Which expression is equivalent to

x2 − 2x − 5

x − 3

?

A) x − 5 −

20

x − 3

B) x − 5 −

10

x − 3

C) x + 1 −

8

x − 3

D) x + 1 −

2

x − 3

Answer:

8.40

Step-by-step explanation:

Since, the simple interest is,

I= p x r x t

     100

Where,

P=Principal

r= rate per periods

t= number of periods

Here, P=140, r=3%, t = 2,

the simple interest will be

I= 140 x 3 x 2

         100

= 840

  100

=8.40

Note that the angle that is vertical to ∠AEB is ∠BEC.

Vertical angles are the angles formed when two lines intersect. The term "vertical" in this context refers to the fact that they share the same Vertex (corner point), rather than the normal connotation of up-down.

Vertical angles of equal measure are always congruent. Both vertical angle pairings (four angles total) always add up to 360 degrees. Angles formed by each pair of vertical angles are referred to as neighboring angles, and they are supplementary (the angles sum up to 180 degrees).

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The line integral [tex]\int_{c}xds[/tex]where c is the straight line segment from point (1, 2) to point (5, 10) is 12√5.

Using the two-point form of the line, we can compute the equation of the line connecting the points and obtain:

y = 2x

Consider x = t. Now the result is:

y = 2t

We can write C as:

[tex]\vec{r}(t)=t\hat{i}+2t\hat{j}[/tex]

Now, t ranges from t = 1 to t = 5 since x = t and x range from x = 1 to x = 5. When we differentiate the C equation, we obtain:

[tex]\vec{r}'(t)=\hat{i}+2\hat{j}[/tex]

Now finding the magnitude

[tex]|\vec{r}'(t)|=\sqrt{(1)^2+(2)^2}[/tex]

[tex]|\vec{r}'(t)|=\sqrt{1+4}[/tex]

[tex]|\vec{r}'(t)|=\sqrt{5}[/tex]

Now, the integral will be:

I = [tex]\int_{t=1}^5t\:(\sqrt 5)dt[/tex]

I = [tex]\sqrt {5}\int_{t=1}^5tdt[/tex]

Now integrating

I = [tex]\sqrt {5}\left(\frac{t^2}{2}\right)_{t=1}^{5}[/tex]

I = [tex]\sqrt {5}\left(\frac{5^2}{2}-\frac{1^2}{2}\right)[/tex]

I = √5(25/2 - 1/2)

I = √5(12.5 - 0.5)

I = 12√5

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The complete question is:

Evaluate the line integral [tex]\int_{c}xds[/tex]where c is the straight line segment from point (1, 2) to point (5, 10).

The total number of twenty questions prepared by Josh can be find by using the expression k = 20 * P(getting a question right) / (1 + 20 * 0.08 / 0.92), where, k = total question prepared by Josh

Let's call the number of questions Josh was prepared for as "k".

The probability of Josh getting a question right given that he knew the answer is 0.92,

So the probability of him getting a question right by guessing is

1 - 0.92 = 0.08.

The overall probability of Josh getting a question right is the sum of the probability of him getting it right given that he was prepared and the probability of him getting it right by guessing:

P(getting a question right) = P(getting a question right | prepared) * P(prepared) + P(getting a question right | guessed) * P(guessed)

Since each question is either prepared or guessed, the sum of P(prepared) and P(guessed) is 1:

P(getting a question right) = 0.92 * P(prepared) + 0.08 * (1 - P(prepared))

Simplifying the expression:

P(getting a question right) = 0.92 * P(prepared) + 0.08

Since the overall probability of Josh getting a question right is the same for every question,

it must be the same for all twenty questions:

So, we can write it as:

0.92 * P(prepared) + 0.08

= 0.92 * k/20 + 0.08 * (20-k)/20

Dividing both sides by 0.92:

P(prepared) = k/20 = (0.92 * k + 0.08 * 20 - 0.08) / 0.92

Isolating k:

k = 20 * P(prepared) / (1 + 20 * 0.08 / 0.92)

Plugging in the overall probability of Josh getting a question right, we have:

k = 20 * P(getting a question right) / (1 + 20 * 0.08 / 0.92)

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The actual weight of a pound is 0.45359237 kilos.

