Fibonacci number. Notice that the coefficients in each successive equation form at least a part of this sequence.

[tex]f(n) = f(n-1) + f(n-2) = F_2f(n-1) + F_1 f(n-2)[/tex]

[tex]f(n) = 2f(n-2) + f(n-3) = F_3 f(n-2) + F_2 f(n-3)[/tex]

[tex]f(n) = 3f(n-3) + 2f(n-4) = F_4 f(n-3) + F_3 f(n-4)[/tex]

[tex]f(n) = 5f(n-4) + 3f(n-5) = F_5 f(n-4) + F_4 f(n-5)[/tex]

[tex]f(n) = 8f(n-5) + 5f(n-6) = F_6 f(n-5) + F_5 f(n-6)[/tex]

and so on. After [tex]k[/tex] iterations of substituting, we would end up with

[tex]f(n) = F_{k+1} f(n - k) + F_k f(n - (k+1))[/tex]

so that after [tex]k=n-2[/tex] iterations,

[tex]f(n) = F_{(n-2)+1} f(n - (n-2)) + F_{n-2} f(n - ((n-2)+1)) \\\\ f(n) = f(2) F_{n-1} + f(1) F_{n-2} \\\\ \boxed{f(n) = -4 F_{n-1} - 6 F_{n-2}}[/tex]

Answer: -221

Step-by-step explanation:

31 is smaller than 252. Therefore, since 31 MINUS x equals 252, x needs to be a negative number in order to complete the equation (recall that a negative number times a negative number equals a positive number).

Therefore, if we subtract 31 on both sides, in other words transpose, we get,

-x = 221

The coefficient of -x is -1, however it is not written as it's implied that if there is not written coefficient in from of a variable then the coefficient of the variable is 1 or -1, depending on its sign.

Therefore, dividing -1 on both sides, we get,

x = -221

Hence, the desired answer is -221.

Answer: [tex]\boxed{30^{\circ}, 60^{\circ}, 90^{\circ}}[/tex]

Step-by-step explanation:

Let the smallest angle be x.

Then, the middle angle is x+30.

The largest angle is 2x+30.

Angles in a triangle add to 180 degrees, so:

[tex]x+x+30+2x+30=180\\\\4x+60=180\\\\4x=120\\\\x=30[/tex]

So, the angles measure [tex]\boxed{30^{\circ}, 60^{\circ}, 90^{\circ}}[/tex]

The measures of all three angles are 30°,60°, and 90°.

A triangle is a 3-sided polygon occasionally (though not frequently) referred to as the trigon. Every triangle has three sides and three angles, some of which may be the same.

Find the measures of all three angles:

The three angles be a, b and c.

Angle “a" = x

Angle “b" = x + 30°

Angle “c" = “a" + “b"

                = x + x + 30°

                = 2x + 30°

Total angle in a triangle = 180°

Therefore;

“a" + “b" + “c" = 180°

Births; (x + (x + 30°) + (2x + 30°)) = 180°

So,

4x + 60° = 180°

x = (180° - 60°) ÷ 4

x = 30°

Plugging this value of x into the earlier equations for angles “a,” “b,” and “c."

“a" = x = 30°

“b" = x + 30° = 60°

“c" = 2x + 30° = 90°

The measures of all three angles are 30°,60°, and 90°.

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Answer: 70 and 60 degrees

Step-by-step explanation:

Angle AOB = 180 - 50 - 60 = 70 degrees so x is 70 degrees

Angle OCD = 180 - 130 = 50 so y = 180 - 70 - 50 = 60 degrees

Answer:

3%, 32%

Step-by-step explanation:

winning 1 game only: two possibilities

a. winning balltoss, losing dart, which is 20%*85% = 17%

b. winning dart, losing ball toss, which is 15%*80% = 12%

so winning 1 game only: 29%

winning both games:

20% * 15% = 3%

winning either one: winning both games+winning 1 game only

29% + 3% = 32%

I'm not sure if the last two apostrophes are part of the quote - "Solve ... " - or if you mean the second derivative [tex]x''[/tex]. I think you mean the first interpretation, but I'll include both cases since they are both solvable.

If the former is correct, separate variables to solve.

[tex]x' = -9tx \implies \dfrac{dx}{dt} = -9tx \implies \dfrac{dx}x = -9t\,dt[/tex]

Integrate both sides to get

[tex]\ln|x| = -\dfrac92 t^2 + C[/tex]

Solve for [tex]x[/tex].

[tex]e^{\ln|x|} = e^{-9/2\,t^2 + C} \implies \boxed{x = Ce^{-9/2\,t^2}}[/tex]

If you meant the latter, then the ODE can be rewritten as

[tex]9t x'' + x' = 0[/tex]

Reduce the order of the equation by substituting [tex]y(t) = x'(t)[/tex] and [tex]y'(t) = x''(t)[/tex].

[tex]9t y' + y = 0[/tex]

Solve for [tex]y'[/tex] and separate variables.

