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ludmilkaskok [199]
2 years ago
7

Which of the following equations has infinitely many solutions? A. 2x + 5(2x + 6) = 12x + 30 B. 2x + 5 = 2 C. 2(x + 5) + 2 = 6x

+ 2(x + 2) D. 5x + 2 = 2x + 6
Mathematics
1 answer:
Nat2105 [25]2 years ago
8 0

Answer:

A. 2x + 5(2x + 6) = 12x + 30

Step-by-step explanation:

Infinitely many solutions means the two sides of the equation will be equal to each other, so B and D can be eliminated. Distribute for A and C to check

A. 2x + 5(2x + 6) = 12x + 30

    2x + 10x + 30 = 12x + 30

    12x + 30 = 12x + 30

C. 2(x + 5) + 2 = 6x + 2(x + 2)

    2x + 10 + 2 = 6x +2x + 4

     2x + 12 = 8x + 4

You only get the same equation for A, therefore A is the correct answer

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3 0
3 years ago
The first term in a geometric series is 64 and the common ratio is 0.75. Find the sum of the first 4 terms in the series.
maks197457 [2]

Answer:

195.25

Step-by-step explanation:

Consider geometric series  S(n) where initial term is a

So S(n)=a+ar^1+...ar^n

Factor out a

S(n)=a(1+r+r^2...+r^n)

Multiply by r

S(n)r=a(r+r^2+r^3...+r^n+r^n+1)

Subtract S(n) from S(n)r

Note that only 1 and rn^1 remain.

S(n)r-S(n)=a(r^n+1  -1)

Factor out S(n)

S(n)(r-1)=a(r^n+1  -1)

The formula now shows S(n)=a(r^n+1  -1)/(r-1)

Now use the formula for the problem

4 0
3 years ago
Use the data set to determine which statements are correct. Check all that apply. 35, 41, 18, 75, 36, 21, 62, 29, 154, 70 The me
stepladder [879]
The median is 38.5

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4 0
3 years ago
Please this is your answer.​
Effectus [21]

Answer:

10. it is correct it is 24

11. 21

Step-by-step explanation:

I could not see the answer but it should look like this:

11.6-3(3-[2-3]-10)-3

6-3(3-[-1]-10)-3

6-3(3-(-1)-10)-3

6-3(4-10)-3

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6-(-18)-3

(6-(-18))-3

24-3

21

4 0
2 years ago
Read 2 more answers
Deepa has 57. She has 4 times as much money as Gaurav. How much
Readme [11.4K]

Answer:

42.75 dollars more

Step-by-step explanation:

57/4= 14.25

Gaurav has 14.25 dollars

57-14.25=42.75

therefore Deepa has 42.75 more dollars than Gaurav

6 0
3 years ago
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