Which expression is equivalent to 2a+1/10a-5 / 10a/4a^2-1

Answer:

[tex]\large\boxed{-9y^5}[/tex]

Step-by-step explanation:

[tex]-3y\cdot3y^4=(-3\cdot3)(y\cdot y^4)\qquad\text{use}\ a^n\cdot a^m=a^{n+m}\\\\=-9y^{1+4}=-9y^5[/tex]

Answer:

9×80=x×15

x=58lbs

Step-by-step explanation:

Considering no frictions applyd, the value of the report of the two forces ( F- the action force and R- the resistance force) equals the value of the report between the value of the distance from the folcrum to the rock and the value of the distance from the fulcrum to the active force

Answer:

8/7

Step-by-step explanation:

8/7 is an improper fraction.

8 > 7

A fraction has to have the numerator less than the denominator, in order to be a proper fraction.

In this case, 8/7 is the only fraction with a numerator more than the denominator.

Therefore, 8/7 is an improper fraction.

Answer:

8/7 is an improper fraction.

Step-by-step explanation:

An improper fraction is just a fraction where the numerator (top number) is greater than the denominator (bottom number)

2<3

6<11

21<25

8>7

Hope this helps!!!

Answer:

2ε + 12 [two trillion (twelve zeros)]

Step-by-step explanation:

[1][2][4][5][10][1000][2000][2500] (1 is unnecessary)

If you are talking about the FIRST 8 factors of 10000 [10⁴], then here you are. Multiply all those numbers to get the calculator notation answer above, or two trillion.

**This question is not specific enough, which was why I improvised and thought that this was what you meant. I apologize if this was not what you meant and I also hope this was what you were looking for.

I am joyous to assist you anytime.

Answer:

[tex]10^{32}[/tex].

Step-by-step explanation:

eight factors of 10^4 are

[tex]10^4*10^4*10^4*10^4*10^4*10^4*10^4*10^4 = (10^4)^8 = 10^{4*8}=10^{32}[/tex]

that is one hundred nonillion.

John's present age is 15 years, and his sister's present age is 10 years.

We have,

Let's denote John's sister's age as x.

According to the given information,

John is five years older than his sister, so John's age would be x + 5.

The product of their present ages is 150, which means:

x * (x + 5) = 150

Expanding the equation:

x² + 5x = 150

Rearranging the equation to standard quadratic form:

x² + 5x - 150 = 0

Now we can solve this quadratic equation.

Factoring or using the quadratic formula will give us the values of x, representing the sister's age.

Factoring the equation:

(x - 10)(x + 15) = 0

Setting each factor to zero:

x - 10 = 0 or x + 15 = 0

Solving for x:

x = 10 or x = -15

Since ages cannot be negative, we can discard the solution x = -15. Therefore, the sister's present age is x = 10.

John's present age is then calculated by adding 5 years to his sister's age:

John's age = x + 5 = 10 + 5 = 15 years.

Thus,

John's present age is 15 years, and his sister's present age is 10 years.

Learn more about expressions here:

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By forming a quadratic equation from the given problem, we determine that John's present age is 15 years and his sister's present age is 10 years.

Let's designate John's age as J and his sister's age as S. We've been given two pieces of information: John is five years older than his sister, and the product of their present ages is 150. Mathematically, these can be expressed as:

Substituting the first equation into the second gives us:

(S + 5) * S = 150

Expanding and rearranging, we get a quadratic equation:

S2 + 5S - 150 = 0

To solve for S, we can either factor the quadratic equation or use the quadratic formula. Factoring seems simpler in this case:

(S + 15)(S - 10) = 0

This gives us two possible values for S: -15 and 10. Since ages cannot be negative, we discard S = -15 and keep S = 10. Now we can find John's age:

J = S + 5 = 10 + 5 = 15

Therefore, John's present age is 15 years, and his sister's present age is 10 years.

The domain of the outer function is all real numbers, because f(x) is an exponential function.

The domain and range of g(x) are all real numbers.

