Answer:
Given that a polynomial has a constant term and at least one other term what is the minimum number of possible zero values based on the rational zero test? a-0 b-2 c-3 d-4
Answer:
Total pictures = 84+12 = 96
roll of films = 96 / 24 = 4
she used 4 rolls
130 you multiply the left side by 13 to get the right side.
Answer:
-51.4
Step-by-step explanation:
Answer:
1)-
How to solve your question
Your question is
4(4−72)−9(5+2)
4(4y-7y^{2})-9(5y+2)4(4y−7y2)−9(5y+2)
Simplify
1
Rearrange terms
4(4−72)−9(5+2)
4({\color{#c92786}{4y-7y^{2}}})-9(5y+2)4(4y−7y2)−9(5y+2)
4(−72+4)−9(5+2)
4({\color{#c92786}{-7y^{2}+4y}})-9(5y+2)4(−7y2+4y)−9(5y+2)
2
Distribute
4(−72+4)−9(5+2)
{\color{#c92786}{4(-7y^{2}+4y)}}-9(5y+2)4(−7y2+4y)−9(5y+2)
−282+16−9(5+2)
{\color{#c92786}{-28y^{2}+16y}}-9(5y+2)−28y2+16y−9(5y+2)
3
Distribute
−282+16−9(5+2)
-28y^{2}+16y{\color{#c92786}{-9(5y+2)}}−28y2+16y−9(5y+2)
−282+16−45−18
-28y^{2}+16y{\color{#c92786}{-45y-18}}−28y2+16y−45y−18
4
Combine like terms
2)
−17y+17z+24
See steps
Step by Step Solution:

STEP1:Equation at the end of step 1
((24 - 4 • (5y - 6z)) + 3y) - 7z
STEP2:
Final result :
-17y + 17z + 24
−282+16−45−18
-28y^{2}+{\color{#c92786}{16y}}{\color{#c92786}{-45y}}-18−28y2+16y−45y−18
−282−29−18
-28y^{2}{\color{#c92786}{-29y}}-18−28y2−29y−18
Solution
−282−29−18
