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dangina [55]
2 years ago
9

Which algebraic expression is equivalent to the expression below? 2(3 + 5x) + 24 A. 10x - 30 B. 10x + 18 C. 5x + 30 D. 10x + 30

Mathematics
1 answer:
tekilochka [14]2 years ago
6 0

Answer:

10x +30

Step-by-step explanation:

2(3 + 5x) + 24

Distribute

2*3 + 2*5x+24

6+10x+24

Combine like terms

10x +30

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Simplify <br> 1/3n (-6+27m-51p)
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3 years ago
A quantity with an initial value of 6200 decays continuously at a rate of 5.5% per month. What is the value of the quantity afte
ELEN [110]

Answer:

410.32

Step-by-step explanation:

Given that the initial quantity, Q= 6200

Decay rate, r = 5.5% per month

So, the value of quantity after 1 month, q_1 = Q- r \times Q

q_1 = Q(1-r)\cdots(i)

The value of quantity after 2 months, q_2 = q_1- r \times q_1

q_2 = q_1(1-r)

From equation (i)

q_2=Q(1-r)(1-r)  \\\\q_2=Q(1-r)^2\cdots(ii)

The value of quantity after 3 months, q_3 = q_2- r \times q_2

q_3 = q_2(1-r)

From equation (ii)

q_3=Q(1-r)^2(1-r)

q_3=Q(1-r)^3

Similarly, the value of quantity after n months,

q_n= Q(1- r)^n

As 4 years = 48 months, so puttion n=48 to get the value of quantity after 4 years, we have,

q_{48}=Q(1-r)^{48}

Putting Q=6200 and r=5.5%=0.055, we have

q_{48}=6200(1-0.055)^{48} \\\\q_{48}=410.32

Hence, the value of quantity after 4 years is 410.32.

4 0
2 years ago
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