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Ber [7]
3 years ago
9

Select the correct answer. Which expression is an equivalent expression of 12x + 10 + 4y? 2(6x + 2y + 6) 3(4x + y + 10) 2(6x + 4

y + 8) 2(6x + 5 + 2y)
Mathematics
1 answer:
Illusion [34]3 years ago
4 0
The answer would be 2(6x+5+2y) you can figure this out through the distributive property. 2(6x)=12x 2(5)=10 and 2(2y)= 4y this would reult in giving you the equation 12x+10+4y
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Answer:

step 1:

2,-5

step 2:

7,1

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6/5

Step-by-step explanation:

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2 years ago
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Step-by-step explanation:

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If f(x)= 3x + 1 what is is the value of x when f(x)= 15
astraxan [27]

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8 0
2 years ago
Solve for x. M=5/dx+e
snow_lady [41]

After solving  M=\frac{5}{dx}+e we get value of x:  x=\frac{5}{(M-e)d}

Step-by-step explanation:

We need to solve the equation M=\frac{5}{dx}+e to find value of x

Solving

M=\frac{5}{dx}+e

Adding -e on both sides

M-e=\frac{5}{dx}+e-e

M-e=\frac{5}{dx}

Multiply both sides by dx

dx(M-e)=5

Divide both sides by M-e

dx=\frac{5}{M-e}

Divide both sides by d

x=\frac{5}{(M-e)d}

So, After solving  M=\frac{5}{dx}+e we get value of x:  x=\frac{5}{(M-e)d}

Keywords: Solving Equations

Learn more about Solving Equations at:

  • brainly.com/question/1563227
  • brainly.com/question/2403985
  • brainly.com/question/11229113

#learnwithBrainly

8 0
3 years ago
The first term of a geometric sequence is equal to a and the common ratio of the sequence is r.
ololo11 [35]

Answer: (a)  {a, ar, ar², ar³, ar⁴, ar⁵...}, (b)  arⁿ⁻¹

For part (a), the question gives us the first term a, and then asks us to apply the common ratio r six times.

In order for ar = a, the nth term of r will have to equal 0 (this implies that n is an exponent; thus giving us the first term a, as r = 1).

Since we use this method on the first term, we must use it for the next five, in which r gains an additional exponent for every consecutive value (nth term) thereafter.  

Ultimately getting: {a, ar, ar², ar³, ar⁴, ar⁵...}

For part (b), we first have to understand that the sequence does not start at 0, but at 1 for n. In order for ar = a, with n = 1, there needs to be subtraction of -1 within the exponent. So that arⁿ⁻¹

If we check and apply this, we can see that:

{ar¹⁻¹, ar²⁻¹, ar³⁻¹, ar⁴⁻¹, ar⁵⁻¹, ar⁶⁻¹...} = {a, ar, ar², ar³, ar⁴, ar⁵...} = arⁿ⁻¹ = Tn

4 0
3 years ago
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