So, the domain of the composition is again all real numbers, because there is no way that an output from g(x) will not be a valid input for f(x).

For this case we have the following functions:

[tex]f (x) = 2 ^ x\\g (x) = x-2[/tex]

We must find [tex](f_ {o} g) (x):[/tex]

By definition we have to:

[tex](f_ {o} g) (x) = f [g (x)][/tex]

So:

[tex](f_ {o} g) (x) = 2 ^ {x-2}[/tex]

By definition, the domain of a function is given by all the values for which the function is defined. Thus, the domain of the composite function is:

In this case, there are no real numbers that make the expression indefinite.

Thus, the domain is given by all the real numbers.

Answer:

All the real numbers

Answer:

The midpoint of the line segment is (9,3).

Step-by-step explanation:

To find the midpoint of a line segment, we use the midpoint formula, ( (x1 + x2)/2, (y1+y2)/2).  This means that to find the midpoint, we must add together both of the x-values of the endpoints and divide by 2 and do the same with the y values, basically finding the average of the two endpoints, or the middle.

x1 + x2 = 10 + 8 = 18

18/2 = 9

y1 + y2 = 2 + 4 = 6

6/2 = 3

Therefore, the midpoint of the line segment is (9,3).

Hope this helps!

Answer:

-18

Step-by-step explanation:

= |3 – 10| - (12 / 4 + 2)^2

= 7 - (3 + 2)^2

= 7 - (5)^2

= 7 - 25

= -18

Answer:

$28.

Step-by-step explanation:

p(x) = -2(x - 9)^2 + 100

If the prices (x) = 15 dollars we work out the profit by substituting x = 15 in the above formula:

p(15) = -2(15 - 9)^2 + 100

= -2 * 6^2 + 100

= $28.

Answer:

28

Step-by-step explanation:

[tex](f+g)(x)=2x^2+3+x-2=2x^2+x+1\\\\(f+g)(2)=2\cdot2^2+2+1=8+3=11[/tex]

Answer:

11

Step-by-step explanation:

(f+g)(2) means f(2)+g(2).

f(2) means we need to replace the x's in f(x)=2x^2+3 with 2.

This gives us f(2)=2(2)^2+3.

Let's simplify f(2)=2(4)+3=8+3=11.

g(2) means we need to replace the x's in g(x)=x-2 with 2.

This gives us g(2)=2-2

Let's simplify g(2)=0.

Now f(2)+g(2) means we just add the result of f(2) to the result of g(2).

So the problem is what is 11+0?

Answer is 11

Answer:

The circumference is a irrational number

Step-by-step explanation:

we know that

The circumference of a circle is equal to

[tex]C=\pi D[/tex]

where

D is the diameter

In this problem we have that

[tex]D=d\ units[/tex] ---> is a rational number

substitute

[tex]C=\pi (d)[/tex]

[tex]C=d\pi\ units[/tex]

Remember that

The number π is a irrational number

and

If you multiply a rational number by a irrational number, the result is a irrational number

therefore

The circumference is a irrational number

Answer: 6 days.

Step-by-step explanation:

Inverse proportion equation has the form:

[tex]y=\frac{k}{x}[/tex]

Where "k" is the constant of proportionality.

Let be "y" the time it takes to do a job (number of days) and "x" the number of workers.

We can find "k" knowing that  it takes 3 workers 16 days to finish a job:

[tex]16=\frac{k}{3}\\\\k=16*3\\\\k=48[/tex]

To find how long will it take 8 workers to finish the job, you must substitute the value of "k" and [tex]x=8[/tex] into [tex]y=\frac{k}{x}[/tex]. Then you get:

 [tex]y=\frac{48}{8}\\\\y=6[/tex]

Answer:

Table P represents a function

Step-by-step explanation:

* Lets explain the meaning of the function

- A function is a relation between a set of inputs and a set of outputs

 in condition of each input has exactly one output

- Ex:

