five million, seven hundred seven thousand, seven hundred nineteen in numbers​

Answer:

24x^5

Step-by-step explanation:

3 x 8 = 24

x multiplied by x^4 = x^5

(the exponents add to each other when multiplied)

In order to calculate the time it will take for Andrew to settle the loan, we can use the formula for compound interest. So, it will take Andrew approximately 5.22 years to settle the loan.

The formula is given as A = P(1 + r/n)^(nt), Where: A = the final amount, P = the principal (initial amount borrowed), R = the annual interest rate, N = the number of times the interest is compounded in a year, T = the time in years.

We know that Andrew borrowed R200 000 from a bank at an annual interest rate of 5.25% compounded quarterly and that he will make repayments of R6 000 at the end of every 3 months.

Since the first repayment will be made 3 months from now, we can consider that the initial loan repayment is made at time t = 0. This means that we need to calculate the value of t when the total amount repaid is equal to the initial amount borrowed.

Using the formula for compound interest: A = P(1 + r/n)^(nt), We can calculate the quarterly interest rate:r = (5.25/100)/4 = 0.013125We also know that the quarterly repayment amount is R6 000, so the amount borrowed minus the first repayment is the present value of the loan: P = R200 000 - R6 000 = R194 000

We can now substitute these values into the formula and solve for t: R194 000(1 + 0.013125/4)^(4t) = R200 000(1 + 0.013125/4)^(4t-1) + R6 000(1 + 0.013125/4)^(4t-2) + R6 000(1 + 0.013125/4)^(4t-3) + R6 000(1 + 0.013125/4)^(4t)

Rearranging the terms gives us: R194 000(1 + 0.013125/4)^(4t) - R6 000(1 + 0.013125/4)^(4t-1) - R6 000(1 + 0.013125/4)^(4t-2) - R6 000(1 + 0.013125/4)^(4t-3) - R200 000(1 + 0.013125/4)^(4t) = 0

Using trial and error, we can solve this equation to find that t = 5.22 years (rounded to 2 decimal places). Therefore, it will take Andrew approximately 5.22 years to settle the loan.

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Triangle AED and triangle BEC have equal areas. The length of AE can be represented as 14 - x (since AC = 14). AE is equal to 10.

To find the length of AE, we can use the fact that triangle AED and triangle BEC have equal areas. Since the areas of two triangles with the same height are proportional to their base lengths, we can set up the following proportion:

Area of triangle AED / Area of triangle BEC = AE / EC

Let's denote EC as x (the length of EC). Then, the length of AE can be represented as 14 - x (since AC = 14).

Now, let's look at the given information about the lengths of the sides of the quadrilateral:

AB = 9

CD = 12

AC = 14

We can use the given information to find the lengths of the other sides of the quadrilateral:

BC = AC - AB = 14 - 9 = 5

AD = AC - CD = 14 - 12 = 2

Next, we can calculate the areas of triangle AED and triangle BEC using the lengths of their sides:

Area of triangle AED = (1/2) * AD * AE

Area of triangle BEC = (1/2) * BC * EC

Since the areas of the two triangles are equal, we can set up the following equation:

(1/2) * AD * AE = (1/2) * BC * EC

Plugging in the values we know:

(1/2) * 2 * (14 - x) = (1/2) * 5 * x

Simplifying the equation:

14 - x = (5/2) * x

Multiplying both sides by 2 to clear the fraction:

28 - 2x = 5x

Adding 2x to both sides:

28 = 7x

Dividing both sides by 7:

x = 4

Therefore, the length of EC is 4.

Since AE = AC - EC, we can calculate AE:

AE = 14 - EC = 14 - 4 = 10

So, AE is equal to 10.

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The complete question is:<Convex quadrilateral ABCD has AB = 9 and CD = 12. Diagonals AC and BD intersect at E, AC = 14 and triangle AED and triangle BEC have equal areas. What is AE?>

If this trend continues, the total books the author would have sold over the first 12 months is 157,810 books.

In this scenario, we would calculate the total books sold by this author over the first 12 months as follows;

First month = 19,000 books.

Second month; 19,000 × (1 - 7)% = 19,000 × 93/100 = 17,670 books.

Third month; 17,670 × (1 - 7)% = 17,670 × 93/100 = 16,433 books.

