Given Q= [2 3]
[1 -2] prove that (3Q)^(t) = 3Q^(t)

Sweety bought enough bottles of sports drink to fill a big cooler for the skateboard team. It toom 25. 5 bottles to fill the cooler and each bottle contained 1. 8 liters. There are 46.8 litres in cooler.

To find the number of liters in the cooler, we need to multiply the number of bottles by the amount of liquid in each bottle. Given that it took 25.5 bottles to fill the cooler and each bottle contains 1.8 liters, we can find the total amount of liquid in the cooler by multiplying these two values together.

First, let's round the number of bottles to the nearest whole number, which is 26.

To calculate the total amount of liquid in the cooler, we multiply the number of bottles by the amount of liquid in each bottle:

26 bottles * 1.8 liters/bottle = 46.8 liters

Therefore, there are 46.8 liters in the cooler.

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Answer:

4.44%

Step-by-step explanation:

%Error = [tex]\frac{E-T}{T}[/tex] x 100

E = experiment

T = Theoretical

E = 86

T = 90

What is the percent error?​

We Take

[tex]\frac{86-90}{90}[/tex] x 100 ≈ 4.44%

So, the percent error is about 4.44%

The linear equation on the graph is:

y = 4x + 20

The general linear equation in slope-intercept form is:

y = ax +b

Where a is the slope and b is the y-intercept.

On the graph we can see that the y-intercept is y = 20, then we can write:

y = ax + 20

We also can see that the line passes through (5, 40), then we can replace these values to get:

40 = 5a + 20

40 - 20 = 5a

20 = 5a

20/5 = a

4 = a

The linear equation is:

y = 4x + 20

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To achieve a future amount of $500,000 in 20 years at a monthly rate of return of 0.5% (6% annually compounded continuously), we need to invest $1,465.68 per month (rounded to the nearest cent).

Given:

Initial amount to be invested = k

Monthly rate of return = 6%/12

                                    = 0.5%/month

Number of months in 20 years = 20 × 12

                                                   = 240

Future amount required = $500,000

First, we will find the formula to calculate future amount as we are given present value, rate of return and time period.

A=P(1 + r/n)nt

where A = future amount

P = present value (initial investment)

r = annual interest rate (as a decimal)

n = number of times the interest is compounded per year

t = number of years

Therefore, here A = future amount, P = 0, r = 6% = 0.06, n = 12, and t = 20 years.

Thus,  A= 0(1 + 0.06/12)^(12×20)

             = 0(1.005)^240

             = 0 × 2.653

             = 0

The future amount is 0 dollars, which means that we cannot achieve our goal of five hundred thousand dollars if we don't invest anything at the beginning of each month.

Now, let's find out how much we need to invest monthly to achieve our target future amount.

500,000 = k[(1 + 0.005)^240 - 1] / (0.005)

k = 500,000 × 0.005 / [(1 + 0.005)^240 - 1]k

   = $1,465.68/month

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a) The interest rate for this problem is given as follows: r = 0.054.

b) The value of the loan after 10 years is given as follows: 12,690.2 pounds.

The amount of money earned, in compound interest, after t years, is given by:

[tex]A(t) = P\left(1 + \frac{r}{n}\right)^{nt}[/tex]

In which:

The interest rate for this problem is obtained as follows:

7905/7500 - 1 = 1.054 - 1 = 0.054.

The parameters are given as follows:

P = 7500, n = 1.

Hence the balance after 10 years is given as follows:

[tex]A(10) = 7500(1.054)^{10} = 12690.2[/tex]

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Answer:

3

Step-by-step explanation:

First find all the factors of 48:

1, 2, 3, 4, 6, 8, 12, 16, 24, 48

These are the only values that x can be.  Try them all and see which results in a whole number:

√48/1 = 6.93  not whole

√48/2 = 4.9  not whole

√48/3 = 4  WHOLE

√48/4 = 3.46  not whole

√48/6 = 2.83  not whole

√48/8 = 2.45  not whole

√48/12 = 2  WHOLE

√48/16 = 1.73  not whole

√48/24 = 1.41  not whole

√48/48 = 1  WHOLE

Therefore, there are 3 values of x for which √48/x = whole number.  The numbers are x = 3, 12, 48

It will take approximately 7 years and 3 months for the quarterly deposits to accumulate to $16,440 at an interest rate of 8.48% compounded quarterly.

