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posledela
3 years ago
14

Y=8x+3 what is the ordered pair is a solution of the equation?

Mathematics
1 answer:
harkovskaia [24]3 years ago
8 0
Here are three ordered pairs, hope this helps :)

(0,3) , (1, 11) , (2, 19)


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Mr. Harris is adding tiles to his patio area. On Monday, he tiled
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What is the elimination of 20x+5y=120 and 10x+7.5=80
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20x+5y=120
10x+7.5y=80    multiply this one by 2
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20x+5y=120
20x+15y=160   subtract 

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        -10y=-40    divide both sides by -10 
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             y=4         

20x+5(4)=120   plug 4 in for y in the first equation 
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A car traveling on the taconic parkway travel 84 miles in two hours. What is the cars speed (a special type of rate) in miles pe
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A caircular fountian has a circumference of 21 feet. Which expression can be used to find the diameter of the fountian in feet?​
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What is the nth term rule of the quadratic sequence below? 4,15,30,49,72,99,130,...
navik [9.2K]

Answer:

The nth term rule of the quadratic sequence is 2n^{2} +5n-3.

Step-by-step explanation:

We are given the following quadratic sequence below;

4, 15, 30, 49, 72, 99, 130,...

As we know that the formula for the nth term of the quadratic sequence is given by =  an^{2}+bn+c

Firstly, we will find the difference between the term of the given sequence;

1st difference of the given sequence;

(2nd term - 1st term), (3rd term - 2nd term), (4th term - 3rd term), (5th term - 4th term), (6th term - 5th term), (7th term - 6th term)

= (15 - 4), (30 - 15), (49 - 30), (72 - 49), (99 - 72), (130 - 99),....

= (11, 15, 19, 23, 27, 31,.....)

Now, we will find the second difference of the given sequence, i.e;

= (15 - 11), (19 - 15), (23 - 19), (27 - 23), (31 - 27),....

= (4, 4, 4, 4, 4)

Since the differences are same now, so to find the value of a we have to divide the value of second difference by 2, i.e;

The value of a  =  \dfrac{4}{2} = 2

SO, the first term of the nth term rule equation is an^{2} = 2n^{2}.

Now, in the term 2n^{2}, put the value of n = 1, 2, 3, 4 and 5 and then form the sequence, i.e;

If n = 1, then 2n^{2} = 2 \times (1)^{2} = 2

If n = 2, then 2n^{2} = 2 \times (2)^{2} = 8

If n = 3, then 2n^{2} = 2 \times (3)^{2} = 18

If n = 4, then 2n^{2} = 2 \times (4)^{2} = 32

If n = 5, then 2n^{2} = 2 \times (5)^{2} = 50

SO, the sequence formed is (2, 8, 18, 32, 50).

Now, find the difference of this sequence and the original quadratic sequence, i.e;

= (4 - 2), (15 - 8), (30 - 18), (49 - 32), (72 - 50)

= (2, 7, 12, 17, 22)

Now, as we can see that the above sequence resembles the general form of (5n - 3) because:

If we put n = 1, then (5n - 3) = 5 - 3 = 2

If we put n = 2, then (5n - 3) = 10 - 3 = 7 and so on....

From this, we concluded that the value of b and c are 5 and (-3) respectively.

Hence, the nth term rule for the given quadratic sequence is 2n^{2} +5n-3.

7 0
3 years ago
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