Answer:
For, "x" greater than 34, the perimeter of the picture frame greater than 152 inches
Solution:
Let "x" be the width of the frame
Given that, The length of a picture frame is 8 inches more than the width
Therefore,
Length = width + 8
Length = x + 8
The perimeter of rectangle is given by formula:
Perimeter = 2(length + width)
Substituting the values we get,
Perimeter = 2(x + 8 + x)
Perimeter = 2(2x + 8)
Perimeter = 4x + 16
The perimeter of the picture frame greater than 152 inches
Perimeter > 152
Therefore, for, "x" greater than 34, the perimeter of the picture frame greater than 152 inches
Step-by-step explanation:
search it up
Start off by distributing the numbers into the parentheses:
5(-3x - 2) - (x - 3) = -4(4x + 5) + 13
-15x - 10 - (x - 3) = -16x - 20 + 13
(Note: It's super important to be careful when opening up negative parentheses! -(x-3) is not just - x - 3, it is actually -x + 3 since the negative is distributed in every number!)
-15x - 10 - x + 3 = -16x - 20 + 13
-16x - 7 = -16x - 7
-16x = -16x
0 = 0
There is an infinite number of solutions in this equation.
(When you get 0=0 when solving for a variable, that means that said variable will have infinite solutions, that is, any number plugged into the equation will work)
Your answer would be 15.6
The sum of the given series can be found by simplification of the number
of terms in the series.
- A is approximately <u>2020.022</u>
Reasons:
The given sequence is presented as follows;
A = 1011 + 337 + 337/2 + 1011/10 + 337/5 + ... + 1/2021
Therefore;
The n + 1 th term of the sequence, 1, 3, 6, 10, 15, ..., 2021 is given as follows;
Therefore, for the last term we have;
2 × 2043231 = n² + 3·n + 2
Which gives;
n² + 3·n + 2 - 2 × 2043231 = n² + 3·n - 4086460 = 0
Which gives, the number of terms, n = 2020
![\displaystyle \frac{A}{2} = \mathbf{ 1011 \cdot \left(\frac{1}{2} +\frac{1}{6} + \frac{1}{12}+...+\frac{1}{4086460} \right)}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7BA%7D%7B2%7D%20%20%3D%20%5Cmathbf%7B%201011%20%5Ccdot%20%20%5Cleft%28%5Cfrac%7B1%7D%7B2%7D%20%2B%5Cfrac%7B1%7D%7B6%7D%20%2B%20%5Cfrac%7B1%7D%7B12%7D%2B...%2B%5Cfrac%7B1%7D%7B4086460%7D%20%20%5Cright%29%7D)
![\displaystyle \frac{A}{2} = 1011 \cdot \left(1 - \frac{1}{2} +\frac{1}{2} - \frac{1}{3} + \frac{1}{3}- \frac{1}{4} +...+\frac{1}{2021}-\frac{1}{2022} \right)](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7BA%7D%7B2%7D%20%20%3D%201011%20%5Ccdot%20%20%5Cleft%281%20-%20%5Cfrac%7B1%7D%7B2%7D%20%2B%5Cfrac%7B1%7D%7B2%7D%20-%20%20%5Cfrac%7B1%7D%7B3%7D%20%2B%20%5Cfrac%7B1%7D%7B3%7D-%20%5Cfrac%7B1%7D%7B4%7D%20%2B...%2B%5Cfrac%7B1%7D%7B2021%7D-%5Cfrac%7B1%7D%7B2022%7D%20%20%5Cright%29)
Which gives;
![\displaystyle \frac{A}{2} = 1011 \cdot \left(1 - \frac{1}{2022} \right)](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7BA%7D%7B2%7D%20%20%3D%201011%20%5Ccdot%20%20%5Cleft%281%20-%20%5Cfrac%7B1%7D%7B2022%7D%20%20%5Cright%29)
![\displaystyle A = 2 \times 1011 \cdot \left(1 - \frac{1}{2022} \right) = \frac{1032231}{511} \approx \mathbf{2020.022}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%20A%20%3D%202%20%5Ctimes%201011%20%5Ccdot%20%20%5Cleft%281%20-%20%5Cfrac%7B1%7D%7B2022%7D%20%20%5Cright%29%20%3D%20%5Cfrac%7B1032231%7D%7B511%7D%20%5Capprox%20%5Cmathbf%7B2020.022%7D)
Learn more about the sum of a series here:
brainly.com/question/190295
Answer:
3 ,5 , and 2
Step-by-step explanation:
Hope it helps :)