To calculate the  time it takes for quarterly deposits of $425 to accumulate to $16,440 at an interest rate of 8.48% compounded quarterly, we can use the formula for compound interest:

A = P(1 + r/n)^(nt).

Where: A = Final amount ($16,440);

P = Quarterly deposit amount ($425);

r = Annual interest rate (8.48% or 0.0848);

n = Number of compounding periods per year (4 for quarterly); t = Time in years.  We need to solve for t. Rearranging the formula, we get:

t = (log(A/P) / log(1 + r/n)) / n.

Substituting the given values into the formula, we have:

t = (log(16440/425) / log(1 + 0.0848/4)) / 4.

Using a calculator, we find that t is approximately 7.27 years. Converting the decimal part to months (0.27 * 12),  we get 3.24 months. Therefore, it will take approximately 7 years and 3 months for the quarterly deposits to accumulate to $16,440 at an interest rate of 8.48% compounded quarterly.

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A. The probability of selecting an even number is P(A) = 2/5.

B. The probability of selecting a multiple of 4 is P(B) = 1/5.

C.  The probability of selecting a prime number is P(C) = 2/5.

To find the indicated probabilities, let's consider the events one by one:

A. The event "the selected number is even":
- Out of the sample space S={1,2,3,4,15}, the even numbers are 2 and 4.

- Therefore, the favorable outcomes for this event are {2,4}, and the total number of outcomes in the sample space is 5.

- The probability of selecting an even number is the ratio of favorable outcomes to the total number of outcomes: P(A) = favorable outcomes / total outcomes = 2/5.

B. The event "the selected number is a multiple of 4":
- From the sample space S={1,2,3,4,15}, the multiples of 4 is only 4.

- The favorable outcomes for this event are {4}, and the total number of outcomes is still 5.

- Therefore, the probability of selecting a multiple of 4 is P(B) = 1/5.

C.The event "the selected number is a prime number":
- Prime numbers are numbers greater than 1 that have no divisors other than 1 and themselves. From the given sample space S={1,2,3,4,15}, the prime numbers are 2 and 3.

- The favorable outcomes for this event are {2,3}, and the total number of outcomes is 5.

- So, the probability of selecting a prime number is P(C) = 2/5.

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Answer:

zeroes of the equations are x= 1 , -3      

Step-by-step explanation:

firstly divide both sides by 2 so new equation will be

x^2+2x-3=0

you can use quadratic formula or simply factor it

its factors will be

x^2 +3x - x -3=0

(x+3)(x-1)=0

are two factors

so

either

x+3=0             or          x-1=0

x=-3                 and        x=1  

so zeroes of the equations are x= 1 , -3      

by the way you can also use quadratic formula which is

[-b+-(b^2 -4ac)]/2a

where a is coefficient of x^2 and b is coefficient of x

and c is constant term

(a) The simultaneous equations representing the given information can be expressed in matrix form as:

3y - x = -6

x + y = 74

In matrix form, this can be written as:

[ 1   1 ] [ x ]   [ 74 ]

(b) To verify that the simultaneous equations have a unique solution, we can check the determinant of the coefficient matrix [ 3 -1 ; 1 1 ]. If the determinant is non-zero, then a unique solution exists.

(c) To solve for x and y using the inverse matrix method, we can represent the system of equations in matrix form:

where A is the coefficient matrix, X is the column vector [ x ; y ], and B is the column vector of constants [ -6 ; 74 ]. By multiplying both sides of the equation by the inverse of matrix A, we can isolate X:

[tex]A^(-1) * (A * X) = A^(-1) * B[/tex]

X = [tex]A^(-1) * B[/tex]

By calculating the inverse of matrix A and multiplying it by matrix B, we can find the values of x and y.

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The IVP has a unique solution defined on some interval I with 0 € I.

here is the  solution to show that there is some interval I with 0 € I such that the IVP has a unique solution defined on I:

The given differential equation is y = y³ + 2.

The initial condition is y(0) = 1.

Let's first show that the differential equation is locally solvable. This means that for any fixed point x0, there is an interval I around x0 such that the IVP has a unique solution defined on I.

To show this, we need to show that the differential equation is differentiable and that the derivative is continuous at x0.

The differential equation is differentiable at x0 because the derivative of y³ + 2 is 3y².

