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andrew11 [14]
3 years ago
12

How do u solve for x in the equation 3x+9=44-2x

Mathematics
2 answers:
Masteriza [31]3 years ago
4 0
3x + 9 = 44 - 2x (add 2x to each side)
5x + 9 = 44 (minus 9 from each side)
5x = 35 (divide each side by 5)
x = 7 
WARRIOR [948]3 years ago
3 0
3 x + 9 = 44 - 2 x

subtract 9 from both sides:

3 x + 9 - 9 =  44 - 2x - 9

3 x = -2 x +35

Add  2 x to both sides:

3 x + 2 x = -2 x + 35 + 2 x

5 x = 35

Divide both sides by 5 :

5x / 5 = 35/ 5

x = 7

hope this helps!




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3 years ago
you currently have 24 credit hours and a 2.8 gpa you need a 3.0 gpa to get into the college. if you are taking a 16 credit hours
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Answer:

\sum_{i=1}^n w_i *X_i = 2.8*24 = 67.2

And for this case we want a gpa of 3.0 taking in count that in this semester he/ she is going to take 16 credits so then the new mean would be given by:

\bar X_f = \frac{\sum_{i=1}^n w_i *X_i+w_f *X_f }{24+16} = 3.0

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3.0 *(24+16) =\sum_{i=1}^n w_i *X_i +\sum_{i=1}^n w_f *X_f

And from the previous result we got:

3.0 *(24+16) =67.2 +\sum_{i=1}^n w_f *X_f

And solving we got:

\sum_{i=1}^n w_f *X_f =120 -67.2= 52.8

And then we can find the mean with this formula:

\bar X_2 = \frac{\sum_{i=1}^n w_f *X_f}{16}= \frac{52.8}{16}=16=3.3

So then we need a 3.3 on this semester in order to get a cumulate gpa of 3.0

Step-by-step explanation:

For this case we know that the currently mean is 2.8 and is given by:

\bar X = \frac{\sum_{i=1}^n w_i *X_i }{24} = 2.8

Where w_i represent the number of credits and X_i the grade for each subject. From this case we can find the following sum:

\sum_{i=1}^n w_i *X_i = 2.8*24 = 67.2

And for this case we want a gpa of 3.0 taking in count that in this semester he/ she is going to take 16 credits so then the new mean would be given by:

\bar X_f = \frac{\sum_{i=1}^n w_i *X_i+w_f *X_f }{24+16} = 3.0

And we can solve for \sum_{i=1}^n w_f *X_f and solving we got:

3.0 *(24+16) =\sum_{i=1}^n w_i *X_i +\sum_{i=1}^n w_f *X_f

And from the previous result we got:

3.0 *(24+16) =67.2 +\sum_{i=1}^n w_f *X_f

And solving we got:

\sum_{i=1}^n w_f *X_f =120 -67.2= 52.8

And then we can find the mean with this formula:

\bar X_2 = \frac{\sum_{i=1}^n w_f *X_f}{16}= \frac{52.8}{16}=16=3.3

So then we need a 3.3 on this semester in order to get a cumulate gpa of 3.0

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