All categories

3 years ago

Can someone help me please

Mathematics

1 answer:

artcher [175]3 years ago

6 0

Answer:

$22.75

Step-by-step explanation:

C = $30.25 - ($0.25/min)x, where x is the number of minutes spent on calls

Here, 30 minutes of calls result in this amount of $ left on the card:

C(30) = $30.25 - ($0.25/min)(30 min) = $22.75

You might be interested in

List x1, x2, x3, x4 where xi is the left endpoint of the four equal intervals used to estimate the area under the curve of f(x)

n200080 [17]

Answer:

Option A. 4, 4.5, 5, 5.5

Step-by-step explanation:

Left point: a=x=4

Right point: b=x=6

Range: r=b-a→r=6-4→r=2

Width of each of the four equal intervals: w=2/4→w=0.5

The first left endpoint is x1=a→x1=4

The second left endpoint is x2=x1+w→x2=4+0.5→x2=4.5

The third left endpoint is x3=x2+w→x3=4.5+0.5→x3=5

The fourth left endpoint is x4=x3+w→x4=5+0.5→x4=5.5

Then, the list is: x1, x2, x3, x4 = 4, 4.5, 5, 5.5

4 0

3 years ago

Write four solution of the equation=2x+y=7

Fantom [35]

Step-by-step explanation:

x = 1, y = 5

x = 2, y = 3

x =3, y= 1

x=4 y=-1

4 0

3 years ago

What is meant mathematically when we say we have an "even" or "odd" function?

Klio2033 [76]

I a function is even the leading term will have an even exponent. If a function is odd than the leading terms exponent is odd.

an even function has a parabola looking graph, while an odd function has min and max graph

6 0

3 years ago

Need help on number 9 plz help

Tcecarenko [31]

27/5 - 10/3
81/15 - 50/15
31/15
2 1/15

8 0

3 years ago

Read 2 more answers

a car purchased for $34,000 is expected to lose value, or depreciate, at a rate of 6% per year.using x for years and y for the v

Vika [28.1K]

Answer:

$y = 34000(1-0.06) ^ t$

After 7.40 years it will be worth less than 21500

Step-by-step explanation:

This problem is solved using a compound interest function.

This function has the following formula:

$y = P(1-n) ^ t$

Where:

P is the initial price = $ 34,000

n is the depreciation rate = 0.06

t is the elapsed time

The equation that models this situation is:

$y = 34000(1-0.06) ^ t$

Now we want to know after how many years the car is worth less than $ 21500.

Then we do y = $ 21,500. and we clear t.

$21500 = 34000(1-0.06) ^ t\\\\log(21500/34000) = tlog(1-0.06)\\\\t = \frac{log(21500/34000)}{log(1-0.06)}\\\\t = 7.40\ years.$

After 7.40 years it will be worth less than 21500

3 0

3 years ago