# The relation {(1 , 2) , (-4 , 5) , (-1 , 5)} is a function because each x in the

   order pair has only one value of y

# The relation {(1 , 2) , (1 , 5) , (3 , 7)} is not a function because there is x

   in the order pairs has two values of y (x = 1 has y = 2 and y = 5)

* Lets solve the problem

# Table P :

- In put    : 8  ,  1  ,  5

- Out put : 3  ,  7  , 4

∵ Each input has only one output

∴ Table P represents a function

# Table Q :

- Input     : 9  ,  9  ,  4

- Out put : 3  ,  5  ,  2

∵ The input 9 has two outputs 5 and 2

∴ Table Q doesn't represent a function

# Table R :

- In put    : 7  ,  8  ,  7

- Out put : 2  ,  6  ,  3

∵ The input 7 has two outputs 2 and 3

∴ Table R doesn't represent a function

# Table S :

- In put    :  1  ,  1  ,  9

- Out put :  7 ,  5  ,  2

∵ The input 1 has two outputs 7 and 5

∴ Table S doesn't represent a function

* Table P represents a function

Answer:

The answer is A I just took the test

Answer

Y= 3 × 2 + 6x + 1 = 6 + 6x +1 = 6x + (6+1) = 6x + 7

Answer:

This is a function, so the answer is a plot, you can see it in the attached picture

Step-by-step explanation:

This is a function, the best way to go here is to graph.

First lest find the roots of 3x2+6x+1, you do this by using the quadratic equation

[tex]x_{1} = \frac{-b + \sqrt{b^{2}-4ac }}{2a}\\x_{2} = \frac{-b - \sqrt{b^{2}-4ac }}{2a}[/tex]

Where a=3, b=6, c=1

Using that, you get, x1 = -1 - sqrt(2/3) which is a negative number, and x2=sqrt(2/3) - 1 which is a negative number.

These number represent values that makes the function equals =0

Now if we make x=0, the result is y=1,

We have a better idea of what is to plot and it is shown on the attached picture

Answer:

The domain is (2,8,12)

Step-by-step explanation:

we have

[tex]f(x)=2x-4[/tex]

we know that

The range is (0,12,20)

Find the domain

1) For f(x)=0 -----> Find the value of x

substitute the value of f(x) in the equation and solve for x

[tex]0=2x-4[/tex]

[tex]2x=4[/tex]

[tex]x=2[/tex]

therefore

For f(x)=0 the domain is x=2

2) For f(x)=12 -----> Find the value of x

substitute the value of f(x) in the equation and solve for x

[tex]12=2x-4[/tex]

[tex]2x=12+4[/tex]

[tex]x=8[/tex]

therefore

For f(x)=12 the domain is x=8

3) For f(x)=20 -----> Find the value of x

substitute the value of f(x) in the equation and solve for x

[tex]20=2x-4[/tex]

[tex]2x=20+4[/tex]

[tex]x=12[/tex]

therefore

For f(x)=20 the domain is x=12

therefore

The domain is (2,8,12)

Answer:

Step-by-step explanation:

[tex]5+20\cdot2^{2-3x}=10\cdot2^{-2x}+5\qquad\text{subtract 5 from both sides}\\\\20\cdot2^{2-3x}=10\cdot2^{-2x}\qquad\text{divide both sides by 10}\\\\2\cdot2^{2-3x}=2^{-2x}\qquad\text{use}\ a^n\cdot a^m=a^{n+m}\\\\2^{1+2-3x}=2^{-2x}\\\\2^{3-3x}=2^{-2x}\iff3-3x=-2x\qquad\text{subteact 3 from both sides}\\\\-3x=-2x-3\qquad\tex\qquad\text{add}\ 2x\ \text{to both sides}\\\\-x=-3\qquad\text{change the signs}\\\\x=3[/tex]

Answer:

Part a) When Marcos purchased the card, the balance on the card was $40

This is the B-intercept

Part b) Mark will have spoken 400 minutes when he runs out of money

This is the n-intercept

part c) The slope of the equation is -0.1 and the units are dollars per minute

Step-by-step explanation:

we have

[tex]B=40-0.1n[/tex]

where

B is balance in dollars on the card

n is the number of minutes

Part a) How much money was on the card when he purchased it? $______. Which intercept is this? B-intercept or N-intercept?

