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solong [7]
1 year ago
15

A gallon of gas has dropped to 2.85 if yesterday was 2.90 what is the percentage of decrease

Mathematics
1 answer:
vodomira [7]1 year ago
3 0

Answer: 5%

Step-by-step explanation:

if it was 2.90 yesterday and 2.85 today there is a 5 cent decrease. to find the percent you convert .05 to a percent which is 5%

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Simplify the following rational expression and express in expanded form.
VladimirAG [237]

Answer:

D m =  -\frac{3}{8}

Step-by-step explanation:

factor the expressions in the denominator and the numerator to simplify the expression:

=> \frac{2(-2m + 1)(2m - 1)}{2(2m - 1)(8m + 3)}

=> -\frac{2m + 1}{8m + 3}

to make a fraction undefined, the numerator should be 0, thus, we substitute the values of m from the options into the denominator to make the denominator equals to 0:

=> -\frac{2m + 1}{8(-\frac{3}{8} ) + 3} = -\frac{2m + 1}{-3 + 3} = -\frac{2m + 1}{0}

in this case, the values of m from option D make the denominator of the fraction equals 0.

6 0
3 years ago
The domain of startfraction f over g endfraction f g consists of numbers x for which g left parenthesis x right parenthesisg(x)
Misha Larkins [42]

The domain of f/g consists of numbers x for which g(x) cannot equal 0 that are in the domains of both f and g.

 

Let’s take this equation as an example:

If f(x) = 3x - 5 and g(x) = square root of x-5, what is the domain of (f/g)x. 


For x to be in the domain of (f/g)(x), it must be in the domain of f and in the domain of g since (f/g)(x) = f(x)/g(x). We also need to ensure that g(x) is not zero since f(x) is divided by g(x). Therefore, there are 3 conditions.

 

x must be in the domain of f: f(x) = 3x -5 are in the domain of x and all real numbers x.

x must be in the domain of g: g(x) = √(x - 5) so x - 5 ≥ 0 so x ≥ 5.

g(x) can not be 0: g(x) = √(x - 5) and √(x - 5) = 0 gives x = 5 so x ≠ 5.

 

Hence to x x ≥ 5 and x ≠ 5 so the domain of (f/g)(x) is all x satisfying x > 5.

 

Thus, satisfying <span>satisfy all three conditions, x x ≥ 5 and x ≠ 5 so the domain of (f/g)(x) is all x satisfying x > 5.</span>

5 0
3 years ago
Answers for part one and part two please!
il63 [147K]

When a customer has a 6 pound Chihuahua, the cost that will be charged is $5.00.

<h3>How to calculate the cost?</h3>

a. If a customer has a 6 pound Chihuahua, how much would you charge?

It should be noted that from the information given, for dogs that weigh 0 to 15 pounds, the amount charged is $5.00.

b. If a customer has a 65 pound Labrador, how much would you charge?

It should be noted that for dogs over 45 pounds, the amount that's charged is $9.00

There, the amount charged will be $9.00.

Learn more about cost on:

brainly.com/question/25109150

#SPJ1

3 0
1 year ago
PLEASE HELP WILL GIVE BRAINLIEST AND LOTS OF POINTS
Tasya [4]

The tiles go in this order: 4,3,2,1

Hope this helps!

3 0
2 years ago
Read 2 more answers
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>

4 0
3 years ago
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