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Let P be the principal Vanilla Ice borrowed. The interest charge is calculated as P * 10.3% * 240/365.

So, 330 = P * 10.3% * 240/365

Solving for P, we get:

P = 330 / (10.3% * 240/365) = $3,116.71

Therefore, Vanilla Ice borrowed a principal of $3,116.71.

Simple Interest (SI) is calculated based on the principal amount only, and the interest rate stays constant throughout the tenure. The formula for Simple Interest is:

SI = P * R * T

Where:

P = Principal amount

R = Interest rate

T = Time (in years)

Compound Interest (CI) is calculated based on the principal amount and the interest earned on it in previous periods. The interest earned in each period is added to the principal, and the new amount becomes the principal for the next period. This leads to an exponential growth of the invested amount. The formula for Compound Interest is:

CI = P * (1 + R/n)^(nt)

Where:

P = Principal amount

R = Interest rate

n = Number of compounding periods per year

t = Time (in years)

In summary, the main difference between simple and compound interest is that simple interest is calculated only on the principal amount, whereas compound interest is calculated on the principal amount and the accumulated interest.

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The slope of line D and line E will be congruent. Then the lines are parallel to each other.

A connection between a number of variables results in a linear model when a graph is displayed. The variable will have a degree of one.

The linear equation is given as,

y = mx + c

Where m is the slope of the line and c is the y-intercept of the line.

The equations are given below.

Line D: y = (3/8)x +3

Line E: y = (3/8)x + 8/3

The slope of line D and line E will be congruent. Then the lines are parallel to each other.

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Among the given, the function that has a percent rate of decrease equal to 5% is: F(x) = 3(0.95)

Hence, option (D) is the correct choice.

For (A):

We have,

F(x) = 3(0.5), Here percentage rate decrease = 50%

For (B):

We have,

F(x) = 3(1.5), Here there is percentage rate increase which will be = 50%

For (C):

We have,

F(x) = 3(0.05), Here percentage rate decrease = 95%

For (D)

F(x) = 3(0.95)

Here, percent rate of decrease equal to 5%

This function represents the relationship between input x and the output y where y = 3(0.95^x).

The number 0.95 can be interpreted as 95% of the original value, which means a 5% decrease.

For (A):

We have,

F(x) = 3(1.05), Here there is percentage rate increase which is = 5%

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The antiderivative of f(x) = tan(eˣ + π/2) on 0≤x≤ln(π/2) is:

⇒  ln |sec([tex]e^{\pi /2}[/tex])| / [tex]e^{\pi /2}[/tex]

Now, For the antiderivative of f(x) = tan(eˣ + π/2) on the interval 0≤x≤ln(π/2), we need to integrate the given function.

Now, using the substitution u = eˣ + π/2.

Then, we have:

du/dx = eˣ

dx = du / eˣ

Using this substitution, we can rewrite the original function as:

f(x) = tan(eˣ + π/2) dx = tan(u) du / eˣ

Now let's integrate this function:

∫tan(u) du / eˣ = - ln |cosec(u)| / eˣ + C

Substituting back u = eˣ+π/2:

ln |cosec(eˣ + π/2)| / eˣ + C

= - ln |cosec(eˣ + π/2)| / eˣ + C

Now we need to evaluate this expression at the limits of integration x = 0 and x = ln(π/2):

F(ln(π/2)) - F(0)

= - ln |cosec([tex]e^{\pi /2 } + \frac{\pi }{2}[/tex])| / [tex]e^{\pi /2}[/tex] + C - (- ln |cos(0)| / [tex]e^{\pi /2}[/tex] + C)

= - ln |sec([tex]e^{\pi /2}[/tex])| / [tex]e^{\pi /2}[/tex] + ln |cos(0)| / [tex]e^{\pi /2}[/tex]

= - ln |sec([tex]e^{\pi /2}[/tex])| / [tex]e^{\pi /2}[/tex] + ln(1) / [tex]e^{\pi /2}[/tex]

=  ln |sec([tex]e^{\pi /2}[/tex])| / [tex]e^{\pi /2}[/tex]

So, the antiderivative of f(x) = tan(eˣ + π/2) on 0≤x≤ln(π/2) is:

⇒  ln |sec([tex]e^{\pi /2}[/tex])| / [tex]e^{\pi /2}[/tex]

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The isolation of variable in the equation will result in the formula - Breadth = (Perimeter - 2length)/2.