we know that

The B-intercept is the value of B when the value of n is equal to zero

so

For n=0

substitute and find the value of B

[tex]B=40-0.1(0)=\$40[/tex]

therefore

When Marcos purchased the card, the balance on the card was $40

This is the B-intercept

Part b) How many minutes will he have talked when he runs out of money? $_____. Which intercept is this? B-intercept or n-intercept ?

we know that

The n-intercept is the value of n when the value of B is equal to zero

so

For B=0

substitute and find the value of n

[tex]0=40-0.1n[/tex]

[tex]0.1n=40[/tex]

[tex]n=400\ minutes[/tex]

therefore

Mark will have spoken 400 minutes when he runs out of money

This is the n-intercept

Part c) What is the slope of this equation?________ . What are the units on the slope? Minute , dollars per minute ,minutes per dollars or dollars ?

we have

[tex]B=40-0.1n[/tex]

This is is the equation of the line into slope intercept form

[tex]m=-0.1\frac{\$}{minute}[/tex] -----> slope of the equation

[tex]b=40[/tex] ------> the B-intercept

The units of the slope are dollars per minute

Marcos purchased a top-up card with an initial balance of $40 (B-intercept). He will run out of money after 400 minutes of call time (n-intercept). The charge rate is $0.1 per minute (slope).

a) The money on the card when he purchased it is given by the constant term in the equation, which is $40. This is the B-intercept, because it is the value of B when n = 0 (meaning no minutes have been used).

b) Marcos will have run out of money when B = 0. To find this, set B = 0 and solve for n: 0 = 40 - 0.1n, which leads to n = 400. So, Marcos will have talked for 400 minutes when he runs out of money. This is the n-intercept. It represents the value of n when B = 0 (meaning there is no money left on the card).

c) The slope of this equation is -0.1. In the context of this problem, the slope represents the rate at which money is deducted from the balance for each minute of talk time. Therefore, the units on the slope are dollars per minute.

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Answer:

Yes, A is correct.

Answer:

Piecewise function model

Step-by-step explanation:

According to the given statement a factory manufactures widgets based on customer orders. If a customer's order is for less than w widgets, the customer's cost per widget is d dollars.

customer order < widgets

If a customer's order is for at least w widgets, the customer's cost per widget is decreased by c cents for each widget ordered over w widgets. A customer's total cost is t.

customer order > widgets

We can notice that we have two different conditions. In this type of question where there are different conditions for different types of domain we use piecewise function.

With different conditions, the cost is different, therefore piecewise function model would be the best option for the situation....

Answer:

9* 3 ^ (x-2)

Step-by-step explanation:

g(x) = 3^x

We know a^ (b) * a^(c) = a^ (b+c)

 9* 3 ^ (x+2) = 3^2 * 3 ^(x+2) = 3^(2+x+2) = 3^x+4  not equal to 3^x

3*(9^(x+2)) = 3*3^2(x+2) = 3^1 * 3^(2x+4) =3^(2x+4+1) = 3^(2x+5) not equal  

9* 3 ^ (x-2) = 3^2 * 3 ^(x-2) = 3^(2+x-2) = 3^x  equal to 3^x    

3*(9^(x-2)) = 3*3^2(x-2) = 3^1 * 3^(2x-4) =3^(2x-4+1) = 3^(2x-3) not equal        

Answer:

D. size and shape

Step-by-step explanation:

Triangles are congruent if they have the same size and shape.

