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ale4655 [162]
3 years ago
7

What is "A number decreased by 4 is less than 14" written as an inequality?

Mathematics
1 answer:
navik [9.2K]3 years ago
6 0

Answer:

"A number decreased by 4 is less than 14" written as an inequality is x - 4 < 14

Step-by-step explanation:

First: Figure out what a number decreased by 4 is.

Well, a number decreased by 4, how would that look like? To find out, we should look for some key words in the term. So, I will bold out the keywords in the term this time:

A number decreased by 4

That word 'decreased' gives us a hint. Out of the four basic operations (addition, subtraction, multiplication and division), 'decreased' fits into the subtraction category best. So we can replace 'decreased' with a minus sign. Since we don't know the specific value of 'A number', we will assign it a variable, the variable x. 4 is just 4 so we can leave that number alone. So the first term would look like this:

A number decreased by 4 = x - 4

Second: Since we know what the first term is, we need to find out what the second term is and the inequality sign.

This step is quite easy. Let's tackle the inequality sign first:

A number decreased by 4 is less than 14

The only words that would describe an inequality sign are the words: 'less than'. The less than sign looks like this: <

Fourteen [just like four] is just a number so we don't need to change that

So our final product is this: x - 4 < 14

Thank you for reading.

Topic: Inequalities

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Which expression represents a negative number? A.-4.5x-22.1 B.4.5x22.1 C.-45x2.21 D.-45x-2.21
nekit [7.7K]

Hey there!

-4.5 * (-22.1)

= -4.5(-22.1)

= 99.45


4.5 * 22.1 
= 99.45


-45 * 2.21

= -99.45

-45(2.21)

= -99


Therefore, your answer is:

Option C.


Good luck on your assignment & enjoy your day!


~Amphitrite1040:)


6 0
3 years ago
Consider the equation below. (If an answer does not exist, enter DNE.) f(x) = x4 ln(x) (a) Find the interval on which f is incre
Ainat [17]

Answer: (a) Interval where f is increasing: (0.78,+∞);

Interval where f is decreasing: (0,0.78);

(b) Local minimum: (0.78, - 0.09)

(c) Inflection point: (0.56,-0.06)

Interval concave up: (0.56,+∞)

Interval concave down: (0,0.56)

Step-by-step explanation:

(a) To determine the interval where function f is increasing or decreasing, first derive the function:

f'(x) = \frac{d}{dx}[x^{4}ln(x)]

Using the product rule of derivative, which is: [u(x).v(x)]' = u'(x)v(x) + u(x).v'(x),

you have:

f'(x) = 4x^{3}ln(x) + x_{4}.\frac{1}{x}

f'(x) = 4x^{3}ln(x) + x^{3}

f'(x) = x^{3}[4ln(x) + 1]

Now, find the critical points: f'(x) = 0

x^{3}[4ln(x) + 1] = 0

x^{3} = 0

x = 0

and

4ln(x) + 1 = 0

ln(x) = \frac{-1}{4}

x = e^{\frac{-1}{4} }

x = 0.78

To determine the interval where f(x) is positive (increasing) or negative (decreasing), evaluate the function at each interval:

interval                 x-value                      f'(x)                       result

0<x<0.78                 0.5                 f'(0.5) = -0.22            decreasing

x>0.78                       1                         f'(1) = 1                  increasing

With the table, it can be concluded that in the interval (0,0.78) the function is decreasing while in the interval (0.78, +∞), f is increasing.

Note: As it is a natural logarithm function, there are no negative x-values.

(b) A extremum point (maximum or minimum) is found where f is defined and f' changes signs. In this case:

  • Between 0 and 0.78, the function decreases and at point and it is defined at point 0.78;
  • After 0.78, it increase (has a change of sign) and f is also defined;

Then, x=0.78 is a point of minimum and its y-value is:

f(x) = x^{4}ln(x)

f(0.78) = 0.78^{4}ln(0.78)

f(0.78) = - 0.092

The point of <u>minimum</u> is (0.78, - 0.092)

(c) To determine the inflection point (IP), calculate the second derivative of the function and solve for x:

f"(x) = \frac{d^{2}}{dx^{2}} [x^{3}[4ln(x) + 1]]

f"(x) = 3x^{2}[4ln(x) + 1] + 4x^{2}

f"(x) = x^{2}[12ln(x) + 7]

x^{2}[12ln(x) + 7] = 0

x^{2} = 0\\x = 0

and

12ln(x) + 7 = 0\\ln(x) = \frac{-7}{12} \\x = e^{\frac{-7}{12} }\\x = 0.56

Substituing x in the function:

f(x) = x^{4}ln(x)

f(0.56) = 0.56^{4} ln(0.56)

f(0.56) = - 0.06

The <u>inflection point</u> will be: (0.56, - 0.06)

In a function, the concave is down when f"(x) < 0 and up when f"(x) > 0, adn knowing that the critical points for that derivative are 0 and 0.56:

f"(x) =  x^{2}[12ln(x) + 7]

f"(0.1) = 0.1^{2}[12ln(0.1)+7]

f"(0.1) = - 0.21, i.e. <u>Concave</u> is <u>DOWN.</u>

f"(0.7) = 0.7^{2}[12ln(0.7)+7]

f"(0.7) = + 1.33, i.e. <u>Concave</u> is <u>UP.</u>

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