The derivative of 3y² is continuous at x0 because y² is continuous at x0.

Therefore, the differential equation is locally solvable.

Now, we need to show that the IVP has a unique solution defined on some interval I with 0 € I.

To show this, we need to show that the solution does not blow up as x approaches infinity.

We can show this by using the fact that y³ + 2 is bounded above by 2.

This means that the solution cannot grow too large as x approaches infinity.

Therefore, the IVP has a unique solution defined on some interval I with 0 € I.

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A basis for the vector space {f(x) ∈ P3[x] | ƒ′(6) = ƒ(1)} is {p(x) = ax^2 + bx + 11a, q(x) = dx}, where a and d can be any real numbers.

To find a basis {p(x), q(x)} for the given vector space {f(x) ∈ P3[x] | ƒ′(6) = ƒ(1)}, we need to find two polynomials p(x) and q(x) that satisfy the condition ƒ′(6) = ƒ(1) and are linearly independent.

Let's start by finding p(x):

We can choose p(x) as a polynomial of degree 2 since we are working with P3[x].

Let p(x) = ax^2 + bx + c.

Taking the derivative of p(x), we have:

p'(x) = 2ax + b.

We need p'(6) to be equal to p(1), so let's evaluate them:

p'(6) = 2a(6) + b = 12a + b

p(1) = a(1)^2 + b(1) + c = a + b + c

For p'(6) = p(1), we have:

12a + b = a + b + c

Simplifying this equation, we get:

11a = c

So, we can choose c = 11a.

Thus, p(x) = ax^2 + bx + 11a.

Now, let's find q(x):

We can choose q(x) as a polynomial of degree 1 since we are working with P3[x].

Let q(x) = dx + e.

Taking the derivative of q(x), we have:

q'(x) = d.

We need q'(6) to be equal to q(1), so let's evaluate them:

q'(6) = d

q(1) = d(1) + e = d + e

For q'(6) = q(1), we have:

d = d + e

Simplifying this equation, we get:

e = 0

Thus, q(x) = dx.

Therefore, a basis for the vector space {f(x) ∈ P3[x] | ƒ′(6) = ƒ(1)} is {p(x) = ax^2 + bx + 11a, q(x) = dx}, where a and d can be any real numbers.

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Answer:

[tex]60cm^{2}[/tex]

Step-by-step explanation:

The area is the total space taken up by a flat (2-D) surface or shape. The area is always measured in square units.

If we look at this shape, we can split it into 3 separate shapes (shown below)

The top rectangle in blue has a length of 2cm and a width of 10cm. We know the width is 10 because if we were to look at the width of the yellow rectangle and add on the original width you would get:

Now that we know that the length is 2 and the width is 10, we can use the following formula to solve for the area of a rectangle:

(Where l = length and h = height)

Inserting 2 in for our length and 10 for our width:

Therefore, the area of the blue rectangle is [tex]20cm^{2}[/tex].

Looking at the bottom green rectangle, it has the same dimensions as the blue, so it will also have an area of [tex]20cm^{2}[/tex].

The same goes for the yellow rectangle. It has a length of 10 and a width of 2. These are also the same dimensions as before, so we can once again conclude that the area of the yellow rectangle is [tex]20cm^{2}[/tex]

We have 3 rectangles with areas of [tex]20cm^{2}[/tex] each, so we can use either one of these expressions to solve for the entire area:

Or we can use:

Therefore the area of the entire shape is [tex]60cm^{2}[/tex]

The elements of the splitting field are:

{0, 1, α, β, α+β, αβ, α+αβ, β+αβ, α+β+αβ}

To find the splitting field of the polynomial p(x) = x² + x + 1 in ℤ/(2ℤ)[x], we need to find the field extension over which the polynomial completely factors into linear factors.

Since we are working with ℤ/(2ℤ), the field consists of only two elements, 0 and 1. We can substitute these values into p(x) and check if they are roots:

p(0) = 0² + 0 + 1 = 1 ≠ 0, so 0 is not a root.

p(1) = 1² + 1 + 1 = 3 ≡ 1 (mod 2), so 1 is not a root.

Since neither 0 nor 1 are roots of p(x), the polynomial does not factor into linear factors over ℤ/(2ℤ)[x].