First of all, let's write this statement in vector form. For the fist vector we have:

Magnitude 3.5 m/s, direction angle 35°:

Let's say this is vector [tex]\vec{A}[/tex], so the magnitude is:

[tex]\left|\vec{A}\right|=3.5m/s[/tex]

And the direction is defined as:

[tex]\theta = 35^{\circ}[/tex]

So the components are:

[tex]Ax=\left|\vec{A}\right| cos\theta \\ \\ Ax=3.5 cos35^{\circ}=2.86m/s \\ \\ \\ Ay=\left|\vec{A}\right| sin\theta \\ \\ Ay=3.5 sin35^{\circ}=2m/s[/tex]

So vector [tex]\vec{A}[/tex] is:

[tex]\vec{A}=2.86i+2j[/tex]

For the second vector:

Magnitude 4 m/s, direction angle 150°:

Let's say this is vector [tex]\vec{B}[/tex], so the magnitude is:

[tex]\left|\vec{B}\right|=4m/s[/tex]

And the direction is defined as:

[tex]\theta = 150^{\circ}[/tex]

So the components are:

[tex]Bx=\left|\vec{B}\right| cos\theta \\ \\ Bx=4 cos150^{\circ}=-2\sqrt{3}m/s \\ \\ \\ By=\left|\vec{B}\right| sin\theta \\ \\ By=4 sin150^{\circ}=2m/s[/tex]

So vector [tex]\vec{B}[/tex] is:

[tex]\vec{B}=-2\sqrt{3}i+2j[/tex]

THE SUM OF THESE TWO VECTORS IS:

[tex]\vec{R}=\vec{A}+\vec{B}=(2.86i+2j)+(-2\sqrt{3}i+2j) \\ \\ \boxed{\vec{R}=-0.60i+4j}[/tex]

THE MAGNITUDE OF THE RESULTANT VECTOR IS:

[tex]\left|\vec{R}\right|=\sqrt{Rx^2+Ry^2} \\ \\ Rx=Ax+Bx \\ \\ Ry=Ay+By \\ \\ \\ \left|\vec{R}\right|=\sqrt{(-0.6^2)+(4)^2} \\ \\ \boxed{\left|\vec{R}\right|=4.05m/s}[/tex]

Magnitude 4.5 m/s, direction angle 55°:

Let's say this is vector [tex]\vec{C}[/tex], so the magnitude is:

[tex]\left|\vec{C}\right|=4.5m/s[/tex]

And the direction is defined as:

[tex]\theta = 55^{\circ}[/tex]

So the components are:

[tex]Cx=\left|\vec{C}\right| cos\theta \\ \\ Cx=4.5 cos55^{\circ}=2.58m/s \\ \\ \\ Cy=\left|\vec{C}\right| sin\theta \\ \\ Cy=4.5 sin55^{\circ}=3.68m/s[/tex]

So vector [tex]\vec{C}[/tex] is:

[tex]\vec{C}=2.58i+3.68j[/tex]

For the second vector:

Magnitude 3 m/s, direction angle 135°:

Let's say this is vector [tex]\vec{D}[/tex], so the magnitude is:

[tex]\left|\vec{D}\right|=3m/s[/tex]

And the direction is defined as:

[tex]\theta = 135^{\circ}[/tex]

So the components are:

[tex]Dx=\left|\vec{D}\right| cos\theta \\ \\ Dx=3 cos135^{\circ}=-\frac{3\sqrt{2}}{2} \\ \\ \\ Dy=\left|\vec{D}\right| sin\theta \\ \\ Dy=3 sin135^{\circ}=\frac{3\sqrt{2}}{2}[/tex]

So vector [tex]\vec{D}[/tex] is:

[tex]\vec{D}=-\frac{3\sqrt{2}}{2}i+\frac{3\sqrt{2}}{2}j[/tex]

THE SUM OF THESE TWO VECTORS IS:

[tex]\vec{R}=\vec{C}+\vec{D}=(2.58i+3.68j)+(-\frac{3\sqrt{2}}{2}i+\frac{3\sqrt{2}}{2}j) \\ \\ \boxed{\vec{R}=0.46i+5.80j}[/tex]

THE MAGNITUDE OF THE RESULTANT VECTOR IS:

[tex]\left|\vec{R}\right|=\sqrt{(0.46)^2+(5.8)^2} \\ \\ \boxed{\left|\vec{R}\right|=5.83m/s}[/tex]

