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pogonyaev
2 years ago
9

PLEASE HELP

Mathematics
1 answer:
ZanzabumX [31]2 years ago
6 0

Answer:

p1=2(3+6)p1=18p2=2(30+60)p2=180p2/p1=180/18=10c. The perimeter will be 10 times the original.  More

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adult tickets to a movie cost $12 while children tickets cost $5. if a total of 500 tickets were sold and if $3550 was collected
svp [43]
The answer to the question

5 0
3 years ago
How many terms of the arithmetic sequence {1,22,43,64,85,…} will give a sum of 2332? Show all steps including the formulas used
MA_775_DIABLO [31]

There's a slight problem with your question, but we'll get to that...

Consecutive terms of the sequence are separated by a fixed difference of 21 (22 = 1 + 21, 43 = 22 + 21, 64 = 43 + 21, and so on), so the <em>n</em>-th term of the sequence, <em>a</em> (<em>n</em>), is given recursively by

• <em>a</em> (1) = 1

• <em>a</em> (<em>n</em>) = <em>a</em> (<em>n</em> - 1) + 21 … … … for <em>n</em> > 1

We can find the explicit rule for the sequence by iterative substitution:

<em>a</em> (2) = <em>a</em> (1) + 21

<em>a</em> (3) = <em>a</em> (2) + 21 = (<em>a</em> (1) + 21) + 21 = <em>a</em> (1) + 2×21

<em>a</em> (4) = <em>a</em> (3) + 21 = (<em>a</em> (1) + 2×21) + 21 = <em>a</em> (1) + 3×21

and so on, with the general pattern

<em>a</em> (<em>n</em>) = <em>a</em> (1) + 21 (<em>n</em> - 1) = 21<em>n</em> - 20

Now, we're told that the sum of some number <em>N</em> of terms in this sequence is 2332. In other words, the <em>N</em>-th partial sum of the sequence is

<em>a</em> (1) + <em>a</em> (2) + <em>a</em> (3) + … + <em>a</em> (<em>N</em> - 1) + <em>a</em> (<em>N</em>) = 2332

or more compactly,

\displaystyle\sum_{n=1}^N a(n) = 2332

It's important to note that <em>N</em> must be some positive integer.

Replace <em>a</em> (<em>n</em>) by the explicit rule:

\displaystyle\sum_{n=1}^N (21n-20) = 2332

Expand the sum on the left as

\displaystyle 21 \sum_{n=1}^N n-20\sum_{n=1}^N1 = 2332

and recall the formulas,

\displaystyle\sum_{k=1}^n1=\underbrace{1+1+\cdots+1}_{n\text{ times}}=n

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

So the sum of the first <em>N</em> terms of <em>a</em> (<em>n</em>) is such that

21 × <em>N</em> (<em>N</em> + 1)/2 - 20<em>N</em> = 2332

Solve for <em>N</em> :

21 (<em>N</em> ² + <em>N</em>) - 40<em>N</em> = 4664

21 <em>N</em> ² - 19 <em>N</em> - 4664 = 0

Now for the problem I mentioned at the start: this polynomial has no rational roots, and instead

<em>N</em> = (19 ± √392,137)/42 ≈ -14.45 or 15.36

so there is no positive integer <em>N</em> for which the first <em>N</em> terms of the sum add up to 2332.

4 0
2 years ago
Solve the system <br> {f (x) = 2x-1<br> {g (x) = x^2-4
Murrr4er [49]

Answer:

2 sets of possible solutions:

x=3, y = 5

and

x=-1, y = -3

Step-by-step explanation:

Using the graphical method, (see attached)

you can graph both equations and find their intersection points.

From the attached plot, you can see that the graphs intersect at (3,5) and (-1,-3)

Alternatively, you can solve this numerically by solving the following system of equations. You will get the same answer.

y = 2x + 1 ------------------- eq. (1)

y = x² - 4 ------------------- eq. (2)

4 0
3 years ago
Can someone help me please
goldenfox [79]

Answer:

1. \quad\dfrac{1}{k^{\frac{2}{3}}}\\\\2. \quad\sqrt[7]{x^5}\\\\3. \quad\dfrac{1}{\sqrt[5]{y^2}}

Step-by-step explanation:

The applicable rule is ...

  x^{\frac{m}{n}}=\sqrt[n]{x^m}

It works both ways, going from radicals to frational exponents and vice versa.

The particular power or root involved can be in either the numerator or the denominator. The transformation applies to the portion of the expression that is the power or root.

7 0
3 years ago
If Figure A is multiplied by a scale factor of .5, reflected over the x-axis, and translated 3 units left and 8 units down, whic
Marina CMI [18]

Answer:

A' has the same shape but not the same size as Figure A.

Step-by-step explanation:

6 0
2 years ago
Read 2 more answers
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