As per the known information, the formula of the perimeter of rectangle is -

Perimeter = 2 (length + breadth)

Expanding the formula we get -

Perimeter = 2length + 2breadth

Rewriting in terms of breadth -

2Breadth = Perimeter - 2length

Breadth = (Perimeter - 2length)/2

The process to isolate the variable will require expanding and rewriting the equation with breadth on Left Hand Side of the equation. Now, move the constant on Right Hand Side and you will get the formula.

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Answer:

A = $2,800 B = $12,600

Step-by-step explanation:

a. How much will the insurance company pay?

The liability insurance will cover the damages to the other car ($1,400) and the personal injury claims ($12,800 + $9,200) up to the limit of $10,000. So the insurance company will pay $10,000.

The collision insurance covers the damage to Rusty's own car, minus the $300 deductible, so the insurance company will pay $3,100 - $300 = $2,800.

Therefore, the insurance company will pay a total of $10,000 + $2,800 = $12,800.

b. What is Rusty’s share of the responsibility?

Rusty is responsible for paying the remaining amount not covered by insurance, which is $1,400 + $3,100 + $12,800 + $9,200 - $12,800 = $12,600.

Therefore, Rusty's share of the responsibility is $12,600.

Answer:66

Step-by-step explanation:

336666

Answer: 10.30

Step-by-step explanation:

Distance formula = [tex]\sqrt{(x_{2} +x_{1})^2+(y_{2} +y_{1})^2}[/tex]

A ([tex]x_{1} ,y_{1}[/tex]) = A(-3,-6) so [tex]x_{1} = -2[/tex] and [tex]y_{1} =-6[/tex]

B ([tex]x_{2} ,y_{2}[/tex]) = B(-6, 1) so [tex]x_{2} = -6[/tex] and [tex]y_{2} = 1[/tex]

Plug in the values. You should get [tex]\sqrt{(-6 +-3)^2+(1 +-6)^2}=\sqrt{(-9)^2+(-5)^2}=\sqrt{81+25}=\sqrt{106}[/tex]

≈10.30

When u, v are nonzero vectors, then Span{u, v} contains only the line through u and the origin, and the line through v and the origin, this statement is false.

The span of two nonzero vectors u and v, Span{u, v}, is the set of all linear combinations of u and v. This set can include not only the lines through u and the origin, and the line through v and the origin, but also all other points on the plane defined by these two lines.

For example, if u = (1, 0) and v = (0, 1), then Span{u, v} would be the set of all points in the x-y plane, since every point in the plane can be represented as a linear combination of u and v.

Therefore, Span{u, v} contains not only the lines through u and the origin, and the line through v and the origin, but all other points on the plane defined by these two lines.

Correct Question :

When u, v are nonzero vectors, then Span{u, v} contains only the line through u and the origin, and the line through v and the origin. True or fasle.

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Answer: -21

Step-by-step explanation: first you multiply 4 x -5 or 4(-5) - 1 4 x -5 would give you -20 and -20 - 1, since both are negative we add to get a total of -21 hope this helps.

Answer: 1.5

Step-by-step explanation:

A statement which best explains if the graph correctly represents the proportional relationship y = −2x include the following: D. Yes, the points shown on the line would be part of y = −2x.

In Mathematics, a proportional relationship can be defined as a type of relationship that generates equivalent ratios, a straight line, and it can be modeled or represented by the following mathematical expression:

y = kx

Where:

Generally speaking, the graph of any proportional relationship such as the linear equation (y = -2x) is characterized by a straight line as shown in the image attached below, with the following ordered pairs (0, 0), (1, -2),(2, -4), and (-1, 2).

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By using permutations and combinations, We can arrange 12 people into 3 teams of 4 in 5775 different ways.