Magnitude 4.5 m/s, direction angle 55°:

This is vector [tex]\vec{E}[/tex], so the magnitude is:

[tex]\left|\vec{E}\right|=3m/s[/tex]

Direction:

[tex]\theta = 70{\circ}[/tex]

Components:

[tex]Ex=\left|\vec{E}\right| cos\theta \\ \\ Ex=3 cos70^{\circ}=1.02m/s \\ \\ \\ Ey=\left|\vec{E}\right| sin\theta \\ \\ Ey=3 sin70^{\circ}=2.82m/s[/tex]

So:

[tex]\vec{E}=1.02i+2.82j[/tex]

For the second vector:

Magnitude 5 m/s, direction angle 210°:

[tex]\vec{F}[/tex]:

[tex]\left|\vec{F}\right|=5m/s[/tex]

Direction:

[tex]\theta = 210^{\circ}[/tex]

Components:

[tex]Fx=5 cos210^{\circ}=-\frac{5\sqrt{3}}{2} \\ \\ \\ Ey=5 sin210^{\circ}=-\frac{5}{2}[/tex]

Then:

[tex]\vec{F}=-\frac{5\sqrt{3}}{2} i-\frac{5}{2}j[/tex]

THE SUM OF THESE TWO VECTORS IS:

[tex]\vec{R}=(1.02i+2.82j)+(-\frac{5\sqrt{3}}{2}i-\frac{5}{2}j) \\ \\ \boxed{\vec{R}=-3.31i+0.32j}[/tex]

THE MAGNITUDE OF THE RESULTANT VECTOR IS:

[tex]\left|\vec{R}\right|=\sqrt{(-3.31)^2+(0.32)^2} \\ \\ \boxed{\left|\vec{R}\right|=3.32m/s}[/tex]

Magnitude 6 m/s, direction angle 120°:

[tex]\left|\vec{W}\right|=6m/s[/tex]

Direction:

[tex]\theta = 120^{\circ}[/tex]

Components:

[tex]Wx=6 cos120^{\circ}=-3m/s \\ \\ \\ Wy=6 sin120^{\circ}=3\sqrt{3}m/s[/tex]

So:

[tex]\vec{W}=-3i+3\sqrt{3}j[/tex]

For the second vector:

Magnitude 2 m/s, direction angle 240°:

[tex]\vec{Z}[/tex]:

[tex]\left|\vec{Z}\right|=2m/s[/tex]

Direction:

[tex]\theta = 240^{\circ}[/tex]

Components:

[tex]Zx=2 cos240^{\circ}=-1 \\ \\ \\ Zy=2 sin240^{\circ}=-\sqrt{3}[/tex]

Then:

[tex]\vec{Z}=-i-\sqrt{3}j[/tex]

THE SUM OF THESE TWO VECTORS IS:

[tex]\vec{R}=(-3i+3\sqrt{3}j)+(-i-\sqrt{3}j) \\ \\ \boxed{\vec{R}=-4i+2\sqrt{3}j}[/tex]

THE MAGNITUDE OF THE RESULTANT VECTOR IS:

[tex]\left|\vec{R}\right|=\sqrt{(-4)^2+(2\sqrt{3})^2} \\ \\ \boxed{\left|\vec{R}\right|=5.29m/s}[/tex]

Answer:

p(x) = 6(x + 2)² - 3

Step-by-step explanation:

This one requires a lot of thinking because since our A is not 1 and how the quadratic equation looks, we need to think of a low number while "completing the square [½B]²". So, let us choose 2. We set it up like this:

6(x + 2)² → 6(x² + 4x + 4) → 6x² + 24x + 24

6x² + 24x + 24 - 3 → 6x² + 24x + 21 [TA DA!]

We know that our vertex formula is correct. Additionally, -h gives you the OPPOSITE terms of what they really are, and k gives you the EXACT terms of what they really are. Therefore, your vertex is [-2, -3].

I am joyous to assist you anytime.

