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Alexus [3.1K]
3 years ago
7

12 less than 7 times a number x

Mathematics
1 answer:
erma4kov [3.2K]3 years ago
8 0
The answer is 5x because when you do the math 12 subtracted by 7 equals 5 then you add the x at the end.
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How do the graphs of the functions f(x)=<img src="https://tex.z-dn.net/?f=%28%5Cfrac%7B3%7D%7B2%7D%29%5Ex" id="TexFormula1" titl
grin007 [14]

Step-by-step explanation:

f(x) = (3/2)ˣ

g(x) = (2/3)ˣ

These are examples of exponential equations:

y = a bˣ

If b > 1, the equation is exponential growth.

If 0 < b < 1, the equation is exponential decay.

So f(x) is an example of exponential growth, and g(x) is an example of exponential decay.

Also, 2/3 is the inverse of 3/2, so:

g(x) = (3/2)^(-x)

So more specifically, f(x) and g(x) are reflections of each other across the y-axis.

5 0
3 years ago
Unit activity: exponential and logarithmic functions
nirvana33 [79]

We will conclude that:

  • The domain of the exponential function is equal to the range of the logarithmic function.
  • The domain of the logarithmic function is equal to the range of the exponential function.

<h3>Comparing the domains and ranges.</h3>

Let's study the two functions.

The exponential function is given by:

f(x) = A*e^x

You can input any value of x in that function, so the domain is the set of all real numbers. And the value of x can't change the sign of the function, so, for example, if A is positive, the range will be:

y > 0.

For the logarithmic function we have:

g(x) = A*ln(x).

As you may know, only positive values can be used as arguments for the logarithmic function, while we know that:

\lim_{x \to \infty} ln(x) = \infty \\\\ \lim_{x \to0} ln(x) = -\infty

So the range of the logarithmic function is the set of all real numbers.

<h3>So what we can conclude?</h3>
  • The domain of the exponential function is equal to the range of the logarithmic function.
  • The domain of the logarithmic function is equal to the range of the exponential function.

If you want to learn more about domains and ranges, you can read:

brainly.com/question/10197594

3 0
1 year ago
Solve the inequality. Show your work. |4r + 8| ≥ 32
S_A_V [24]

|4r + 8| ≥ 32

Split this expression into two expressions:

First ⇒ 4r + 8 ≥ 32 and second ⇒ 4r + 8 ≤ - 32

---

First expression: 4r + 8 ≥ 32

Subtract 8 from both sides.

4r ≥ 24

Divide both sides by 4.

r ≥ 6

---

Second expression: 4r + 8 ≤ - 32

Subtract 8 from both sides.

4r ≤ -40

Divide both sides by 4.

r ≤ -10

---

Your answer is \boxed {r \geq 6~or~ r \leq  -10}

7 0
3 years ago
Read 2 more answers
Someone help plzzzzz
slavikrds [6]

Answer:

8.5

Step-by-step explanation:

4 0
3 years ago
Emmett gathered some data to compare the monthly cost of renting a one-bedroom apartment in Dallas and Austin. The data collecte
Goryan [66]

Answer:

The correct options are;

a) The city's data has the higher median monthly apartment cost of the two cities by $155

b) The city's data had the greater variability among monthly apartment costs because the standard deviation was $49.13 more than that of the other set of data

Step-by-step explanation:

The given data for the cost of renting a one-bedroom apartment in Dallas is presented as follows;

$994, $1,322, $1,075, $1,189, $1,172, $1,465, $1,215, $930, $1,090, $1,288

Which can be arranged in increasing order and analyzed to find the interquartile range, mean and standard deviation using Microsoft Excel as follows;

$930, $994, $1,075, $1,090, $1,172, $1,189, $1,215, $1,288, $1,322, and $ 1,465

The first quartile, Q₁ = $1,054.75

The third quartile, Q₃ = $1,296.5

The interquartile range = Q₃ - Q₁ = $241.75

The median = $1,180.5

The average monthly cost = $1,174

The standard deviation of the sample = $159.9597

The given data for the cost of renting a one-bedroom apartment in Austin is presented in increasing order using Microsoft Excel as follows;

$900, $950, $1,100, $1,250, $1,296, $1,375, $1,389, $1,400, $1,450, $1,495

The interquartile range, mean and standard deviation are found using Microsoft Excel as follows;

The first quartile, Q₁ = $1,062.5

The third quartile, Q₃ = $1,412.5

The interquartile range = Q₃ - Q₁ = $1,412.5 - $1,062.5 = $350

The median = $1,335.5

The average monthly cost = $1,260.5

The standard deviation of the sample = $209.0945

The difference in the median cost of the two cities is $1,335.5 - $1,180.5 = $155

Therefore, the Austin's city data has the higher median monthly apartment cost of the two cities by $155

b) The difference in the sample standard deviation of the two cities is $209.0945 - $159.9597 = $49.13476

Therefore, the Austin city data had the greater variability among monthly apartment costs because the standard deviation was $49.13 more than that of the other set of data.

7 0
3 years ago
Read 2 more answers
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