Here, we will use the concept of permutations and combinations to solve the question.

We have to arrange 12 people into 3 teams of 4.

We know that ⁿCr = n! / [(n - r)! × r!]

No. of ways to select the first 4 people in the first group = ¹²C₄

                                                                                         = 12! / [(12 - 4)! × 4!]

                                                                                             = 12! / [(8! × 4!)]

                                                                                             = 495

No. of ways to select 4 people from the remaining 8 for the second group = ⁸C₄   = 8! / [(8 - 4)! × 4!] = 8! / [(4! × 4!)] = 70

No. of ways to select 4 people from 4 for third group = ⁴C₄  = 4! / [(4 - 4)!] × 4!] = 4! / [(0! ×4!)] = 1

Total no. of ways to select people for group = (495 x 70 x 1) / 3! = 34650 /6 = 5775

Hence, we can split 12 people into 3 teams of 4 in 5775 different ways.

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By using the properties of similar triangles, the values of x and y are respectively, 15 & 7

Triangles with the same shape but different sizes are said to be similar triangles. Squares with any side length and all equilateral triangles are examples of related objects. In other words, if two triangles are similar, their corresponding sides are proportionately equal and their corresponding angles are congruent. Triangle resemblance is shown here by the symbol "≈"

Given that,

Side lengths of triangle 1,  y + 9, x, x -2

Side lengths of triangle 2,  30, 26, 5y - 3

Both the triangles are similar,

So, we can write

(x - 2)/x = 26/30

After solving it, x = 15

Similarly,

(y + 9)/x = (5y - 3)/30

y = 7

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For the given function; Domain: -4 ≤ x ≤ 4, Range: 0 ≤ y ≤ 3, Relative max: x = 0, y=3, and Relative min: x = -2, y=0 and x = 2, y =0.

The range of values that we are permitted to enter into our function is known as the domain of a function.

The x values for a function like f make up this set (x).

A function's range is the collection of values it can take as input. The values in this set

Hence, once we input an x value, the function returns. The y values are those.

The domain of the function is the input values, that is the x-coordinates thus,

Domain: -4 ≤ x ≤ 4

Range of the function is the output values, that is the y-coordinate thus,

Range: 0 ≤ y ≤ 3

Relative max is a point where the function changes from increasing to decreasing thus,

Relative max: x = 0, y=3

Relative min is a point where the function changes from decreasing to increasing, thus,

Relative min: x = -2, y=0 and x = 2, y =0

Hence, the Domain: -4 ≤ x ≤ 4, Range: 0 ≤ y ≤ 3, Relative max: x = 0, y=3, and Relative min: x = -2, y=0 and x = 2, y =0.

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The general solution for given equation is y = c₁e^5t + c₂e^-3t - (1/7)e^4t.

A differential equation is an equation that contains at least one derivative of an unknown function, either an ordinary derivative or a partial derivative. Suppose the rate of change of a function y with respect to x is inversely proportional to y, we express it as dy/dx = k/y.

In calculus, a differential equation is an equation that involves the derivative (derivatives) of the dependent variable with respect to the independent variable (variables). The derivative represents nothing but a rate of change, and the differential equation helps us present a relationship between the changing quantity with respect to the change in another quantity. y=f(x) be a function where y is a dependent variable, f is an unknown function, x is an independent variable.

We have to find the general solution for d²y/dt² - 2 dy/dt - 15y = e^4t

Solving the above ordinary differential equation, we get:

y = c₁e^5t + c₂e^-3t - (1/7)e^4t

Thus, the general solution for given equation is y = c₁e^5t + c₂e^-3t - (1/7)e^4t

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xplanation:

a b c your way out the picture

The CEO should allocate $144,000 to labor and $216,000 to capital equipment.

Step-by-step explanation given below

The CEO should allocate $144,000 to labor and $216,000 to capital equipment. This allocation maximizes the output of the factory, which can be found by taking the partial derivatives of the function Q with respect to K and L and setting them equal to 0.