Answer: the length is 13

Step-by-step explanation:

one side is 26 ft. and y is half of one side so by dividing 26 by 2 you would get 13

Final answer:

Only options b (74) and c (80) could have their square roots between 8 and 9, as their values lie between 64 (8 squared) and 81 (9 squared).

Explanation:

To find which numbers could have their square roots lying between 8 and 9, we need to consider the squares of these two numbers. 8 squared is 64 and 9 squared is 81. Therefore, any number that has a square root between 8 and 9 must be greater than 64 and less than 81.

Thus, the values that could be the number with the square root between 8 and 9 are 74 and 80.

Answer:

C

Step-by-step explanation:

The general rule for the quadratic function is

[tex]y=ax^2+bx+c[/tex]

Use the data from the table:

[tex]y(50)=130\Rightarrow 130=a\cdot 50^2+b\cdot 50+c\\ \\y(70)=130\Rightarrow 130=a\cdot 70^2+b\cdot 70+c\\ \\y(90)=200\Rightarrow 200=a\cdot 90^2+b\cdot 90+c[/tex]

We get the system of three equations:

[tex]\left\{\begin{array}{l}2500a+50b+c=130\\ \\4900a+70b+c=130\\ \\8100a+90b+c=200\end{array}\right.[/tex]

Subtract these equations:

[tex]\left\{\begin{array}{l}4900a+70b+c-2500a-50b-c=130-130\\ \\8100a+90b+c-2500a-50b-c=200-130\end{array}\right.\Rightarrow \left\{\begin{array}{l}2400a+20b=0\\ \\5600a+40b=70\end{array}\right.[/tex]

From the first equation

[tex]b=-120a[/tex]

Substitute it into the second equation:

[tex]5600a+40\cdot (-120a)=70\Rightarrow 800a=70,\\ \\ a=\dfrac{7}{80},\\ \\ b=-120\cdot \dfrac{7}{80}=-\dfrac{21}{2}=-10.5[/tex]

So,

[tex]2500\cdot \dfrac{7}{80}+50\cdot (-10.5)+c=130\Rightarrow 218.75-525+c=130\\ \\c=130-218.75+525=436.25[/tex]

The quadratic function is

[tex]y=\dfrac{7}{80}x^2-10.5x+436.25\\ \\y=0.0875x^2-10.5x+436.25[/tex]

Answer:

f(x) = (x+2)(x-8)/(x-6)(x+4) <-> x=6,x=-4

i(x) = (x-4)(x-6)/(x-2)(x+8)  <-> x=2,x=-8

k(x) = (x-2)(x+8)/(x+6)(x-4) <-> x=-6,x=4

m(x) = (x+4)(x-6)/(x+2)(x-8)  <-> x=-2,x=8

Step-by-step explanation:

The function is discontinuous if the denominator is zero.

We will check for which function the values are given

1) f(x) = (x+2)(x-8)/(x-6)(x+4)

if x = 6 and x = -4 the  denominator  is zero

So, x=6 and x=-4 given

2) g(x) = (x+4)(x-8)/(x+2)(x-6)

if x = -2 and x = 6 the  denominator  is zero

So, x= -2 and x= 6 not given so, g(x) will not be considered

3) h(x)= (x+2)(x-6)/(x-8)(x+4)

if x = 8 and x = -4 the  denominator  is zero

So, x= 8 and x= -4 not given so, h(x) will not be considered

4) i(x) = (x-4)(x-6)/(x-2)(x+8)

if x = 2 and x = -8 the  denominator  is zero

So, x= 2 and x= -8 given

5) j(x) = (x-2)(x+6)/(x-4)(x+8)

if x = 4 and x = -8 the  denominator  is zero

So, x= 4 and x= -8 not given so, j(x) will not be considered

6) k(x) = (x-2)(x+8)/(x+6)(x-4)

if x = -6 and x = 4 the  denominator  is zero

So, x= -6 and x= 4  given

7) l(x) = (x-4)(x+8)/(x+6)(x-2)

if x = -6 and x = 2 the  denominator  is zero

So, x= -6 and x= 2  not given so, l(x) will not be considered

8) m(x) = (x+4)(x-6)/(x+2)(x-8)

if x = -2 and x = 8 the  denominator  is zero

So, x= -2 and x= 8 given

Answer:

