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3 years ago

Indicate whether the following statement is true or false: An exponential growth function

Mathematics

1 answer:

barxatty [35]3 years ago

3 0

Answer:

The given statement is true.

In general, the growth rate of exponential functions is far greater than the quadratic function. This can be illustrated through an example.

Consider an exponential function and a quadratic function .

Note that the leading coefficient of the quadratic function is far greater than the base of the exponential function but still the exponential function will exceeds the quadratic function after x = 15.664 as shown in the graph below.

Therefore, we can conclude that the given statement is correct.

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If 40% of a number is 14, find 20% of that number.

Tom [10]

Answer:

Step-by-step explanation:

Because half of 40% is 20%

So 14%2 = 7

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3 years ago

The two plots compare the top 25 home run hitters in 2007 with those in 2008. Which choice is NOT true?

Fed [463]

Answer:

The median number of home runs in 2007 is equal to the minimum value for 2008.

Step-by-step explanation:

i got it correct

8 0

3 years ago

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Find the area of the following figure. Show your work.

Deffense [45]

Answer:

I think the answer is 46 because 9 + 23 + 7 + 7 = 46.

Step-by-step explanation:

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3 years ago

Read 2 more answers

The median player salary for a professional football team was $457,200 in 2000 and $1,342,676 in 2008. Write a linear equation g

Vesna [10]

Answer:

The linear equation giving the median salary y in terms of the year x is:

y = 110,684.5x + 457,200

Step-by-step explanation:

For calculating the linear equation first we need to calculate the slope of the equation represented with an m, the formula is the following:

$m=\frac{y_{2}- y_{1} }{x_{2}-x_{1}}$

As in the problem it is said, x will represent the time so:

x1 will be the initial year so x1=2000

x2 will be the final year so x2=2008

Now the salary will be represented with the y so:

y1 will be the salary for the year 2000 then y1 = $457,200

y2 will be the salary for the year 2008 then y2 = $1,342,576

Then replacing the values in the equation we will have the following expresión:

$m=\frac{1,342,676-457,200}{2008-2000} = \frac{885,476}{8}=110,684.5$

Now we have to replace this values in the following formula for linear equation

y-y1 = m(x-x1)

As they say that x represent 2000 as the initial year then when we replace we will have the following expresion

y - 457,200 = 110,684.5(x - 0)

y = 110,684.5x + 457,200

To verify our equation we can check for the salary in 2008 as it is 8 years after of 2000 x=8 then

y = 110,684.5 (8) + 457,200

y = 885,476 + 457,200 = 1,342,676

6 0

3 years ago

Math pls help!! answers?

statuscvo [17]

Answer: Choice B) Infinitely many solutions

one solution: x = 8, y = -7/2, z = 0
another solution: x = -12, y = 13/2, z = 10

=======================================================

Explanation:

Here's the starting original augmented matrix.

$\left[\begin{array}{ccc|c} 1 & 0 & 2 & 8\\5 & 1 & 9 & 73/2\\-4 & 0 & -8 & -32\\\end{array}\right]$

We'll multiply everything in row 3 (abbreviated R3) by the value -1/4 or -0.25, which will make that -4 in the first column turn into a 1.

We use this notation to indicate what's going on: $(-1/4)*R3 \to R3$

That notation says "multiply everything in R3 by -1/4, then replace the old R3 with the new corresponding values".

So we have this next step:

$\left[\begin{array}{ccc|c} 1 & 0 & 2 & 8\\5 & 1 & 9 & 73/2\\1 & 0 & 2 & 8\\\end{array}\right]\begin{array}{l} \ \\\ \\(-1/4)*R3 \to R3\\\end{array}$

Notice that the new R3 is perfectly identical to R1.

So we can subtract rows R1 and R3, and replace R3 with the result of nothing but 0's

$\left[\begin{array}{ccc|c} 1 & 0 & 2 & 8\\5 & 1 & 9 & 73/2\\0 & 0 & 0 & 0\\\end{array}\right]\begin{array}{l} \ \\\ \\R3-R1 \to R3\\\end{array}$

Whenever you get an entire row of 0's, it always means there are infinitely many solutions.

-------------------

Now let's handle the second row. That 5 needs to turn into a 0. We can multiply R1 by 5, and subtract that from R2.

So we need to compute 5*R1-R2 and have that replace R2.

$\left[\begin{array}{ccc|c} 1 & 0 & 2 & 8\\0 & 1 & -1 & -7/2\\0 & 0 & 0 & 0\\\end{array}\right]\begin{array}{l} \ \\5*R1-R2 \to R2\ \\\ \\\end{array}$

Notice that in the third column of R2, we have 9-5*2 = 9-10 = -1. So we have -1 replace the 9. In the fourth column of R2, we have 73/2 - 5*8 = -7/2. So the -7/2 replaces the 73/2.

--------------------

At this point, the augmented matrix is in RREF form. RREF stands for Reduced Row Echelon Form. It seems a bit odd that the "F" of "RREF" stands for "form" even though we say "form" right after "RREF", but I digress.

Because the matrix is in RREF form, this means R1 and R2 lead to these equations:

$R1 : 1x+0y+2z = 8\\ R2: 0z+1y-1z = -7/2$

which simplify to

$R1: x+2z = 8\\R2: y-z = -7/2$

Let's get the z terms to each side like so:

$x+2z = 8\\x = -2z+8\\\text{ and }\\y-z = -7/2\\y = z-7/2\\$

Therefore, all of the solutions are of the form $(x,y,z) = (-2z+8, z-7/2, z)$ where z is any real number.

If z is allowed to be any real number, then we can simply pick any number we want to replace it. We consider z to be the "free variable", in that it's free to be whatever it wants. The values of x and y will depend on what we pick for z.

So the concept of "infinitely many solutions" doesn't exactly mean we can pick just any triple for x,y,z (admittedly it would be nice to randomly pick any 3 numbers off the top of my head and be done right away). Instead, we can pick anything we want for z, and whatever we picked, will directly determine x and y. The x and y are locked into place so to speak.

Let's say we picked z = 0.

That would lead to...

$x = -2z+8\\x = -2(0)+8\\x = 8\\\text{ and }\\y = z-7/2\\y = 0-7/2\\y = -7/2\\$

So z = 0 would lead to x = 8 and y = -7/2

Rearranging the items in alphabetical order gets us:

x = 8, y = -7/2, z = 0

We have one solution of (x,y,z) = (8, -7/2, 0)

Now let's say we picked z = 10

$x = -2z+8\\x = -2(10)+8\\x = -12\\\text{ and }\\y = z-7/2\\y = 10-7/2\\y = 13/2\\$

So we have x = -12, y = -13/2, z = 10

Another solution is (x,y,z) = (-12, 13/2, 10)

There's nothing special about z = 0 or z = 10. You can pick any two real numbers you want for z. Just be sure to recalculate the x and y values of course.

To verify each solution, you'll need to plug them back into the original equations formed by the original augmented matrix. After simplifying, you should get the same thing on both sides.

8 0

3 years ago