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melamori03 [73]
3 years ago
12

What value of x satisfies the equation 2x 5 − 3/5 = 4/5

Mathematics
2 answers:
kondor19780726 [428]3 years ago
4 0
I hope this helps you

mr Goodwill [35]3 years ago
3 0
I think it is x+7/2
 i will look it up and tell u more about it
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!!!!!!!!!!!!!!!!!!!!!!!!!PLEASE HELP!!!!!!!!!!!!!!!!!!!!
denis-greek [22]

Answer:

-t2+10t-h+8  = 0

Step-by-step explanation:

5 0
3 years ago
What is 2(5)+3(4)+5^2? (The ^ means the exponent)
castortr0y [4]
I believe the answer is 47.
6 0
3 years ago
Weiming, Joyce and Siti divide cookies to sell in a ratio of 5: 12: 9. After Weiming sold 28 cookies, the ratio becomes 1: 8: 6.
Strike441 [17]

Answer:

216 cookies

Step-by-step explanation:

From the question,

Let the total cookies in the first place be x cookies.

Their ratio = 5:12:9

Total  = 5+12+9 = 26.

Weiming share = (5/26)x = 5x/26

If weiming sold 28 cookies

⇒ (5x/26)-28

⇒ (5x-728)/28

Hence Weiming new share = (5x-728)/28 cookies

The new ratio is 1:8:6

The total cookies becomes = x-28

Therefore,

(1/15)(x-28) = (5x-728)/28

(x-28)/15 = (5x-728)/28

Crossmultiply

28(x-28) = 15(5x-728)

28x-784 = 75x-10920

Collect like terms

28x-75x = -10920+784

-47x = -10136

x = -10136/-47

x = 215.7

x ≈ 216 cookies

8 0
3 years ago
The steps below show the incomplete solution to find the value of x for the equation
mafiozo [28]
Step 1: 6x - 2x - 5 = -6 + 21 
Step 2: 6x - 2x - 5 = 15 
Step 3: 4x - 5 = 15 

The next step would be to add 5 to both sides of the equation. 
so Step 3: 4x - 5 = 15 .......add 5 to both sides would make 
Step 4: 4x = 20 
 
So after the next step you would have 4x = 20 

Hope this helps! :) 

    
7 0
3 years ago
Suppose the scores of students on an exam are normally distributed with a mean of 244 and a standard deviation of 79. According
Ahat [919]
244 - 165 = 79 . . . . one standard deviation
323 - 244 = 79 . . . . one standard deviation

The range of scores is ±1 standard deviation from the mean. The empirical rule says 
   68% of scores lie in that range.
3 0
3 years ago
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