700

Step-by-step explanation:

he originally has 500 songs which is the constant. He adds 10 songs a week which is the slope. and 20 will be the number of weeks. You can make the equation y=10x+500 and replace the x with 20. 20 times 10 is 200 and 200 plus 500 is 700 which is your answer

namely, what is the leftover amount when the decay rate is 2% for an original amount of 75 grams after 10 years?

[tex]\bf \qquad \textit{Amount for Exponential Decay} \\\\ A=P(1 - r)^t\qquad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{initial amount}\dotfill &75\\ r=rate\to 2\%\to \frac{2}{100}\dotfill &0.02\\ t=\textit{elapsed time}\dotfill &10\\ \end{cases} \\\\\\ A=75(1-0.02)^{10}\implies A=75(0.98)^{10}\implies A\approx 61.28\implies \stackrel{\textit{rounded up}}{A=61~grams}[/tex]

Answer with explanation:

The exponential decay function is written as :-

[tex]f(x)=A(1-r)^x[/tex], where f (x) is the amount of material left after x years , A is the initial amount of material and r is the rate of decay.

Given : The amount of original item remained now = 75 grams

The rate of decay = 2% = 0.02

Now, the amount of original artifact would there be left if they had not discovered it for another 10 years is given by :-

[tex]f(10)=75(1-0.02)^{10}[/tex]

Solving the above exponential equation , we get

[tex]=61.2804605166\approx61[/tex]

Hence only 61 grams original artifact would there be left if they had not discovered it for another 10 years .

Answer:

f(x) = (x+2)(x+3)(x-(3-6i))(x-(3+6i))

f(x) = 270 + 189 x + 21 x^2 - x^3 + x^4

Step-by-step explanation:

First of all, we must know that complex roots come in conjugate pairs.

So the zeros of your equation would be

x = -2

x = -3

x = 3 - 6i

x = 3 + 6i

Your polynomial is of fourth degree.

f(x) = (x-(-2))(x-(-3))(x-(3-6i))(x-(3+6i))

f(x) = (x+2)(x+3)(x-(3-6i))(x-(3+6i))

Please , see attached image below for full expression

f(x) = 270 + 189 x + 21 x^2 - x^3 + x^4

Answer:

The required polynomial is [tex]P(x)=a\left(x^4-x^3+21x^2+189x+270\right)[/tex].

Step-by-step explanation:

The general form of a polynomial is

[tex]P(x)=a(x-c_1)^{m_1}(x-c_2)^{m_2}...(x-c_n)^{m_n}[/tex]

where, a is a constant, [tex]c_1,c_2,..c_n[/tex] are zeroes with multiplicity [tex]m_1,m_2,..m_n[/tex] respectively.

It is given that  –2, –3,3 – 6i are three zeroes of a polynomial.

According to complex conjugate root theorem, if a+ib is a zero of a polynomial, then a-ib is also the zero of that polynomial.

3 – 6i is a zero. By using complex conjugate root theorem 3+6i is also a zero.

The required polynomial is

[tex]P(x)=a(x-(-2))(x-(-3))(x-(3-6i))(x-(3+6i))[/tex]

[tex]P(x)=a(x+2)(x+3)(x-3+6i)(x-3-6i)[/tex]

[tex]P(x)=a\left(x^2+5x+6\right)\left(x-3+6i\right)\left(x-3-6i\right)[/tex]

On further simplification, we get

[tex]P(x)=a\left(x^3+6ix^2+2x^2+30ix-9x+36i-18\right)\left(x-3-6i\right)[/tex]

[tex]P(x)=a\left(x^4-x^3+21x^2+189x+270\right)[/tex]

Therefore the required polynomial is [tex]P(x)=a\left(x^4-x^3+21x^2+189x+270\right)[/tex].