To find the splitting field, we need to extend the field to include the roots of p(x). In this case, the roots are complex numbers, namely:

α = (-1 + √3i)/2

β = (-1 - √3i)/2

The splitting field will include these two roots α and β, as well as all their linear combinations with coefficients in ℤ/(2ℤ).

The elements of the splitting field are:

{0, 1, α, β, α+β, αβ, α+αβ, β+αβ, α+β+αβ}

These elements form the splitting field of p(x) = x² + x + 1 in ℤ/(2ℤ)[x].

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The value of J from the given Fourier transform of the function f(t) is 5/6.

Fourier Transform of f(t):

F(ω) = 2∫1+t(sin(ωt))dt + 2∫1-t(sin(ωt))dt

= -2cos(ω) + 2∫cos(ωt)dt

= -2cos(ω) + (2/ω)sin(ω)                

J = ∫π/2-0sin(x/2)(x²-1)dx

J = [-sin(x/2)x²/2 - cos(x/2)]π/2-0

J = [2/3 +cos (π/2) - sin(π/2)]/2

J = 1/3 + 1/2

J = 5/6

Therefore, the value of J from the given Fourier transform of the function f(t) is 5/6.

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If a ≠ 0 and b = 0: The solution set is {0}. If a ≠ 0 and b ≠ 0: The solution set is {b/a}. If a = 0 and b ≠ 0: There are no solutions. If a = 0 and b = 0: The solution set is all real numbers.

The possible solution sets of the linear equation ax = b, where a and b are real numbers, depend on the values of a and b.

If a ≠ 0:

If b = 0, the solution is x = 0. This is a single solution.

If b ≠ 0, the solution is x = b/a. This is a unique solution.

If a = 0 and b ≠ 0:

In this case, the equation becomes 0x = b, which is not possible since any number multiplied by 0 is always 0. Therefore, there are no solutions.

If a = 0 and b = 0:

In this case, the equation becomes 0x = 0, which is true for all real numbers x. Therefore, the solution set is all real numbers.

In summary, the possible solution sets of the linear equation ax = b are as follows:

If a ≠ 0 and b = 0: The solution set is {0}.

If a ≠ 0 and b ≠ 0: The solution set is {b/a}.

If a = 0 and b ≠ 0: There are no solutions.

If a = 0 and b = 0: The solution set is all real numbers.

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Answer:

The solutions are,

x=0 and x= 5

(I don't know if you have to write both of these or only one, sorry)

Step-by-step explanation:

[tex]x^2-3x+6=2x+6\\solving,\\x^2-3x-2x+6-6=0\\x^2-5x+0=0\\x^2-5x=0\\x(x-5)=0\\\\x=0, x-5=0\\x=0,x=5[/tex]

So, the solutions are,

x=0 and x= 5

The equation that relates x and y is:

y = 100x + 500

In this equation, y is the total amount of money in the checking account (in dollars), and x is the number of weeks Deshaun has been adding money. The coefficient of x, 100, represents the rate at which Deshaun is adding money to the account. So, each week, Deshaun adds $100 to the account. The y-intercept, 500, represents the initial amount of money in the account. So, when Deshaun starts adding money to the account, the account already has $500 in it.

To see how this equation works, let's say that Deshaun has been adding money to the account for 5 weeks. In this case, x = 5. Substituting this value into the equation, we get:

y = 100 * 5 + 500 = 1000

This means that after 5 weeks, the total amount of money in the account is $1000.

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The expression `cos⁻¹(-2.35)` is undefined.

What is the inverse cosine function?

The inverse cosine function, denoted as `cos⁻¹(x)` or `arccos(x)`, is the inverse function of the cosine function.

The inverse cosine function, cos⁻¹(x), is only defined for values of x between -1 and 1, inclusive. The range of the cosine function is [-1, 1], so any value outside of this range will not have a corresponding inverse cosine value.

In this case, -2.35 is outside the valid range for the input of the inverse cosine function.

The result of `cos⁻¹(x)` is the angle θ such that `cos(θ) = x` and `0 ≤ θ ≤ π`.

When `x < -1` or `x > 1`, `cos⁻¹(x)` is undefined.

Therefore, the expression cos⁻¹(-2.35) is undefined.

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The y-intercept of the function P(z) is 0.

The z-intercepts are z₁ = -2, z₂ = 7, and z₃ = -5.

