Jacoob Has 7 Bags Of Choclates He Eats 5 Of Them And Gave 1 To Tom, And Buys 9 More Packs How Many Does He Have Now?

Jacoob has 7 bags of choclates he eats 5 of them and gave 1 to tom

and buys 9 more packs how many does he have now?

 

Answer:

10 packs of chocolates

Step-by-step explanation:

Equation: x = 7 - (5 + 1) + 9

Solving for x:

7 - 6 + 9 = x

1 + 9 = x

x = 10

Hope this helps!

~~DeanGD20


Comments

Post a Comment

Popular posts from this blog

Which Species Is Most Likely To Exhibit Clumped Dispersion?, A. , Dandelions With Seeds Dispersed By The Wind, B. , Tigers That Defend Their Territori

Which species is most likely to exhibit clumped dispersion? A. dandelions with seeds dispersed by the wind B. tigers that defend their territories from one another C. a bird that nests in many different types of trees D. a tree that grows only on the banks of streams   Which species is most likely to exhibit clumped dispersion? A.dandelions with seeds dispersed by the wind B.tigers that defend their territories from one another C.a bird that nests in many different types of trees D.a tree that grows only on the banks of streams The answer is letter B.  tigers that defend their territories from one another. Here are the explanation. Clumped dispersion is when there is a clustering of individuals in a  group, producing some patches with several individuals and some patches with no individuals.   Individuals are spaced evenly across  an area, in uniform dispersion.  And the individuals are distributed in  random dispersion without any apparent pattern. ...
Read more

9. What Is The 6th Term Of The Geometric Sequence 2/25, 2/5, 2, 10,2026?, A. 25 , B. 250 , C. 1250 , D. 2500

9. What is the 6th term of the geometric sequence 2/25, 2/5, 2, 10,…? a. 25 b. 250 c. 1250 d. 2500   Geometric Sequence The 6th term of the geometric sequence 2/25, 2/5, 2, 10 ,.... is 250. Solution: Geometric Sequence Formula: a_n = a₁r ⁿ ⁻ ¹ Given: a₁ = 2/25           a₂ = 2/5           a₃ = 2           a₄ = 10 1. Find the common ratio.        a_n = a₁r ⁿ ⁻ ¹         a₂  = a₁r ⁿ ⁻ ¹        2/5 = 2/25 r  ² ⁻ ¹        2/5 = 2/25 r 2. Divide both sides of the equation by 2/25 to find r.     2/5/2/25 = 2/25 r/2/25   (2/5)(25/2) = r           50/10 = r                  5 = r 3. Using r = 5, find the 6th term of the geometric sequence.             a_n = a₁r ⁿ ⁻ ¹               a₆ = a₁r ⁿ ⁻ ¹               a₆ = 2/25(5) ⁶ ⁻ ¹               a₆ = 2/25 (5)⁵               a₆ = 2/25 (3, 125)               a₆ = 6, 250/25              a₆ = 250 4. Therefore, the 6th term of the geometric sequence 2/25, 2/5, 2, 10… .is 250. Definition: ...
Read more