i cant see the pic that good
So just your normal equation.
We need to isolate x alone.
First lets write out our equation
3 +7x = 12
Lets do this step by step.
first subtract 3 from both sides.
3 + 7x = 12
-3 = -3 it cancels out on the left side and we solve on the right side.
*************
7x = 9
second we divide both sides by 7
7x = 9
÷7 = 7 we got the x alone and we just write 9/7 as our answer.
x = 9/7
Or as a decimal.
9/7 could be <span>1.28571428571429
</span>
Either will work.
x = 9/7
or
x = <span>1.28571428571429
</span>
Have a nice day. :)
Answer:
1 x 10^21
Step-by-step explanation:
The initial number (2 x 10^4) is in scientific notation. Scientific notation must be written as: a X 10^n, where 0<a<10 and n is an integer. In this case, the number in scientific notation is being raised to the power of 5. When we raise a power by another power, we must multiply the power of all the terms by the additional power. In this case, we would get 2^5 x 10^20, this would be 10 x 10^20. Since the value of 'a' needs to be greater than 0 and less than 10, we need to move the decimal to the left and add another power to our 10: 1 x 10^21.
This answer maybe long, but it includes all the essential steps to make it easy
This problem shows that the fraction of the time worked by three individuals need to be = to one person working full time.
It just means that the fractions need to add up to one. Since we got to know that one person works 1/2 the time on the project and the other works 1/3 of the time, let us use a variable to represent the amount of time the last person needs to work to have a total of 1.
This is quite easy:
variable = t(representative of the last person)
step 1: 1=1/2+1/3+t
step 2: keep the variable alone at one side
1-1/2-1/3=t
this step on a calculator will be easy, but dont have a calculator? Thats fine!
Lets have a common denominator
2 and 3 can be 6
1/1*(6/6)-1/2*(3/3)-1/3*(2/2)=t
6/6-3/6-2/6=t
step 3: just subtracts and get the answer
1/6=t
In conclusion, the third person must be working 1/6 the time on the project
456.......................