To find the y-intercept of the function P(z), we need to evaluate P(0), which gives us the value of the function when z = 0.

For P(z) = z(z - 7)(z + 5), substituting z = 0:

P(0) = 0(0 - 7)(0 + 5) = 0

To find the z-intercepts of the function P(z), we need to find the values of z for which P(z) = 0. These are the values of z that make each factor of P(z) equal to zero.

Given:

z₁ = -2

z₂ = 7

z₃ = -5

The z-intercepts are the values of z that make P(z) equal to zero:

P(z₁) = (-2)(-2 - 7)(-2 + 5) = 0

P(z₂) = (7)(7 - 7)(7 + 5) = 0

P(z₃) = (-5)(-5 - 7)(-5 + 5) = 0

As for the behavior of the function as z approaches positive or negative infinity:

When z goes to positive infinity (z → +∞), the function P(z) also goes to positive infinity (y → +∞).

When z goes to negative infinity (z → -∞), the function P(z) goes to negative infinity (y → -∞).

Please note that the information provided in the question about T2 and c-intercepts for the second function (P(z) = (z-1)²(z-9)) is incomplete or unclear. If you can provide additional information or clarify the question, I will be happy to help further.

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The set S = {-1, 1, 1} forms a basis of ℝ³, and the orthonormal basis T = {1, 0, 0} is obtained through the Gram-Schmidt orthonormalization process.

To show that the set S = {X₁, X₂, X₃} = {-1, 1, 1} forms a basis of ℝ³ and find an orthonormal basis T = {Y₁, Y₂, Y₃} using the Gram-Schmidt orthonormalization process, we'll follow the steps of the process.

Step 1:

Verify linear independence of S:

We need to check if the vectors in S are linearly independent. If they are linearly independent, then S will form a basis of ℝ³.

Set up a linear combination equation:

a₁X₁ + a₂X₂ + a₃X₃ = 0

Substituting the values of X₁, X₂, and X₃:

-a₁ + a₂ + a₃ = 0

We can observe that for a₁ = 1, a₂ = 1, and a₃ = 1, the equation is satisfied. Therefore, the only solution to the linear combination equation is the trivial solution a₁ = a₂ = a₃ = 0. Hence, the vectors in S are linearly independent.

Step 2:

Normalize the vectors:

To find an orthonormal basis using Gram-Schmidt, we need to normalize the vectors in S.

Y₁ = X₁ / ||X₁||

= X₁ / √(X₁ · X₁)

= X₁ / √((-1)²)

= -X₁

Y₂ = X₂ - projₙ(Y₁)

= X₂ - ((X₂ · Y₁) / (Y₁ · Y₁)) Y₁

Calculating the projection:

X₂ · Y₁ = (1) · (-1) = -1

Y₁ · Y₁ = (-1) · (-1) = 1

Y₂ = X₂ - (-1 / 1) (-X₁)

= X₂ + X₁

= 1 + (-1)

= 0

Y₃ = X₃ - projₙ(Y₁) - projₙ(Y₂)

= X₃ - ((X₃ · Y₁) / (Y₁ · Y₁)) Y₁ - ((X₃ · Y₂) / (Y₂ · Y₂)) Y₂

Calculating the projections:

X₃ · Y₁ = (1) · (-1) = -1

X₃ · Y₂ = (1) · (0) = 0

Y₃ = X₃ - (-1 / 1) (-X₁) - (0 / 0) Y₂

= X₃ + X₁

= 1 + (-1)

= 0

Now, we have the orthonormal basis T = {Y₁, Y₂, Y₃} = {-X₁, 0, 0} = {1, 0, 0}.

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Answer:

B

Step-by-step explanation:

the 'dot' between 11 and x represents multiplication.

two numbers being multiplied are referred to as a product.

11 • x ← is the product of 11 and x

The function f(z) = |z|² is differentiable only along the y-axis (where x = 0), but not along any other line. It is not holomorphic anywhere in the complex plane, and its derivative at points along the y-axis is 0.

The function f(z) = |z|² is defined as the modulus squared of z, where z = x + iy and x, y are real numbers.

