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kupik [55]
3 years ago
15

Megan walks 1.3 miles to school and 1.3 miles home from school every day. How far does Megan walk after 5 days

Mathematics
1 answer:
elixir [45]3 years ago
5 0

Answer:

13 miles after 5 days

Step-by-step explanation:

1.3x2= 2.6 miles per day

2.6x5= 13.0 miles after 5 days

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1) Let f(x)=6x+6/x. Find the open intervals on which f is increasing (decreasing). Then determine the x-coordinates of all relat
brilliants [131]

Answer:

1) increasing on (-∞,-1] ∪ [1,∞), decreasing on [-1,0) ∪ (0,1]

x = -1 is local maximum, x = 1 is local minimum

2) increasing on [1,∞), decreasing on (-∞,0) ∪ (0,1]

x = 1 is absolute minimum

3) increasing on (-∞,0] ∪ [8,∞), decreasing on [0,4) ∪ (4,8]

x = 0 is local maximum, x = 8 is local minimum

4) increasing on [2,∞), decreasing on (-∞,2]

x = 2 is absolute minimum

5) increasing on the interval (0,4/9], decreasing on the interval [4/9,∞)

x = 0 is local minimum, x = 4/9 is absolute maximum

Step-by-step explanation:

To find minima and maxima the of the function, we must take the derivative and equalize it to zero to find the roots.

1) f(x) = 6x + 6/x

f\prime(x) = 6 - 6/x^2 = 0 and x \neq 0

So, the roots are x = -1 and x = 1

The function is increasing on the interval (-∞,-1] ∪ [1,∞)

The function is decreasing on the interval [-1,0) ∪ (0,1]

x = -1 is local maximum, x = 1 is local minimum.

2) f(x)=6-4/x+2/x^2

f\prime(x)=4/x^2-4/x^3=0 and x \neq 0

So the root is x = 1

The function is increasing on the interval [1,∞)

The function is decreasing on the interval (-∞,0) ∪ (0,1]

x = 1 is absolute minimum.

3) f(x) = 8x^2/(x-4)

f\prime(x) = (8x^2-64x)/(x-4)^2=0 and x \neq 4

So the roots are x = 0 and x = 8

The function is increasing on the interval (-∞,0] ∪ [8,∞)

The function is decreasing on the interval [0,4) ∪ (4,8]

x = 0 is local maximum, x = 8 is local minimum.

4) f(x)=6(x-2)^{2/3} +4=0

f\prime(x) = 4/(x-2)^{1/3} has no solution and x = 2 is crtitical point.

The function is increasing on the interval [2,∞)

The function is decreasing on the interval (-∞,2]

x = 2 is absolute minimum.

5) f(x)=8\sqrt x - 6x for x>0

f\prime(x) = (4/\sqrt x)-6 = 0

So the root is x = 4/9

The function is increasing on the interval (0,4/9]

The function is decreasing on the interval [4/9,∞)

x = 0 is local minimum, x = 4/9 is absolute maximum.

5 0
3 years ago
Help help help help help
Semenov [28]
Your answers are a and d hope this helps
4 0
3 years ago
Determine the solution to the system of equations below.<br> x-3y=1<br> 3x-5y=11
EastWind [94]
The answer you are looking for is (7, 2).
5 0
3 years ago
An object that weighs 5 pounds on earth will weigh 2 pounds on mercury. The Statue of Liberty weighs 2,250 pounds. What would it
murzikaleks [220]

2/5=x/2250

2250x2=4500

4500/5=900 pounds

7 0
3 years ago
Read 2 more answers
The balance on a credit card, that charges a 15.5%APR interest rate, over a 1 month period is given inthe following table:Days 1
Karo-lina-s [1.5K]

First step is to get the average daily balance.

Take the sum of each day's balances.

If a payment has been made, the sign will be negative.

From Days 1 - 5 :

Day 1 = $200

Day 2 = $200

...

Day 5 = $200

That's $200 x 5 days = $1000

From Days 6 - 20 :

Day 6 = $350

Day 7 = $350

...

Days 20 = $350

That's $350 x 15 days = $5250

From Days 21 - 30 :

Day 21 = $150

Day 22 = $150

...

Day 30 = $150

That's $150 x 10 days = $1500

A total of :

$1000 + $5250 + $1500 = $7750

Now divide this total to the number of days to get the ADB or Average Daily Balance.

\text{ADB}=\frac{7750}{30}=258.33

Next Step is to calculate for the finance charge using the formula :

\text{Charge}=\text{ADB}\times r\times d

where ADB is the average daily balance

r is the interest rate per day

d is the number of days or period given

From the problem, APR is 15.5%, it means that we need to divide the APR by 365 days.

So r = 0.155/365 = 0.000425

And we have d = 30 days

The charge will be :

Charge = 258.33 x 0.000425 x 30 = $3.29

The answer is $3.29

5 0
1 year ago
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