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NARA [144]
4 years ago
11

8 - 3x = 14 what is the value of x

Mathematics
2 answers:
wel4 years ago
3 0

Answer: it is 2

Step-by-step explanation: because 7 x 2 is 14

Karo-lina-s [1.5K]4 years ago
3 0

Answer:

x= -2

Step-by-step explanation:

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m = (7-(-2))/(3-1)

m = 9/2

b = -6.5

y = 9/2x -6.5

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500 tickets were sold for a game for a total of $862.50. If adult tickets sold for $2.00 and children's tickets sold for $1.50,
Alexxandr [17]
You can work these in your head if you consider the revenue generated by the least contributor (children's tickets) and the difference in revenue between that and the larger contributor ($2.00 -1.50 = 0.50).

If all were children's tickets, the revenue would be 500*$1.50 = $750.00. The actual revenue exceeded that amount by $862.50 -750.00 = 112.50. This difference in revenue is made up by the sale of $112.50/$0.50 = 225 adult tickets. Then the number of children's tickets is 500 -225 = 275.

225 adult tickets were sold
275 children's tickets were sold.

_____
If you need an equation, you can write an equation using a variable for the number of adult tickets sold (the largest contributor).
.. (500 -a)*1.50 +a*2.00 = 862.50
.. 0.50a = (862.50 -750.00) . . . . . does this look familiar, yet?
.. a = 112.50/0.50 = 225
6 0
4 years ago
Write out the first four terms of the series to show how the series starts. Then find the sum of the series or show that it dive
Nostrana [21]

Answer:

The first four terms of the series are

(9+3),(\frac97+\frac35),(\frac9{7^2}+\frac3{5^2}),(\frac9{7^3}+\frac3{5^3})

\sum_{n=0}^\infty \frac9{7^n}+\frac{3}{5^n} = 14.25

Step-by-step explanation:

We know that

Sum of convergent series is also a convergent series.

We know that,

\sum_{k=0}^\infty a(r)^k

If the common ratio of a sequence |r| <1 then it is a convergent series.

The sum of the series is \sum_{k=0}^\infty a(r)^k=\frac{a}{1-r}

Given series,

\sum_{n=0}^\infty \frac9{7^n}+\frac{3}{5^n}

=(9+3)+(\frac97+\frac35)+(\frac9{7^2}+\frac3{5^2})+(\frac9{7^3}+\frac3{5^3})+.......

The first four terms of the series are

(9+3),(\frac97+\frac35),(\frac9{7^2}+\frac3{5^2}),(\frac9{7^3}+\frac3{5^3})

Let

S_n=\sum_{n=0}^\infty \frac{9}{7^n}    and     t_n=\sum_{n=0}^\infty \frac{3}{5^n}

Now for S_n,

S_n=9+\frac97+\frac{9}{7^2}+\frac9{7^3}+.......

    =\sum_{n=0}^\infty9(\frac 17)^n

It is a geometric series.

The common ratio of S_n is \frac17

The sum of the series

S_n=\sum_{n=0}^\infty \frac{9}{7^n}

    =\frac{9}{1-\frac17}

    =\frac{9}{\frac67}

    =\frac{9\times 7}{6}

    =10.5

Now for t_n

t_n= 3+\frac35+\frac{3}{5^2}+\frac3{5^3}+.......

    =\sum_{n=0}^\infty3(\frac 15)^n

It is a geometric series.

The common ratio of t_n is \frac15

The sum of the series

t_n=\sum_{n=0}^\infty \frac{3}{5^n}

    =\frac{3}{1-\frac15}

    =\frac{3}{\frac45}

    =\frac{3\times 5}{4}

    =3.75

The sum of the series is \sum_{n=0}^\infty \frac9{7^n}+\frac{3}{5^n}

                                        = S_n+t_n

                                       =10.5+3.75

                                       =14.25

4 0
4 years ago
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