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AveGali [126]
3 years ago
13

A cereal box says that now it contains 30 ounces of cereal. Originally, it came with 18.5 ounces of cereal Calculate the percent

change and round to the
nearest percent

62% increase

38% increase

62% increase

38% decrease
Mathematics
1 answer:
MakcuM [25]3 years ago
7 0

Answer:62% increase

Step-by-step explanation:

old price of cereal box = 18.5 ounces

New price of cereal box =30 ounces

Percent change = (New Price - old price / Old price ) x 100

30 - 18.5 / 18.5 x 100

11.5/18.5 x100

0.621 x100

= 62% increase

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Tom Thomas took tennis lessons. He traveled 12 miles one way at an average cost of $0.42 per mile. His racket cost $94, balls co
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Racket : 94
balls : 12.50
warm ups : 34
shorts : 19.95
shoes : 44.50
total with tax : 204.95 + .05(204.95) = 204.95 + 10.25 = 215.20

20 per lesson for 12 lessons : 20 * 12 = 240
12 miles 1 way....thats 24 miles round trip, at 0.42 per mile : 24(0.42) = 
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Read 2 more answers
F(x) = x2 − x − ln(x) (a) find the interval on which f is increasing. (enter your answer using interval notation.) find the inte
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Answer:

(a) Decreasing on (0, 1) and increasing on (1, ∞)

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(c) No inflection point; concave up on (0, ∞)

Step-by-step explanation:

ƒ(x) = x² - x – lnx

(a) Intervals in which ƒ(x) is increasing and decreasing.

Step 1. Find the zeros of the first derivative of the function

ƒ'(x) = 2x – 1 - 1/x = 0

           2x² - x  -1 = 0

     ( x - 1) (2x + 1) = 0

         x = 1 or x = -½

We reject the negative root, because the argument of lnx cannot be negative.

There is one zero at (1, 0). This is your critical point.

Step 2. Apply the first derivative test.

Test all intervals to the left and to the right of the critical value to determine if the derivative is positive or negative.

(1) x = ½

ƒ'(½) = 2(½) - 1 - 1/(½) = 1 - 1 - 2 = -1

ƒ'(x) < 0 so the function is decreasing on (0, 1).

(2) x = 2

ƒ'(0) = 2(2) -1 – 1/2 = 4 - 1 – ½  = ⁵/₂

ƒ'(x) > 0 so the function is increasing on (1, ∞).

(b) Local extremum

ƒ(x) is decreasing when x < 1 and increasing when x >1.

Thus, (1, 0) is a local minimum, and ƒ(x) = 0 when x = 1.

(c) Inflection point

(1) Set the second derivative equal to zero

ƒ''(x) = 2 + 2/x² = 0

             x² + 2 = 0

                   x² = -2

There is no inflection point.

(2). Concavity

Apply the second derivative test on either side of the extremum.

\begin{array}{lccc}\text{Test} & x < 1 & x = 1 & x > 1\\\text{Sign of f''} & + & 0 & +\\\text{Concavity} & \text{up} & &\text{up}\\\end{array}

The function is concave up on (0, ∞).

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