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3 years ago

Please help me create a function,and if you can,please explain how

Mathematics

1 answer:

qaws [65]3 years ago

6 0

Current volume = 10 glns
Miles he can drive currently = 28miles/ gln * 10= 280miles
Cost of gas= 1.40/gln
Function= s /1.40 *28 +280= 28s/1.40 +280=20s + 280
Final answer would be 20s + 280

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I NEED HELP!!!

sp2606 [1]

Answer:

36 seconds

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

2,17,82,257,626,1297 next one please ?

In-s [12.5K]

The easy thing to do is notice that 1^4 = 1, 2^4 = 16, 3^4 = 81, and so on, so the sequence follows the rule $n^4+1$ . The next number would then be fourth power of 7 plus 1, or 2402.

And the harder way: Denote the n-th term in this sequence by $a_n$ , and denote the given sequence by $\{a_n\}_{n\ge1}$ .

Let $b_n$ denote the n-th term in the sequence of forward differences of $\{a_n\}$ , defined by

$b_n=a_{n+1}-a_n$

for n ≥ 1. That is, $\{b_n\}$ is the sequence with

$b_1=a_2-a_1=17-2=15$

$b_2=a_3-a_2=82-17=65$

$b_3=a_4-a_3=175$

$b_4=a_5-a_4=369$

$b_5=a_6-a_5=671$

and so on.

Next, let $c_n$ denote the n-th term of the differences of $\{b_n\}$ , i.e. for n ≥ 1,

$c_n=b_{n+1}-b_n$

so that

$c_1=b_2-b_1=65-15=50$

$c_2=110$

$c_3=194$

$c_4=302$

etc.

Again: let $d_n$ denote the n-th difference of $\{c_n\}$ :

$d_n=c_{n+1}-c_n$

$d_1=c_2-c_1=60$

$d_2=84$

$d_3=108$

etc.

One more time: let $e_n$ denote the n-th difference of $\{d_n\}$ :

$e_n=d_{n+1}-d_n$

$e_1=d_2-d_1=24$

$e_2=24$

etc.

The fact that these last differences are constant is a good sign that $e_n=24$ for all n ≥ 1. Assuming this, we would see that $\{d_n\}$ is an arithmetic sequence given recursively by

$\begin{cases}d_1=60\\d_{n+1}=d_n+24&\text{for }n>1\end{cases}$

and we can easily find the explicit rule:

$d_2=d_1+24$

$d_3=d_2+24=d_1+24\cdot2$

$d_4=d_3+24=d_1+24\cdot3$

and so on, up to

$d_n=d_1+24(n-1)$

$d_n=24n+36$

Use the same strategy to find a closed form for $\{c_n\}$ , then for $\{b_n\}$ , and finally $\{a_n\}$ .

$\begin{cases}c_1=50\\c_{n+1}=c_n+24n+36&\text{for }n>1\end{cases}$

$c_2=c_1+24\cdot1+36$

$c_3=c_2+24\cdot2+36=c_1+24(1+2)+36\cdot2$

$c_4=c_3+24\cdot3+36=c_1+24(1+2+3)+36\cdot3$

and so on, up to

$c_n=c_1+24(1+2+3+\cdots+(n-1))+36(n-1)$

Recall the formula for the sum of consecutive integers:

$1+2+3+\cdots+n=\displaystyle\sum_{k=1}^nk=\frac{n(n+1)}2$

$\implies c_n=c_1+\dfrac{24(n-1)n}2+36(n-1)$

$\implies c_n=12n^2+24n+14$

$\begin{cases}b_1=15\\b_{n+1}=b_n+12n^2+24n+14&\text{for }n>1\end{cases}$

$b_2=b_1+12\cdot1^2+24\cdot1+14$

$b_3=b_2+12\cdot2^2+24\cdot2+14=b_1+12(1^2+2^2)+24(1+2)+14\cdot2$

$b_4=b_3+12\cdot3^2+24\cdot3+14=b_1+12(1^2+2^2+3^2)+24(1+2+3)+14\cdot3$

and so on, up to

$b_n=b_1+12(1^2+2^2+3^2+\cdots+(n-1)^2)+24(1+2+3+\cdots+(n-1))+14(n-1)$

Recall the formula for the sum of squares of consecutive integers:

$1^2+2^2+3^2+\cdots+n^2=\displaystyle\sum_{k=1}^nk^2=\frac{n(n+1)(2n+1)}6$

$\implies b_n=15+\dfrac{12(n-1)n(2(n-1)+1)}6+\dfrac{24(n-1)n}2+14(n-1)$

$\implies b_n=4n^3+6n^2+4n+1$

$\begin{cases}a_1=2\\a_{n+1}=a_n+4n^3+6n^2+4n+1&\text{for }n>1\end{cases}$

$a_2=a_1+4\cdot1^3+6\cdot1^2+4\cdot1+1$

$a_3=a_2+4(1^3+2^3)+6(1^2+2^2)+4(1+2)+1\cdot2$

$a_4=a_3+4(1^3+2^3+3^3)+6(1^2+2^2+3^2)+4(1+2+3)+1\cdot3$

$\implies a_n=a_1+4\displaystyle\sum_{k=1}^3k^3+6\sum_{k=1}^3k^2+4\sum_{k=1}^3k+\sum_{k=1}^{n-1}1$

$\displaystyle\sum_{k=1}^nk^3=\frac{n^2(n+1)^2}4$

$\implies a_n=2+\dfrac{4(n-1)^2n^2}4+\dfrac{6(n-1)n(2n)}6+\dfrac{4(n-1)n}2+(n-1)$

$\implies a_n=n^4+1$

4 0

3 years ago

At the movies, we had 286 people buy tickets. The adult tickets were $12 and the children's tickets were $4. The total sold was

Elina [12.6K]

103 adults and 183 children's attended

Solution:

Let "a" be the number of adults attended

Let "c" be the number of children attended

Cost of 1 adult ticket = $ 12

Cost of 1 children ticket = $ 4

At the movies, we had 286 people buy tickets

Therefore,

number of adults attended + number of children attended = 286

a + c = 286 ---- eqn 1

The total sold was $1968

Therefore,

number of adults attended x Cost of 1 adult ticket + number of children attended x Cost of 1 children ticket = 1968

$a \times 12 + c \times 4 = 1968$

12a + 4c = 1968 ------- eqn 2

Let us solve eqn 1 and eqn 2

From eqn 1,

a = 286 - c -- eqn 3

Substitute eqn 3 in eqn 2

12(286 - c) + 4c = 1968

3432 -12c + 4c = 1968

-8c = 1968 - 3432

-8c = -1464

8c = 1464

Substitute c = 183 in eqn 3

a = 286 - 183

Thus 103 adults and 183 children's attended

6 0

4 years ago

What value of z makes the equation true? 0.2z -8 =0.4z -7

alukav5142 [94]

Answer:

X= -0.4545

Step-by-step explanation:

YOU HAVE TO SUBTRACT .2Z FROM .4Z WHICH GETS YOU .2 ON THE RIGHT SIDE

THAT WOULD MEAN YOU HAVE TO CLEAR OUT THE SEVEN BY ADDING IT ON BOTH SIDES GETTING -1

THEN YOU WOULD HAVE -1=.2Z

THEN DIVIDE BY .2 GETTING YOU A TOTAL OF -0.4545 WHEN YOU FINISH DIVIDING

4 0

3 years ago

A watering can dispenses water at the rate of 0.25 gallon per minute. The original volume of water in the can was 6 gallons. Whi

Nikolay [14]

Out of all of these, I believe your best bet is A :)

4 0

3 years ago

Read 2 more answers