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daser333 [38]
3 years ago
13

Which ordered pair describes a translation of 6 units up up and 4 units left

Mathematics
1 answer:
Aleksandr-060686 [28]3 years ago
4 0

Answer:

The correct answer for this is (-4,6).

Step-by-step explanation:

As it can be seen from the attached graph image:

  • The <em>y-axis </em><em>increases in positive</em> values when we move in up direction i.e. we have +<em>y-axis </em>in upward direction.
  • The <em>y-axis </em><em>decreases and keeps on becoming more and more negative</em> values when we move in down direction i.e. we have -<em>y-axis </em>in downward direction.
  • The <em>x-axis </em><em>increases in positive</em> values when we move in right direction i.e. we have <em>+x-axis </em>in right direction.
  • The x<em>-axis </em><em>decreases and keeps on becoming more and more negative</em> values when we move in left direction i.e. we have <em>-x-axis </em>in left direction.

Now, according to question, the translation is performed 6 points up i.e. towards positive y-axis and 4 points towards left i.e. towards negative x-axis.

And, as per rule, the co-ordinates are represented as (x,y) where x is the x-co-ordinate and y is y-co-ordinate.

So, the answer here is (-4,6).

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A sequence consists of 20102010 terms. Each term after the first is 11 larger than the previous term. The sum of the 20102010 te
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You're considering a sequence of in which consecutive terms differ by 1, meaning

<em>a(n)</em> = <em>a</em> (<em>n</em> - 1) + 1

so <em>a(n)</em> is an arithmetic sequence. (I'm guessing 20102010 should actually be 2010, and 53075307 should be 5307, so 11 should probably be just 1.)

The sum of the first 2010 terms is 5307, or

\displaystyle\sum_{n=1}^{2010}a(n)=5307

Find the value of the first term in the sequence, <em>a</em>(1).

We can write <em>a(n)</em> in terms of <em>a</em>(1) by iterative substitution:

<em>a(n)</em> = <em>a</em>(<em>n</em> - 1) + 1

<em>a(n)</em> = (<em>a</em>(<em>n</em> - 2) + 1) + 1 = <em>a</em>(<em>n</em> - 2) + 2

<em>a(n)</em> = (<em>a</em>(<em>n</em> - 3) + 1) + 2 = <em>a</em>(<em>n</em> - 3) + 3

and so on, down to

<em>a(n)</em> = <em>a</em>(1) + <em>n</em> - 1

So the sum of the first 2010 terms is

\displaystyle\sum_{n=1}^{2010}a(n)=\sum_{n=1}^{2010}\left(a(1)+n-1\right)=(a(1)-1)\sum_{n=1}^{2010}1+\sum_{n=1}^{2010}n=5307

Recall that

\displaystyle\sum_{n=1}^N1=\underbrace{1+1+\cdots+1}_{N\text{ times}}=N

and

\displaystyle\sum_{n=1}^Nn=1+2+\cdots+N=\dfrac{N(N+1)}2

So we have

\displaystyle\sum_{n=1}^{2010}a(n)=2010(a(1)-1)+\frac{2010\cdot2011}2=5307

Solve for <em>a</em>(1) :

2010 (<em>a</em>(1) - 1) + 2,021,055 = 5307

2010 (<em>a</em>(1) - 1) = -2,015,748

<em>a</em>(1) - 1 = - 335,958/335

<em>a</em>(1) = - 335,623/335

Now, every second term, starting with <em>a</em>(1), differs by 2, so they form another arithmetic sequence <em>b(n)</em> given by

<em>b(n)</em> = <em>b</em>(<em>n</em> - 1) + 2

or, using the same method as before,

<em>b(n)</em> = <em>b</em>(1) + 2 (<em>n</em> - 1) = <em>a</em>(1) + 2<em>n</em> - 2

The sum of the 1005 terms in this sequence is

\displaystyle\sum_{n=1}^{1005}b(n)=(a(1)-2)\sum_{n=1}^{1005}1+2\sum_{n=1}^{1005}n

= (- 335,623/335 - 2)•1005 + 2•1005•1006/2

= 1146

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