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Stells [14]
3 years ago
12

Which represents the solution(s) of the graphed system of equations, y = x2 + x – 2 and y = 2x – 2? (–2, 0) and (0, 1) (0, –2) a

nd (1, 0) (–2, 0) and (1, 0) (0, –2) and (0, 1) Mark this and return
Mathematics
2 answers:
MA_775_DIABLO [31]3 years ago
5 0

Answer:

<h2>(0, -2) (1, 0)</h2>

Step-by-step explanation:

I got it right on the test... Have a great day :)

Luda [366]3 years ago
3 0

ANSWER

(0,-2), (1,0)

EXPLANATION

The first equation is

y =  {x}^{2}  + x - 2

The second equation is

y = 2x - 2

We equate both equations to get,

{x}^{2}  + x - 2 = 2x - 2

{x}^{2}  + x - 2x - 2 + 2 = 0

Simplify

{x}^{2}  - x = 0

Factor

x(x - 1) = 0

Either

x = 0

Or

x - 1 = 0

x = 1

Put x=0 or x=1 into the second equation to get,

y =   2(0) - 2 =  - 2

Or

y =   2(1) - 2 =  0

Therefore the solutions are;

(0,-2), (1,0)

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X^2 + y^2 = 25 <br> x^2 + y^2 = 19
olga55 [171]
What do you need help on.
7 0
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
Juliette [100K]

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

6 0
3 years ago
Help, adding extra text so that I can post this question
GREYUIT [131]
The answer 0.1, 0.136, 0.652, 0.87
8 0
3 years ago
Read 2 more answers
Log2x (2x^2+8x-6) =2
irinina [24]

Answer:

2x^3+8x^2-3x=0

Step-by-step explanation:

2x (2x^2+8x-6)=2\\multiply-all-in-bracets-by-2x\\4x^3+16x^2-6x=2\\devide-all-by-2\\2x^3+8x^2-3x=0

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Simplify log(16x²) ± 2㏒(1÷×)
sp2606 [1]

Answer: just simplify condense it then your answer will be log(16)

8 0
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