Help please I need it please please help me in it

Solve -11 - x + = - 5 (2x-3) + 7

garri49 [273]

Answer:

Let's see what to do buddy...

Step-by-step explanation:

_________________________________

$- 11 - x = - 5(2x - 3) + 7$

$- 11 - x = - 10x + 15 + 7$

$- 11 - x = - 10x + 22$

Subtract the sides of the equation plus 11 :

$- x = - 10x + 22 + 11$

$- x = - 10x + 33$

Subtract the sides of the equation plus 10x

$10x - x = 10x - 10x + 33$

$9x = 33$

Divided the sides of the equation by 9 

$\frac{9}{9}x = \frac{33}{9} \\$

$x = \frac{33}{9} = \frac{3 \times 11}{3 \times 3} = \frac{11}{3} \\$

And we're done.

Thanks for watching buddy good luck.

♥️♥️♥️♥️♥️

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

3 years ago

Else's dad is a policeman. One afternoon her dad stopped 19 people for speeding. This was 12 more than the previous day and 7 mo

vekshin1

I'm pretty sure that the answer is 71, hope it helped!

6 0

3 years ago

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A coin is thrown at random into the red triangle below.

Yakvenalex [24]

it is impossible

Step-by-step explanation:

8 0

3 years ago

To drive to work, Dave has to drive 20 miles east and then 15 miles north. If there were a direct road going northeast, how many

suter [353]

By the Pythagorean theorem:

$x= \sqrt{20^2+15^2}= \sqrt{400+225}= \sqrt{625}=25 \ miles$

7 0

3 years ago