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xxTIMURxx [149]
3 years ago
6

... Helppppp!! will mark brainiest

Mathematics
1 answer:
Marianna [84]3 years ago
4 0

Answer: The answer is C

Step-by-step explanation:

C is the answer because the base is +2 and the coefficent  is  1/2 which is rise over run so you go up one space and to the right two spaces.          

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Please helppppp!!!!!
BaLLatris [955]

Answer:

C

Step-by-step explanation:

Let g(x) be y

From the (1,4)

When x =1 y =?

These (2x)^2

(2 × 1)^2 = 2 ^2 = 4

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RUDIKE [14]

The answer is reduction. This can be found by the size and number differences in the two quadrilaterals.

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How to put 63 over 84 in simplest form
julia-pushkina [17]
63 and 84 are both divisible by 21 63/21=3 84/21=4 the reduced form would be 3/4

8 0
4 years ago
Find the slope of (3,5) and (-2,1)
Sonbull [250]

m= y2-y1/x2-x1

m= 1-5/-2-3

m = -4/-5

m = 4/5

Thus the slope is 4/5

7 0
3 years ago
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wolverine [178]

Answer for the first box:  20

Answer for the second box: 6

====================================================

Explanation:

The notation a(1) represents the first term of the sequence given. The first term is 20, so thats why a(1) = 20.

The notation a(n) = a(n-1) + ____ means we add some term to the previous term to get the next. This process happens over and over. In this case, we add 6 to each term to get the next

  • term1 = 20
  • term2 = term1 + 6 = 20 + 6 = 26
  • term3 = term2 + 6 = 26 + 6 = 32
  • term4 = term3 + 6 = 32 + 6 = 38

and so on...

So we write a(n) = a(n-1) + 6

This value of 6 can be found by selecting any term and subtracting off the previous term, so we could say any of the following below

  • term2 - term1 = 26 - 20 = 6
  • term3 - term2 = 32 - 26 = 6
  • term4 - term3 = 38 - 26 = 6

That's how we end up with

\begin{cases}a(1) = 20\\a(n) = a(n-1) + 6\end{cases}

-------------------------

Extra info:

The common difference is d = 6 and the first term is a = 20

We can use these two facts to find the nth term a(n)

a(n) = a + d*(n-1)

a(n) = 20 + 6*(n-1)

a(n) = 20 + 6n - 6

a(n) = 6n + 14

Then note how plugging in n = 1 leads to

a(n) = 6*n + 14

a(1) = 6*1 + 14

a(1) = 20

and n = 2 leads to

a(n) = 6*n + 14

a(2) = 6*2 + 14

a(2) = 26

and so on

This allows us to plug in any value of n we want, without having to generate the previous terms before it. So for instance, you could plug in n = 100 to jump directly to the 100th term. If you use the recursive definition in the last section, then you have to generate the first 99 terms first before you can get to the 100th term.

7 0
3 years ago
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