the problem would be set up as 6+(7*x)=41
first u take off the number 6 by subtracting it from itself to cancel it out then u do the same to the number 41
7x =41 now u need to cancel out the number 7 attatched to the x so u divikde it by itself and then do the same to the other side so 41 divided by 7 =5.9
so x =5.9
Answer: 60
Step-by-step explanation:
First one:
you can add -10m and -13m but you can't add -10m and 2m^4 becuase the powers aren't the same so
when adding the like terms
look at the:
powers, (x^3 adds with x^3)
placehloder letter (x adds with x and y adds with y and so on)
-10m+2m^4-13m-20m^4
powers: m^1 and M^4
placeholders: all m
add
-10m-13m+2m^4-20m^4
-23m-18m^4
second one:
when multiplying exponents, you add with like
so if you multipliy
x^2yz^3 times x^4y^2z^2 thne you would get x^6y^3z^5
when multiply with coeficients
2x^2yz^3 times 4x^4y^2z^2=8x^6y^3z^5
so using associative property a(bc)=(ab)c
2/3 times p^4 times y^3 times y^4 times s^5 times 6 times p^2 times s^3
group like terms
(2/3 times 6) times (p^4 times p^2) times (y^3 times y^4) times (s^5 times s^3)
(4) times (p^6) times (y^7) times (s^8)
4p^6y^7s^8
(14,-1) is the answer to the question
I also think the answer is D, who else agrees with me? Btw, thank you for posting this question. I enjoy answering fun things sometimes.