Fourth month; 16,433 × (1 - 7)% = 16,433 × 93/100 = 15,283 books.

Fifth month; 15,283 × (1 - 7)% = 15,283 × 93/100 = 14,213 books.

Sixth month; 14,213 × (1 - 7)% = 14,213 × 93/100 = 13,218 books.

Seventh month; 13,218 × (1 - 7)% = 13,218 × 93/100 = 12,293 books.

Eigth month; 12,293 × (1 - 7)% = 12,293 × 93/100 = 11,432 books.

Ninth month; 11,432 × (1 - 7)% = 11,432 × 93/100 = 10,632 books.

Tenth month; 10,632 × (1 - 7)% = 13,218 × 93/100 = 9,888 books.

Eleventh month; 11,432 × (1 - 7)% = 11,432 × 93/100 = 9,196 books.

Twelveth month; 9,196 × (1 - 7)% = 9,196 × 93/100 = 8,552 books.

Next, we would add all of the books sold in each month together;

Total books sold = 19,000 + 17,670 + 16,433 + 15,283 + 14,213 + 13,218 + 12,293 + 11,432 + 10,632 + 9,888 + 9,196 + 8,552

Total books sold = 157,810 books.

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Answer:

y=2/3

Step-by-step explanation:

Answer:

(-2,3) x= -2 y=3

Step-by-step explanation:

The first thing you should do is convert both into standard y=mx+b form

For the first equation you subtract 5y and add 11 to both sides to get

13x+11= -5y

you divide by negative 5 to get

-13x/5-11/5=y

For the second equation we subtract 11y and 7 from both sides to get

13x-7= -11y

divide both sides by -11 to get

-13x/11 + 7/11

We then set both equations equal to each other and solve for x

-13x/5-11/5= -13x/11 + 7/11

Add 11/5 to both sides to get

-13x/5 = -13x/11 + 156/55

Add 13x/11 to both sides to get

-78/55x=156/55

Divide and solve for x to get -2

Then plug x = -2 into the original equations to find y

13*-2 + 5y = -11

-26 + 5y = -11

Add 26 to both sides

5y= 15

y =3

The point of diminishing returns is (20.98, 21247.3).

The point of diminishing returns occurs when the marginal cost of producing an extra unit of output exceeds the marginal revenue generated from selling that unit. Mathematically, it is the point at which the derivative of the production function equals zero and the second derivative is negative.

Given the polynomial function N(x) of degree 3, we can find the point of diminishing returns by finding the critical points where the first derivative equals zero and evaluating the second derivative at those points.

The derivative of N(x) is N'(x) = -18x² + 360x + 2250. To find the critical points, we set N'(x) = 0:

0 = -18x² + 360x + 2250

Dividing by -18 simplifies the equation:

0 = x² - 20x - 125

Using the quadratic formula, we find the solutions to the equation:

x₁,₂ = (20 ± √(20² - 4(1)(-125))) / 2(1)

x₁,₂ = 10 ± 5√5

Thus, the two critical points of N(x) are at x = 10 - 5√5 and x = 10 + 5√5.

To determine the point of diminishing returns, we evaluate the second derivative N''(x) = -36x + 360 at these critical points:

N''(10 - 5√5) = -36(10 - 5√5) + 360 ≈ -264.8

N''(10 + 5√5) = -36(10 + 5√5) + 360 ≈ 144.8

From the evaluations, we find that N''(10 + 5√5) is negative while N''(10 - 5√5) is positive. Therefore, the point of diminishing returns corresponds to x = 10 + 5√5.

To find the corresponding y-coordinate (N(10 + 5√5)), we can substitute the value of x into the original function N(x).

Hence, the point of diminishing returns is approximately (20.98, 21247.3).

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Answer:

5 Hours

Step-by-step explanation:

8 x 5 = 40.

Heather wants to spend $35 dollars at most. She has a discount for $5 dollars off, so basically we do 40 -5, which equals $35.

