5^(x+7)=(1/625)^(2x-13)
We move all terms to the left:
5^(x+7)-((1/625)^(2x-13))=0
Domain of the equation: 625)^(2x-13))!=0
x∈R
We add all the numbers together, and all the variables
5^(x+7)-((+1/625)^(2x-13))=0
We multiply all the terms by the denominator
(5^(x+7))*625)^(2x+1-13))-((=0
We add all the numbers together, and all the variables
(5^(x+7))*625)^(2x-12))-((=0
We add all the numbers together, and all the variables
(5^(x+7))*625)^(2x=0
not sure if this is right :/
Answer:
Latest time he can leave to be home by a quarter before 5 is 4:13
Step-by-step explanation:
Given Max's trip home takes 32 minutes. we have to find the time at which he can leave to be home by a quarter before 5.
quarter before 5 means 4:45
Max's takes 32 min to come to home so he has to leave 32 minutes before the given time.
Hence, latest time he can leave to be home by a quarter before 5 is 4:45-32 = 4:13
Answer:
Step-by-step explanation:
7x + 3= 24
7x= 24-3
7x= 21
x= 21/7
x= 3
Then we have that
5-4x
5-4(3)
5-12
Answer= -7
Lets say a=2 and b=5 you now plug in you numbers to get 3x2x5 we know 3x2=6 so now take 6 and multiply it by 5 so 6x5=30 you can also set this equal to 0 3ab=0 then divide by either a or b ill show you both if you divide by a you are left with 3b=a if you divide by b than you get 3a=b
That is false because you have to subtract the exponents not divide
