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uhhhhhhhhh we can just plug in those values and solve?!!!
YA I KNOW BUT EVERY TIME I MESS UP -_- AND I NEED HELP PLEASE!
well ok.. we can combine like terms first to make it easier \[2r^2-r^2-r^2+3s^3+4t^2\] \[2r^2-2r^2+3s^3+4t^2\] \[3s^3+4t^2\] s=-3 t=5 \[3(-3)^3+4(5)^2\] \[3(-27)+4(25)\] \[-81+100=19\]
yup checks out on the first one I got 19 too, but at least I showed work :P
The second question too i am confused on can you help with that too?
Find −3a2 − b3 + 3c2 − 2b3 if a = 2, b = −1, and c = 3.
\[-3a^2-3b^3+3c^2\] a=2, b=-1, c = 3 \[-3(2)^2-3(-1)^3+3(3)^2\] \[-3(4)-3(-1)+3(9)\] \[-12+3+27 =18\]
I would thank the creator of the algebra calculator.
^which you shouldn't depend on.. because that's not really learning how to solve these problems.
It shows you the steps!
True, but if OP has a closed book closed note test... then what? The only way to solve it is the manual way.
these sites that automatically solve the problem for you are only supposed to be used to check your answer... it shouldn't be a replacement to do the math the pencil and paper way.
@cutiiepie7272 you must learn to replace each variable with a given value and simplify the resulting numerical expression.
This is called the substitution principle: An expression maybe replaced by another expression that has the same value.
4 + 2 can be replaced by 6 because they have the same value 4 + 2 = 6. Similarly, if x = 6, then x can be replaced by 6 because they have the same value 6 = 6.
Also, please use the "^" symbol for exponentiation :-)