bearing in mind that an absolute value expression is in effect a piece-wise expression, because it has a ± version.
![\bf 3|x|+7=28\implies 3|x|=21\implies |x|=\cfrac{21}{3}\implies |x|=7\implies \begin{cases} +(x)=7\\ -(x)=7 \end{cases} \\\\[-0.35em] ~\dotfill\\\\ +(x)=7\implies \boxed{x=7}~\hfill -x=7\implies \boxed{x=-7}](https://tex.z-dn.net/?f=%20%5Cbf%203%7Cx%7C%2B7%3D28%5Cimplies%203%7Cx%7C%3D21%5Cimplies%20%7Cx%7C%3D%5Ccfrac%7B21%7D%7B3%7D%5Cimplies%20%7Cx%7C%3D7%5Cimplies%20%20%5Cbegin%7Bcases%7D%20%2B%28x%29%3D7%5C%5C%20-%28x%29%3D7%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%2B%28x%29%3D7%5Cimplies%20%5Cboxed%7Bx%3D7%7D~%5Chfill%20%20-x%3D7%5Cimplies%20%5Cboxed%7Bx%3D-7%7D%20)
Non zero digits are always significant. any zeros between two significant digits are significant. a final zero or trailing zero in the decimal portion only are significant.
Answer:
12 pigs and 20 chickens
Step-by-step explanation:
Chicken = 2 legs
Pig = 4 legs
(15 * 4) + (17 * 2) = 88 NO
(12 * 4) + (20 * 2) = 88 Yes, and 12 + 20 = 32
Yes
factor out the 2z^2 in each term
(2z^2)(z^2-5z+4)
factor some more
z^2-5z+4
find what 2 numbers multiply to get 4 and add to get -5
the numbers are -1 and -4
(z-1)(z-4)
the factored form is
(2z^2)(z-1)(z-4)
Answer: A B and C
Step-by-step explanation:
I did the usatestprep