Answer:
It takes 6 hours for both to reach an equal paid.
Step-by-step explanation:
You have to form an expressions in terms of x where x determine the number of hours :
Option 1,
$(30 + 9x)
Option 2,
$(14x)
Next, you have to compare it since they want to find out how many hours they work until they reach the same amount :
30 + 9x = 14x
30 = 14x - 9x
30 = 5x
5x = 30
x = 30÷5
x = 6
Answer: (why is it 5 points?) Thats ok.
5 times what X is - 24 meaning whatever X is also by sry i don't make a lot of sense but i'm smart and know the answer. I will change it later after i find out which shouldn't take too long.
Step-by-step explanation:
Thanks, stay safe, have a great life! Byes! Brainliest then i will give answer! ;) :) :P XD :3 (: hehe :)
Answer: v = -3
=========================================
Work Shown:
![5 = \sqrt{-2-6v}+1\\\\5-1 = \sqrt{-2-6v}\\\\4 = \sqrt{-2-6v}\\\\4^2 = \left(\sqrt{-2-6v}\right)^2 \ \text{ .... square both sides}\\\\16 = -2-6v\\\\16+2 = -6v\\\\18 = -6v\\\\-6v = 18\\\\v = 18 \div (-6)\\\\v = -3](https://tex.z-dn.net/?f=5%20%3D%20%5Csqrt%7B-2-6v%7D%2B1%5C%5C%5C%5C5-1%20%3D%20%5Csqrt%7B-2-6v%7D%5C%5C%5C%5C4%20%3D%20%5Csqrt%7B-2-6v%7D%5C%5C%5C%5C4%5E2%20%3D%20%5Cleft%28%5Csqrt%7B-2-6v%7D%5Cright%29%5E2%20%5C%20%5Ctext%7B%20....%20square%20both%20sides%7D%5C%5C%5C%5C16%20%3D%20-2-6v%5C%5C%5C%5C16%2B2%20%3D%20-6v%5C%5C%5C%5C18%20%3D%20-6v%5C%5C%5C%5C-6v%20%3D%2018%5C%5C%5C%5Cv%20%3D%2018%20%5Cdiv%20%28-6%29%5C%5C%5C%5Cv%20%3D%20-3)
As you can see, we don't square both sides until the square root portion is fully isolated or on its own side. So it happens after we subtract 1 from both sides.
Technically, you are able to square both sides without first isolating the square root. But that would mean you'd have to use the FOIL rule and things would get a bit messier than they have to be. Not to mention that the square root term wouldn't fully go away (so you'd have to square again later down the line).
------------------
Checking the answer:
Replace every copy of v with -3. Simplify both sides. We should end up with the same number on each side
![5 = \sqrt{-2-6v}+1\\\\5 = \sqrt{-2-6(-3)}+1\\\\5 = \sqrt{-2+18}+1\\\\5 = \sqrt{16}+1\\\\5 = 4+1\\\\5 = 5\\\\](https://tex.z-dn.net/?f=5%20%3D%20%5Csqrt%7B-2-6v%7D%2B1%5C%5C%5C%5C5%20%3D%20%5Csqrt%7B-2-6%28-3%29%7D%2B1%5C%5C%5C%5C5%20%3D%20%5Csqrt%7B-2%2B18%7D%2B1%5C%5C%5C%5C5%20%3D%20%5Csqrt%7B16%7D%2B1%5C%5C%5C%5C5%20%3D%204%2B1%5C%5C%5C%5C5%20%3D%205%5C%5C%5C%5C)
The answer is confirmed.
Answer:
The answer is 25064.
Step-by-step explanation:
<u>There is a mistake in the question so the correct is below:</u>
What is 6 tenths 20 thousandths 4 ones and 50 hundreds.
Now, to solve it:
So, we arrange it in sequence:
20 thousandths 50 hundreds 6 tenths and 4 ones.
Now,
20 thousandths = 20 <u>000</u>
50 hundreds = 50 <u>00</u>
6 tenths = 6 <u>0</u>
4 ones = 4
Thus, 20 thousandths 50 hundreds 6 tenths and 4 ones is :
20,000 + 5000 + 60 + 4.
= 25064.
Therefore, the answer is 25064.
Answer:
7.5%
Step-by-step explanation:
one year has 12 months...and <em>in</em><em> </em><em>one</em><em> </em><em>mo</em><em>nth</em><em> </em><em>the</em><em> </em><em>person</em><em> </em><em>saves</em><em> </em><em>2</em><em>0</em><em>0</em><em> </em>
<em>so</em><em> </em><em>in</em><em> </em><em>one</em><em> </em><em>yea</em><em>r</em><em> </em><em>it</em><em> </em><em>wi</em><em>ll</em><em> </em><em>be</em><em> </em><em>2</em><em>0</em><em>0</em><em>×</em><em>1</em><em>2</em><em> </em><em>=</em><em> </em><em>2</em><em>4</em><em>0</em><em>0</em>
<em>%</em><em>s</em><em>a</em><em>v</em><em>e</em><em>=</em><em>tota</em><em>l</em><em> </em><em>saves</em><em>/</em><em>tota</em><em>l</em><em> </em><em>earni</em><em>ng</em><em>×</em><em>1</em><em>0</em><em>0</em>
<em>=</em><em> </em><em>(</em><em>2</em><em>4</em><em>0</em><em>0</em><em>/</em><em>3</em><em>2</em><em>0</em><em>0</em><em>0</em><em>)</em><em>×</em><em>1</em><em>0</em><em>0</em><em> </em><em> </em><em>=</em><em> </em><em>7</em><em>.</em><em>5</em><em>%</em>
<em>I </em><em>hop</em><em>e</em><em> this</em><em> will</em><em> help</em><em>!</em>