Answer:

1.8 pounds

Step-by-step explanation:

All you have to do is multiply 0.1 * 4, and then 0.1 * 14. Once you get the products, you can just add them together to get 1.8 pounds

Answer:

(2x+3y)^2

= (2x)^2 + 2(2x)(3y) + (3y)^2

= 4x^2 + 12xy + 9y^2

Answer:

4x^2 12xy +9y^2

Step-by-step explanation:

(2x+3y)^2

(2x+3y)(2x+3y)

FOIL

4x^2 + 6xy+6xy + 9y^2

4x^2 12xy +9y^2

The equation for the hyperbola with foci (0,±5) and asymptotes y=± 3/4 x is:

y^2 / 25 - x^2 / a^2 = 1

where a is the distance from the center to a vertex and is related to the slope of the asymptotes by a = 5 / (3/4) = 20/3.

Thus, the equation for the hyperbola is:

y^2 / 25 - x^2 / (400/9) = 1

or

9y^2 - 400x^2 = 900

The center of the hyperbola is at the origin, since the foci have y-coordinates of ±5 and the asymptotes have y-intercepts of 0.

To graph the hyperbola, we can plot the foci at (0,±5) and draw the asymptotes y=± 3/4 x. Then, we can sketch the branches of the hyperbola by drawing a rectangle with sides of length 2a and centered at the origin. The vertices of the hyperbola will lie on the corners of this rectangle. Finally, we can sketch the hyperbola by drawing the two branches that pass through the vertices and are tangent to the asymptotes.

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Step-by-step explanation:

question 1

\(6 {x}^{4} - 36 {x}^{3} \\ 6 {x}^{3} (x - 6)\)

question 2

\(64 {x}^{2} + 48x + 9 \\ {(8x + 3)}^{2} \)

It is proved that, (A,<0​) and (A,<1​) have the same order type.

The relation ∼ satisfies reflexivity, symmetry, and transitivity, it is an equivalence relation.

Here, we have,

To prove that the relation ∼ on the set A of orderings is an equivalence relation, we need to show that it satisfies three properties: reflexivity, symmetry, and transitivity.

Reflexivity: For any ordering (A, <) in A, it must be the case that (A, <) ∼ (A, <). This means that any ordering (A, <) has the same order type as itself. Therefore, the relation ∼ is reflexive.

Symmetry: If (A, <0) ∼ (A, <1), it implies that the order types of (A, <0) and (A, <1) are the same. To show symmetry, we need to prove that if (A, <0) ∼ (A, <1), then (A, <1) ∼ (A, <0).

Assume (A, <0) ∼ (A, <1), which means that (A, <0) and (A, <1) have the same order type.

Let's denote the order type of (A, <0) and (A, <1) as OT(<0) and OT(<1), respectively.

Since the order types are the same, we can find order-preserving bijections from A to itself:

f: A → A (which preserves the order <0) and

g: A → A (which preserves the order <1).

We can define a new ordering <' on A such that for any elements x, y in A, x <' y if and only if f(x) <1 g(y).

It's important to note that f and g are invertible and order-preserving, which guarantees that <' is a valid order relation on A.

Now, we need to show that (A, <0) and (A, <') have the same order type. Let's define the function h: A → A such that h(x) = g^(-1)(f(x)) for all x in A. This function is well-defined and invertible since f and g are invertible. We can also show that h preserves the order <0 since:

For any x, y in A, if x <0 y, then f(x) <1 g(y) (by the order-preserving property of f and g).

If f(x) <1 g(y), then h(x) = g^(-1)(f(x)) <' g^(-1)(g(y)) = y (by the invertibility of g).

Thus, h preserves the order <0, and since h is invertible, it's an order-preserving bijection from (A, <0) to (A, <').

Therefore, (A, <0) and (A, <') have the same order type.

We have shown that if (A, <0) ∼ (A, <1), then (A, <1) ∼ (A, <0).

Hence, the relation ∼ is symmetric.

Transitivity: To prove transitivity, we need to show that if (A, <0) ∼ (A, <1) and (A, <1) ∼ (A, <2), then (A, <0) ∼ (A, <2). Assume (A, <0) ∼ (A, <1) and (A, <1) ∼ (A, <2), meaning that (A, <0), (A, <1), and (A, <2) have the same order type.

Let OT(<0), OT(<1), and OT(<2) denote the order types of (A, <0), (A, <1), and (A, <2), respectively. We can find order-preserving bijections:

f: A → A (which preserves the order <0) and

g: A → A (which preserves the order <1).

Similarly, we can find another order-preserving bijection:

h: A → A (which preserves the order <2).

Now, let's define a new ordering <' on A such that for any elements x, y in A, x <' y if and only if h(f(x)) <2 h(g(y)).

