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sladkih [1.3K]
2 years ago
13

Given the functions f(x)=1x−3+1 and g(x)=1x+4+3 .

Mathematics
2 answers:
SVETLANKA909090 [29]2 years ago
6 0

Answer:

The graph shifts 7 units left and 2 units up.

Step-by-step explanation:

In transformations, when a number is added or subtracted to "x" directly, that is the left/right shift.

When a number is added or subtracted outside of the "rest" of the equation, that is the up/down shift.

I learned transformations with these variables:

f(x) = \frac{a}{k(x-c)} + d

"a" - vertical stretch (a > 1) or compression (0 < a < 1)

"k" - horizontal stretch (0 < k < 1) or compression (k > 1)

"c" - translate left (+) or right (-)

"d" - translation up (+) or down (-)

The untranslated graph is centred around the origin f(x) = 1/x

f(x) begins 3 units right and 1 unit up of the untranslated graph.

g(x) is 4 units left and 3 units up of the untranslated graph.

            y

g           |

            |           f

-----------|------------ x

There is a 7 units translation left and 2 units translation up.

Papessa [141]2 years ago
3 0
The graphs shifts 7 units up and 2 units up
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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

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

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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