To determine where this function is differentiable, we can apply the Cauchy-Riemann equations. The Cauchy-Riemann equations state that a function f(z) = u(x, y) + iv(x, y) is differentiable at a point z = x + iy if and only if its partial derivatives satisfy the following conditions:

1. ∂u/∂x = ∂v/∂y
2. ∂u/∂y = -∂v/∂x

Let's find the partial derivatives of f(z) = |z|²:

u(x, y) = |z|² = (x² + y²)
v(x, y) = 0 (since there is no imaginary part)

Taking the partial derivatives:
∂u/∂x = 2x
∂u/∂y = 2y
∂v/∂x = 0
∂v/∂y = 0

The first condition is satisfied: ∂u/∂x = ∂v/∂y = 2x = 0. This implies that the function f(z) = |z|² is differentiable at all points where x = 0. In other words, f(z) is differentiable along the y-axis.

However, the second condition is not satisfied: ∂u/∂y ≠ -∂v/∂x. Therefore, the function f(z) = |z|² is not differentiable at any point where y ≠ 0. In other words, f(z) is not differentiable along the x-axis or any other line that is not parallel to the y-axis.

Next, let's determine where the function f(z) = |z|² is holomorphic. For a function to be holomorphic, it must be complex differentiable in a region, meaning it must be differentiable at every point within that region. Since the function f(z) = |z|² is not differentiable at any point where y ≠ 0, it is not holomorphic anywhere in the complex plane.

Finally, let's find the derivatives of f(z) at points where it is differentiable. Since f(z) = |z|² is differentiable along the y-axis (where x = 0), we can calculate its derivative using the definition of the derivative:

f'(z) = lim(h -> 0) [f(z + h) - f(z)] / h

Substituting z = iy, we have:

f'(iy) = lim(h -> 0) [f(iy + h) - f(iy)] / h
       = lim(h -> 0) [h² + y² - y²] / h
       = lim(h -> 0) h
       = 0

Therefore, the derivative of f(z) = |z|² at points where it is differentiable (along the y-axis) is 0.

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The correct interpretation of the point (2, 4) in this context is:

2. After 2 minutes, the depth of the pool is 4 feet.

In the given model y = -2x + 8, the equation represents the relationship between the time in minutes (x) and the depth of the pool in feet (y) after draining. The equation is in the form of a linear function, where the coefficient of x (-2) represents the rate of change of the depth of the pool over time.

To determine the meaning of the point (2, 4) in this context, we need to substitute the value of x as 2 into the equation and solve for y.

When x = 2:

y = -2(2) + 8

y = -4 + 8

y = 4

Therefore, when 2 minutes have passed, the depth of the pool is 4 feet. This means that after 2 minutes of draining, the water level in the pool has decreased to 4 feet.

It is important to note that in this model, the coefficient -2 indicates that the depth of the pool decreases by 2 feet for every minute that passes. As time increases, the depth of the pool will continue to decrease at a constant rate of 2 feet per minute.

The given point (2, 4) provides a specific example that illustrates the relationship between time and the depth of the pool. It confirms that after 2 minutes of draining, the pool's depth is indeed 4 feet.

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The resulted inequality is y > (9 + x) / 7.

To make y the subject of the inequality x < -9/y - 7, we need to isolate y on one side of the inequality.

Let's start by subtracting x from both sides of the inequality:

x + 9/y < 7

Next, let's multiply both sides of the inequality by y to get rid of the fraction:

y(x + 9/y) < 7y

This simplifies to:

x + 9 < 7y

Finally, let's isolate y by subtracting x from both sides:

x + 9 - x < 7y - x

9 < 7y - x

Now, we can rearrange the inequality to make y the subject:

7y > 9 + x

Divide both sides by 7:

y > (9 + x) / 7

So, the inequality x < -9/y - 7 can be rewritten as y > (9 + x) / 7.


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a) The regular selling price for the refrigerator is approximately $1,471.43.

b) The amount of profit based on the selling price is approximately $441.43.

a) To calculate the regular selling price, we need to consider the expenses and the desired profit.

Let's denote the regular selling price as "P."

Expenses average 8% of the regular selling price, which means expenses amount to 0.08P.

The desired profit based on selling price is 22% of the regular selling price, which means profit amounts to 0.22P.

The total cost of the refrigerator, including expenses and profit, is the purchase price plus expenses plus profit: $1,030 + 0.08P + 0.22P.

To find the regular selling price, we set the total cost equal to the regular selling price:

$1,030 + 0.08P + 0.22P = P.

