S+m+l=28
4s+2m+l=58
6s+5m+4l=135
Eliminate the variable l from the first 2 equations
s+m+l=28
-4s-2m-l=-58
-3s-m=-30
Elminiate the variable l from the last 2 equations
6s+5m+4l=135
-16s-8m-4l=-232
-10s-3m=-97
Now solve for s and m using the 2 equations without l
-3s-m=-30
-10s-3m=-97
9s+3m=90
-10s-3m=-97
-s=-7
s=7
Then plug in s into one of the equations without l
-3(7)-m=-30
-21-m=-30
-m=-9
m=9
Now plus in s and m into one of the original 3 equations
(7)+(9)+l=28
16+l=28
l=12
Final answer:
Small=$7
Medium=$9
Large=$12
I know it only asks for large but I wanted to show you how to find them all for future reference. :)
Your answer:1/5 Hope this helps!
Please, post the instructions along with your question, and if possible share the question in symbolic form, not in words.
Do you mean Question #5? By (2) x cube, do you mean 2x^3?
I strongly suggest that you use lots of parentheses ( ) to show how your numbers are grouped, and not to use " x " to denote multiplication (use " * " for that, please.
If only you'll clear this up, I'd be happy to help.
I will assume that your post is 2x^3 - 3(9x-5)^2.
Then 2x^3 - 3(81x - 90x + 25). Does this have any resemblance to what you wanted me to see in your post?
Answer:
Step-by-step explanation:
17) HI ≅ UH ; GH ≅ TU ; GI ≅ TH
ΔHGI ≅ ΔUTH by Side Side Side congruent
∠G ≅ ∠T ; GI ≅ TH ; ∠GIH ≅ ∠THU
ΔHGI ≅ ΔUTH by Angel Side Angle congruent
19) IJ ≅ KD ; IK ≅ KC ; KJ ≅ CD
ΔIJK ≅ ΔKDC by Side Side Side congruent
∠J ≅ ∠D ; IJ ≅ KD ; ∠I ≅ ∠DKC
ΔIJK ≅ ΔKDC by Angle Side Angle congruent
This is my answer. Hope that it will help you.