Partial derivative of Q with respect to K = -20K^-4/3L^2/3 = 0

20K^-4/3L^2/3 = 0

K^4/3L^2/3 = 20

K = (20L^2/3)^1/4

Partial derivative of Q with respect to L = -40K^-1/3L^-1/3 = 0

40K^-1/3L^-1/3 = 0

K/L = 40

K = 40L

Solving for K and L, we get:

K = (20*L^2/3)^1/4

L = 40K

Substituting K into the equation for L, we get:

L = (40*(20L^2/3)^1/4)

L = 80(20L^2/3)^1/4

L^3/4 = 160L^2/3

L^3 = 1280L^2

L^2 = 1280L

L = 1280

Substituting L into the equation for K, we get:

K = (20*1280^2/3)^1/4

K = (307200/3)^1/4

K = 144

Therefore, the CEO should allocate $144,000 to labor and $216,000 to capital equipment.

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The time taken by Edson to complete his workout routine is 13 t

The unitary method is a method in which you find the value of a unit and then the value of a required number of units.

Given here:  Edson exercises in sets that are each t minutes long. He does 6 sets of push-ups, 3 sets of pull-ups, and 4 sets of sit-ups

Now Edson does 6 sets of pushups time taken to finish the sets are

Time ( pushups)=6t

similarly, the time it takes to complete the other exercises are

Time(pull ups)=3t

Time(sit ups)=4t

Thus total time taken is equal to 3t+4t+6t=13t

Hence, The time taken by Edson to complete his workout routine is 13 t

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EF is parallel to GH by using the slope formula. Both the slope is equal to -4/5.

Given that the information below:

E(2,5), F(7,1), G(2.-3) and H(-8,5) is EF parallel to line GH.

By using slope formula method we can compare the slopes of line EF & line GH.

Where m = (y2 - y1 ) / (x2 - x1) slope of segment become (x1,y1) & (x2,y2)

EF Slope = (1-5)/(7-2) = -4/5

GH Slope = (5-(-3)/(-8-2) = 8/-10 = -4/5

By using slope formula method EF  slope is -4/5 & GH slope is -4/5. Both the slopes of the equation is equal. In this way we can prove that the EF is parallel to GH.

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Answer:

y = 64

x = 99

Step-by-step explanation:

180 - 116 = y

y = 64

A parallelogram is 360 degrees.

360 - 72 -125 - 64 = x

x = 99

I hope this helps!

On the right side, we can just simplify the expression to 1+tanθ+cotθ

Yes, this is true.

The left side of the equation can be simplified to 1, which is equal to the right side.

Left side:

(cotθ/1-tanθ)+(tanθ/1-cotθ)

= (cotθ + tanθ)/(1 - tanθ - cotθ)

= (tanθ + cotθ)/(1 - tanθ - cotθ)

= 1/(1 - tanθ - cotθ)

= 1

Right side:

1 + tanθ + cotθ = 1

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Answer:

See proof below

Step-by-step explanation:

(Please excuse any typos and point them out to me. It is a pain typing in LateX)

We are asked to prove:

[tex]\dfrac{\cot\theta}{1-\text{\ensuremath{\tan\theta}}}+\dfrac{\tan\theta}{1-\cot\theta}=1+\tan\theta+\cot\theta[/tex]

Since
[tex]\tan\theta =\dfrac {\sin\theta}{\cos\theta}\\\\\cot\theta = \dfrac {\cos\theta}{\sin\theta}\\\\[/tex]

the left side becomes

[tex]\dfrac{\cos\theta/\sin\theta}{1-\sin\theta/\cos\theta}+ \dfrac{{\sin\theta/\cos\theta}}{1-\cos\theta/\sin\theta} \cdots \cdots \cdots(1)[/tex]

For ease of expressing this let

[tex]a = \sin\theta\\b = \cos\theta\\[/tex]

Substituting the above in expression (1) gets us
[tex]\dfrac{b/a}{1- a/b} + \dfrac{a/b}{1-b/a}[/tex]

Let's take the individual terms and simplify each
[tex]\dfrac{b/a}{1- a/b} = \dfrac{b}{a}\cdot \dfrac{1}{1-\dfrac{a}{b}}[/tex]

[tex]\dfrac{1}{1-\dfrac{a}{b}}= \dfrac{1}{\dfrac{b-a}{b}}=\dfrac{b}{b-a}[/tex]