Combining like terms, we have:

$1,030 + 0.30P = P.

0.30P - P = -$1,030.

-0.70P = -$1,030.

Dividing both sides by -0.70:

P = -$1,030 / -0.70.

P ≈ $1,471.43.

Therefore, the regular selling price is approximately $1,471.43.

b) To calculate the amount of profit, we can subtract the cost from the regular selling price:

Profit = Regular selling price - Cost.

Profit = $1,471.43 - $1,030.

Profit ≈ $441.43.

Therefore, the amount of profit is approximately $441.43.

Please note that the values are rounded to the nearest cent.

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To find the daily mileage for which the Ultimo charge is twice the Primo charge, we can set up an equation and solve for the unknown value.

Let's start by defining some variables:
- Let x be the daily mileage.
- The Primo car rental agency charges $45 per day plus $0.40 per mile, so the Primo charge can be expressed as 45 + 0.40x.
- The Ultimo car rental agency charges $26 per day plus $0.85 per mile, so the Ultimo charge can be expressed as 26 + 0.85x.
According to the question, we need to find the value of x for which the Ultimo charge is twice the Primo charge. Mathematically, we can write this as:
26 + 0.85x = 2(45 + 0.40x)
Now, let's solve this equation step-by-step:
1. Distribute the 2 to the terms inside the parentheses on the right side of the equation:
26 + 0.85x = 90 + 0.80x
2. Move all the x terms to one side of the equation and all the constant terms to the other side:
0.85x - 0.80x = 90 - 26
3. Simplify and solve for x:
0.05x = 64
x = 64 / 0.05
x = 1280
Therefore, the daily mileage for which the Ultimo charge is twice the Primo charge is 1280 miles.

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Answer:

Width=10 hileight 5cm length 18

(a) The number of ways to choose nine po-boys from twelve options is 220.

(b) The number of ways to choose 20 po-boys with at least one of each kind is 36,300.

The number of ways to choose po-boys can be found using combinations.

a) To determine the number of ways to choose nine po-boys, we can use the concept of combinations. In this case, we have twelve different types of po-boys to choose from. We want to choose nine po-boys, without any restrictions on repetition or order.

The formula to calculate combinations is given by C(n, r) = n! / (r!(n-r)!), where n is the total number of items and r is the number of items to be chosen.

Using this formula, we can calculate the number of ways to choose nine po-boys from twelve options:

C(12, 9) = 12! / (9!(12-9)!) = 12! / (9!3!) = (12 × 11 × 10) / (3 × 2 × 1) = 220.

Therefore, there are 220 ways to choose nine po-boys from the twelve available options.

b) To determine the number of ways to choose 20 po-boys with at least one of each kind, we can approach this problem using combinations as well.

We have twelve different types of po-boys to choose from, and we want to choose a total of twenty po-boys. To ensure that we have at least one of each kind, we can choose one of each kind first, and then choose the remaining po-boys from the remaining options.

Let's calculate the number of ways to choose the remaining 20-12 = 8 po-boys from the remaining options:

C(11, 8) = 11! / (8!(11-8)!) = 11! / (8!3!) = (11 × 10 × 9) / (3 × 2 × 1) = 165.

Therefore, there are 165 ways to choose the remaining eight po-boys from the eleven available options.

Since we chose one of each kind first, we need to multiply the number of ways to choose the remaining po-boys by the number of ways to choose one of each kind.

So the total number of ways to choose 20 po-boys with at least one of each kind is 220 × 165 = 36300.

Therefore, there are 36,300 ways to choose 20 po-boys with at least one of each kind.

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To join the points (2, 16) and (8, 4), we can use the slope-intercept form of a linear equation, which is y = mx + b, where m is the slope and b is the y-intercept.

First, let's calculate the slope (m) using the formula:

m = (y2 - y1) / (x2 - x1)

Substituting the coordinates of the two points:

m = (4 - 16) / (8 - 2)

m = -12 / 6

m = -2

Now that we have the slope, we can choose either of the two points and substitute its coordinates into the slope-intercept form to find the y-intercept (b).

Let's choose the point (2, 16):

16 = -2(2) + b

16 = -4 + b

b = 20

Now we have the slope (m = -2) and the y-intercept (b = 20), we can write the equation of the line:

y = -2x + 20

This equation represents the line passing through the points (2, 16) and (8, 4).

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