All categories

3 years ago

Terry Wade's variable costs for his car

Mathematics

1 answer:

Pie3 years ago

8 0

Answer:

id guess if i were u but thats just me

Step-by-step explanation:

You might be interested in

What is the range of the following function: f(x)=-x^2+x^-1 if the domain is {-2, -1, 1}

postnew [5]

f(x)= -x^2 + x^-1

f(-2)= - (-2)^2 + (-2)^-1 = -4 - 1/2 = -9/2

f(-1)= - (-1)^2 + (-1)^-1 = -1 -1 = -2

f(1) = - (1)^2 + (1)^-1 = -1 +1 = 0

The range is { -9/2, -2, 0}

8 0

3 years ago

You are going on a road trip over a distance of 3000 kilometers with three friends. The car consumes 6 liters of gas per 100 kil

Effectus [21]

180 liters and $216.00 for gas.

3 0

3 years ago

How do you know if the factors will have negatives or positives in them?

harina [27]

When using factoring by grouping, the replacement terms will both have the same sign of the original middle term. If the last term is negative, then the signs of the factors will have one of each sign with the “stronger” number (the larger absolute value) matching the middle term. hope this helps

8 0

3 years ago

If there are 100 tickets to a concert and 200 fans who would like to go to the concert, each placing a slightly different value

Setler79 [48]

It's more efficient to hold a random drawing because the probability is higher.
$\frac{100}{200} = \frac{1}{2}$
For every 2 person, 1 person gets to go to the concert.

Answer is 1/2 aka 0.5 aka 50%

Hope this helps. - M

3 0

3 years ago

Match each function to its corresponding sequence.

givi [52]

The sequence and their functions are:

f(n) = f(n - 1) + 7 ⇒ 3, 10, 17, 24, 31.....
f(n) = f(n - 1) + 4n - 2 ⇒ 3, 9, 19, 33, 51.....
f(n) = 2[f(n - 1)] ⇒ 3, 6, 12, 24, 48.....

<h3>How to match the functions?</h3>

To do this, we simply set values for n and calculate the function values using the current value of n.

So, we have:

Function 1: f(n) = f(n - 1) + 6n - 5 where f(1) = 3

Let n = 2

f(2) = f(1) + 6(2) - 5 = 3 + 12 - 5 = 10

Let n = 3

f(3) = f(2) + 6(3) - 5 = 10 + 18 - 5 = 23

None of the sequence follows the pattern 3, 10, 23....

Function 2: f(n) = f(n - 1) + 2n - 1 where f(1) = 3

Let n = 2

f(2) = f(1) + 2(2) - 1 = 3 + 4 - 1 = 6

Let n = 3

f(3) = f(2) + 2(3) - 1 = 6 + 6 - 1 = 11

None of the sequence follows the pattern 3, 6, 11....

Function 3: f(n) = f(n - 1) + 6 where f(1) = 3

Let n = 2

f(2) = f(1) + 6 = 3 + 6 = 9

Let n = 3

f(3) = f(2) + 6 = 9 + 6 = 15

None of the sequence follows the pattern 3, 9, 15....

Function 4: f(n) = f(n - 1) + 7 where f(1) = 3

Let n = 2

f(2) = f(1) + 7 = 3 + 7 = 10

Let n = 3, 4 and 5

f(3) = f(2) + 7 = 10 + 7 = 17

f(4) = f(3) + 7 = 17 + 7 = 24

f(5) = f(4) + 7 = 24 + 7 = 31

So, we have:

f(n) = f(n - 1) + 7 ⇒ 3, 10, 17, 24, 31.....

Function 5: f(n) = f(n - 1) + 4n - 2 where f(1) = 3

Let n = 2

f(2) = f(1) + 4(2) - 2 = 3 + 8 - 2 = 9

Let n = 3, 4 and 5

f(3) = f(2) + 4(3) - 2 = 9 + 12 - 2 = 19

f(4) = f(3) + 4(4) - 2 = 19 + 16 - 2 = 33

f(5) = f(4) + 4(5) - 2 = 33 + 20 - 2 = 51

So, we have:

f(n) = f(n - 1) + 4n - 2 ⇒ 3, 9, 19, 33, 51.....

Function 6: f(n) = 2[f(n - 1)] where f(1) = 3

Let n = 2

f(2) = 2 * f(1) = 2 * 3 = 6

Let n = 3, 4 and 5

f(3) = 2 * f(2) = 2 * 6 = 12

f(4) = 2 * f(3) = 2 * 12 = 24

f(5) = 2 * f(4) = 2 * 24 = 48

So, we have:

f(n) = 2[f(n - 1)] ⇒ 3, 6, 12, 24, 48.....