As before, we can show that <' is a valid order relation on A.

We need to demonstrate that (A, <0) and (A, <') have the same order type.

Consider the function k: A → A defined as k(x) = h(g(f(x))) for all x in A.

This function is well-defined and invertible since f, g, and h are invertible. We can also show that k preserves the order <0 since:

For any x, y in A, if x <0 y, then f(x) <1 g(y) (by the order-preserving property of f and g).

If f(x) <1 g(y), then h(f(x)) <2 h(g(y)) (by the order-preserving property of h).

If h(f(x)) <2 h(g(y)), then k(x) = h(g(f(x))) <' h(g(h(g(y)))) = y (by the invertibility of g and h).

Thus, k preserves the order <0, and since k is invertible, it's an order-preserving bijection from (A, <0) to (A, <').

Therefore, (A, <0) and (A, <') have the same order type.

We have shown that if (A, <0) ∼ (A, <1) and (A, <1) ∼ (A, <2), then (A, <0) ∼ (A, <2).

Hence, the relation ∼ is transitive.

Since the relation ∼ satisfies reflexivity, symmetry, and transitivity, it is an equivalence relation.

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Answer:

32 $

Step-by-step explanation:

Answer:

(5/6)*(2 dozen) = 20

Step-by-step explanation:

Answer:

Step-by-step explanation:

2 dozen = 24

5/6(24)= 5*4= 20

Answer:

lateral area and area without one side: 20.8

Answer:

m= 6/7

decimal form m= 0.857

Step-by-step explanation:

hope this helps

Answer:

m=6/7

m=0.857...

Step-by-step explanation:

-4m - 6 + 6-m = -5 + 2m - 1​

-5m-6+6=-5+2m-1

-5m=-5+2m-1

-5m=2m-5-1

-5m=2m-6

-5m-2m=2m-6-2m

-7m=-6

divide both sides by -7

x=6/7

A) The price per share three years later is $43.29, B) The dividend three years later is $1.9425 and C) The yield three years later is approximately 4.49%.

a. To find the price per share three years later, we need to calculate the increase in value and add it to the original price.
First, let's calculate the increase in value. The stock has increased 17% in value, so we can calculate it as: 17% of $37 = $6.29 (rounded to two decimal places).
Next, we add the increase in value to the original price: $37 + $6.29 = $43.29 (rounded to two decimal places).
Therefore, the price per share three years later is $43.29.
b. To find the dividend three years later, we need to calculate the increase in the dividend and add it to the original dividend.
The dividend has increased 5%, so we can calculate it as: 5% of $1.85 = $0.0925 (rounded to four decimal places).
Next, we add the increase in dividend to the original dividend: $1.85 + $0.0925 = $1.9425 (rounded to four decimal places).
Therefore, the dividend three years later is $1.9425.
c. To find the yield three years later, we divide the dividend by the price per share three years later and multiply by 100 to express it as a percentage.
Yield = (Dividend / Price per share) x 100
Yield = ($1.9425 / $43.29) x 100 ≈ 4.49% (rounded to two decimal places).
Therefore, the yield three years later is approximately 4.49%.

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The given differential equation is:

\(e^x y \frac{dy}{dx}\\e^{-Y} + e^{-3x} - y\)

We are supposed to solve the given differential equation by separating the variables.

To do that, we need to rearrange the equation such that all the y terms are on one side and all the x terms are on the other side.

\(e^x y \frac{dy}{dx} + y\\e^{-Y} + e^{-3x} - y\)

We can then group all the y terms on one side of the equation and all the x terms on the other side. This gives:

\(\frac{dy}{dx} + \frac{1}{e^x y}\frac{e^{-Y} + e^{-3x}}{e^x y + y}\)

Now we can multiply both sides of the equation by e^x y:

\(e^x y \frac{dy}{dx} + y e^x y = 0\\e^{-Y} e^x y + e^{-3x} e^x y + y e^x y = 0\\e^{2x} y \frac{dy}{dx} + y e^x = 0\\e^{x-Y} + 1 = 0\end{equation}\)

We can now integrate both sides of the equation:

=∫ e^(2x) y dy + ∫ y e^x dx

\(\int e^{2x} y dy + \int y e^x dx .\\= \int (e^{x-Y} + 1) dx\)

Integrating the left side of the equation by parts:

u = y,

\(dv = e^{2x} \, dx\)

\(du = dy\),

\(v = \frac{1}{2} e^{2x} \int y e^{2x} \, dx\)

\(v = \frac{1}{2} y e^{2x} - \frac{1}{2} \int e^{2x} \, dy\)

Substituting these results back into the equation gives:

\((1/2) e^{2x} y^2 - (1/2) y e^{2x} + (1/2) e^x = 0\\y = e^{x-Y} + x + C\)

where C is the constant of integration.

