How can graphics and/or statistics be used to misrepresent data? where have you seen this done?

Slav-nsk [51]

First you need to combine like terms.

9x + 11y + 13

A coefficient is the number that comes before the variable (which is y).

The coefficient for y is 11.

Constants are the numbers that do not have variables attached to them.

The constant is 13.

Hope this helps!

5 0

3 years ago

Read 2 more answers

Show work please!!! Solve the equation by completing the square. If necessary, round to the nearest hundredth.

weeeeeb [17]

Answer:

1. B 2. C

Step-by-step explanation:

1. x^2-18x+(18/2)^2=19+(18/2)^2

(x-18/2)^2=100

(x-9)^2=100

x1= -1 x2= 19

2. x^2=81

x=+-9

x1=-9 x2=9

5 0

3 years ago

Dennis spent $5 less on lunch than he spent on dinner. Dennis spent at most $12 on dinner. Write and solve an inequality to find

Vinil7 [7]

12>x-5 is the answer your looking for. Make sure to put a line under the more than sign, my computer wouldn't allow me to plug that part in.

7 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 n-th term of the sequence, a (n), is given recursively by

• a (1) = 1

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

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

a (2) = a (1) + 21

a (3) = a (2) + 21 = (a (1) + 21) + 21 = a (1) + 2×21

a (4) = a (3) + 21 = (a (1) + 2×21) + 21 = a (1) + 3×21

and so on, with the general pattern

a (n) = a (1) + 21 (n - 1) = 21n - 20

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

a (1) + a (2) + a (3) + … + a (N - 1) + a (N) = 2332

or more compactly,

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

It's important to note that N must be some positive integer.

Replace a (n) 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 N terms of a (n) is such that

21 × N (N + 1)/2 - 20N = 2332

Solve for N :

21 (N ² + N) - 40N = 4664

21 N ² - 19 N - 4664 = 0

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

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

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

4 0

2 years ago

A = 9 m, b = 4 m and c = 13 m for the triangle shown below.

maw [93]

The answer is 9 shown below a b c triangle shown below ⬇️

7 0

2 years ago