Therefore

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

For the second term we get

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

Noting that (a - b) = - (b - a)

the entire expression becomes

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

Multiply this expression throughout by the LCM which is
[tex]ab(b-a)[/tex]

[tex]\dfrac{b^2 \cdot b - {a^2}\cdot a}{ab(b-a)}\\\\= \dfrac{b^3 - a^3}{ab(b-a)}\cdots\cdots(2)\\[/tex]

(b -a) cancels in numerator and denominator on both sides giving

For the numerator we have the identity
[tex]b^3 - a^3 = (b-a)(b^2 + ab + b^2)[/tex]

So the expression in (2) becomes
[tex]\dfrac{(b-a)(b^2 + ab + a^2)}{ab(b-a)}\\\\\textrm{(b-a) cancels from the numerator and denominator}\\\\[/tex]

This leaves us with:

[tex]\dfrac{(b^2 + ab + a^2)}{ab}[/tex]

Substituting for a = sinθ and b = cosθ we get

[tex]\dfrac{\cos^2\theta + \sin^2\theta+ \cos\theta\sin\theta}{\sin\theta\cos\theta}[/tex]

Divide each term in the numerator by [tex]\sin\theta \cos\theta[/tex]

==> [tex]\dfrac{cos^2\theta}{\sin\theta\cos\theta} + \dfrac{\sin^2\theta}{\sin\theta\cos\theta}+ \dfrac{\sin\theta\cos\theta}{\sin\theta\cos\theta}}[/tex]

==> [tex]\dfrac{\cos\theta}{\sin\theta} + \dfrac{\sin\theta}{\cos\theta} + 1[/tex]

[tex]\cot\theta + \tan\theta + 1\\== > 1 + \tan\theta + \cot\theta[/tex]

Hence PROVED

Small market orders copies of a certain magazine for its magazine rack each week. The number of copies to order to generate more revenue is 4.

Let the profit y, the demand for the magazine be x , the number owner ordered k , then we get :

y = 4x - 2m ( m > x)

y = 2m        ( m ≤ x)

If m = 3 we get

E (y) = 1/14 (4 - 2x3) + 2/14 (4x2 - 2x3) + 3/14 (2x3) + 3/14 (2x3) + 2/14 (2x3) + 3/14 (2x3)

E(y) = 68/14

If m = 4 get,

E (y) = 1/14 (4 - 2x4) + 2/14 (4x2 - 2x4) + 3/14 (4x3 -2x4) + 3/14 (2x4) + 2/14 (2x4) + 3/14 (2x4)

E(y) = 72/14

E (y)m=4 > E(y)m=3

So order number is 4.

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The range of the function lies between  -∞<y<∞ and the domain of the function is given by -1<x<3

Function, in mathematics, is an expression, rule, or law that defines a relationship between one variable (the independent variable) and another variable (the dependent variable).

Given here: The graph of a function

We know  The domain of a function is the set of values that we are allowed to plug into our function. This set is the x values in a function such as f(x). The range of a function is the set of values that the function assumes. This set is the values that the function shoots out after we plug an x value in.

It is clear from the graph that y ∈ (-∞,∞) and x∈ (-1,3)

we know a function is given by y=f(x)

where y is the range values and x is the domain.

Hence, The range of the function lies between  -∞<y<∞ and the domain of the function is given by -1<x<3

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An interest rate of 2.5 % is required to obtained a current capital of $ 143,000 after a period of 16 years.

In this problem we find that Jayden deposits an amount of money for a period of 16 years and such account is under composite interest, that is, that the account increases its capital continously in time. Compound interest model is shown:

C' = C · (1 + r / 100)ˣ

Where:

If we know that C = 97,000, C' = 143,000 and x = 16, then the interest rate is:

143,000 = 97,000 · (1 + r / 100)¹⁶

143 / 97 = (1 + r / 100)¹⁶

1.025 = 1 + r / 100

0.025 = r / 100

r = 2.5

The needed interest rate needed for a capital of $ 143,000 after a period of 16 years is equal to 2.5 %.

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