Finally, we can solve for y by rearranging the equation:

\(\frac{1}{2} e^{2x} y^2 - \frac{1}{2} y e^{2x} - e^{x-Y} + \frac{1}{2} e^x = 0\)

= x + C

This gives the solution to the differential equation in terms of y.

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Answer:

x₁ = 3,29296875

x₂ = 3,276659786

x₃ = 3,279420685

Step-by-step explanation:

This is a recurrent relationship between xₙ₊₁ and xₙ

To get x₁, write the formula with n=0:

\(x_1 = 3+ \frac{3}{x_0^2}\)

Then fill in what you know, x₀=3.2. This gives you x₁.

Repeat for n=1,... and so forth.

Excel can do this effectively, see picture.

The statement "biblical account of creation is a metaphor' is a contentious one, as it challenges a traditional and fundamental belief within Christianity.

One argument on how biblical account of creation is a metaphor is because it is written in a symbolic and metaphorical language. Many scholars and theologians argue that the creation story in the Bible should not be taken literally, but rather as a symbolic representation of God's power and authority.

Another argument is that the scientific evidence for the origins of the universe and the evolution of life on Earth is incompatible with the literal interpretation of the biblical account of creation. Therefore, the statement that the "biblical account of creation is a metaphor" can be seen as a way of reconciling scientific understanding with religious belief.

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The dimension of the row space is m - n, where m is the number of columns in matrix A.

If the null space of a matrix A is n-dimensional, then the dimension of the row space is equal to the number of columns minus the dimension of the null space. In other words, if there are m columns in matrix A, then the dimension of the row space is m - n.

To understand this relationship, it is helpful to consider the fundamental theorem of linear algebra, which states that the dimension of the row space plus the dimension of the null space equals the number of columns.

This means that if we know the dimension of one of these subspaces, we can easily calculate the dimension of the other using the equation:

Dimension of row space + Dimension of null space = Number of Columns

In this case, we know that the dimension of the null space is n, so we can substitute this value into the equation:

Dimension of row space + n = m

Solving for the dimension of the row space, we get:

Dimension of row space = m - n

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Answer:

2.4%

Step-by-step explanation:

Answer:

2.4 percent

Step-by-step explanation:

The solution is given below.

An equation is a  mathematical statement that is made up of two expressions connected by an equal sign.  In its simplest form in algebra, the definition of an equation is a mathematical statement that shows that two mathematical expressions are equal. For instance, 3x + 5 = 14 is an equation, in which 3x + 5 and 14 are two expressions separated by an 'equal' sign.

here, we have,

The perimeter of a rectangle is 40 cm. The length is 14 cm.

Let x = width of the rectangle.

Ravi says he can find the width using the equation 2(x + 14) = 40.

Fran says she can find the width using the equation 2x + 28 = 40.

now, we get,

1. Divide both sides by 2

 2(x+14) = 40

 x+14 = 20

2. Isolate the x term by subtracting 14 from both sides

3. x = 6. The width of the triangle is 6 cm.

4. Isolate the x term by subtracting 28 from both sides

 2x + 28 = 40

 2x = 12

5. Divide both sides by 2

6. x = 6

7. The two equations have the same solution, because by the distributive rule, 2(x+14) = 2x+28.

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Answer:

a

c

c

a

Step-by-step explanation:

Answer:

Below

Step-by-step explanation:

As I posted ....looks to be 147°

Answer:

Step-by-step explanation:

800-160=640

640/800=64/80

64/80=16/20

16/20=4/5

Answer=4/5

The fraction of his original money does he have left = 4/5 .

What is a fraction in math?

A boy has = 800 and spends = 160

800-160=640

640/800=64/80

64/80=16/20

16/20=4/5

Therefore, his original money left =4/5

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