[tex]y' = -\dfrac y{9t} \implies \dfrac{dy}{dt} = -\dfrac y{9t} \implies \dfrac{dy}y = -\dfrac{dt}{9t}[/tex]

Integrate.

[tex]\ln|y| = -\dfrac19 \ln|t| + C[/tex]

Solve for [tex]y[/tex].

[tex]e^{\ln|y|} = e^{-1/9 \,\ln|t| + C} \implies y = Ct^{-1/9}[/tex]

Solve for [tex]x[/tex] by integrating.

[tex]x' = Ct^{-1/9} \implies x = C_1 t^{8/9} + C_2[/tex]

The two different ways of finding the area are,
Case 1 = assume horizontal rectangles,
Case 2 = assume vertical rectangles.

the rectangle is a four-sided polygon whose opposites sides are equal and has an angle of 90° between its sides.

Here,

case 1,
As shown in the image
Area = sum of horizontal rectangles
Area = 10 * 3 + 2 * 5 + 2 * 1
Area  = 30 + 10 + 2
Area = 42    

Case II,
As shown in right figure,
Area of the vertical rectangles
Area = 3 * 5 + 5 * 3 + 2 * 6
Area = 15 + 15 + 12
Area = 42
Here, the area in case 1 is equal to case 2.

Thus, the two different ways of finding the area have shown above.

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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]

The missing value in the logarithm are as follows:

Using logarithm rule,

logₐ b - logₐ c = logₐ (b / c)

logₐ b + logₐ c = logₐ (b × c)

Therefore,

log₃ 7 - log₃ 2 = log₃ (7 / 2)

log₉ 7 + log₉ 4 = log₉ (7 × 4) =  log₉ 28

log₆ 1 / 81 = log₆ 81⁻¹ = log₆ 3⁻⁴ =  - 4 log₆ 3

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The relationship of the graph g(x) = (x − 2)^3 + 6 compare to the parent function of f(x) = x^3 is that g(x) is shifted 2 units to the right and 6 units up.

Translations is a transformation technique that changes the position of an object from one point on the plane to another.

Given the function below

g(x) = (x − 2)^3 + 6

The function compared to f(x) = x^3, shows a translation of f(x) by 2 unit to the right along the horizontal and vertical translation of the function 6 units up

Hence the relationship of the graph g(x) = (x − 2)^3 + 6 compare to the parent function of f(x) = x^3 is that g(x) is shifted 2 units to the right and 6 units up.

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Answer:

g(x) is shifted 6 units to the left and 2 units down.

Step-by-step explanation:

took the test

Answer:

Step-by-step explanation:

I am going to be honest here. I know the answer is 22 but I cant really explain it you kinda just have to trust I'm right.

The Gauss-Jordan elimination method different from the Gaussian elimination method in that unlike the Gauss-Jordan approach, which reduces the matrix to a diagonal matrix, the Gauss elimination method reduces the matrix to an upper-triangular matrix.

Gauss-Jordan Elimination is a technique that may be used to discover the inverse of any invertible matrix as well as to resolve systems of linear equations.

It is based on the following three basic row operations that one may apply to a matrix: Two of the rows should be switched around. Multiply a nonzero scalar by one of the rows.

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Answer: [tex]y= x-8[/tex]

Step-by-step explanation:

[tex]y-x=-8[/tex]

[tex]y-x+x=-8+x[/tex]

Final Answer: [tex]y=x-8[/tex]

Answer:

1.60

Step-by-step explanation:

[tex]~\hspace{7em}\textit{rational exponents} \\\\ a^{\frac{ n}{ m}} \implies \sqrt[ m]{a^ n} ~\hspace{10em} a^{-\frac{ n}{ m}} \implies \cfrac{1}{a^{\frac{ n}{ m}}} \implies \cfrac{1}{\sqrt[ m]{a^ n}} \\\\[-0.35em] ~\dotfill\\\\ a^{\frac{1}{5}}\implies \sqrt[5]{a^1}\implies \sqrt[5]{a}[/tex]

Using translation concepts, it is found that pi is the phase shift of the graph, and since it is negative, the graph is shifted right pi units.

A translation is represented by a change in the function graph, according to operations such as multiplication or sum/subtraction either in it’s definition or in it’s domain. Examples are shift left/right or bottom/up, vertical or horizontal stretching or compression, and reflections over the x-axis or the y-axis.

In this problem, the change is given as follows:

x -> x - pi

It means that the change is in the domain, in which pi is the phase shift of the graph, and since it is negative, the graph is shifted right pi units.

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since there are 365 days in a year, so 180 days is really just 180/365 of a year, so

[tex]~~~~~~ \textit{Simple Interest Earned} \\\\ I = Prt\qquad \begin{cases} I=\textit{interest earned}\\ P=\textit{original amount deposited}\dotfill & \$1295\\ r=rate\to 7\%\to \frac{7}{100}\dotfill &0.07\\ t=years\to \frac{180}{365}\dotfill &\frac{36}{73} \end{cases} \\\\\\ I = (1295)(0.07)(\frac{36}{73})\implies I\approx 44.7[/tex]