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Eduardwww [97]
2 years ago
10

Answer the following questions. Make sure to show all your work.

Mathematics
1 answer:
KengaRu [80]2 years ago
5 0

a) Since the corresponding y-value is -0.6, hence the point (-0.8, -0.6) is a solution to the system of  equations

b) since  the corresponding x-value is not 1/3, hence the point (1/3, 2) is not a solution to the system of  equation

In order to show if the given point corresponds to the given function, we will have to substitute the value of x into the function to see if we will have its corresponding y-value

For the point (-0.8, -0.6), substitute x = -0.8 into both functions as shown:

f(x)  = 2x + 1

f(-0.8) = 2(-0.8) + 1

f(-0.8) = -1.6  + 1

f(-0.8) = -0.6

Simiarly;

y = -3(-0.8)- 3

y = 2.4 - 3

y = -0.6

Since the corresponding y-value is -0.6, hence the point (-0.8, -0.6) is a solution to the system of  equations

For the point (1/3, 2), substitute x = 1/3 into both functions as shown:

x = (y+2)/2

x = (2+2)/2

x = 4/2

x = 2

Simiarly;

x + 2 = 3

x = 3-2

x = 1

Since the corresponding x-value is not 1/3, hence the point (1/3, 2) is not a solution to the system of  equations

Learn more on systems of equation here: brainly.com/question/847634

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Answer: +23

Step-by-step explanation:

It would be 23 because the difference between -19 and 4 is 23

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3 years ago
How much time I spend on the phone affects how much studying I get done what is the independent and the dependent variable in th
stepladder [879]

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3 0
3 years ago
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Find the midpoint of the line segment joining the points (9,2) and (-3,-2)
FromTheMoon [43]

Answer:

(3,0)

Step-by-step explanation:

The midpoint formula is as follows:

( (x1+x2) /2, (y1 + y2) /2 )

x1 would be 9, x2 would be -3, y1 would be 2, y2 would be -2. When plugged in, you should get 6/2, and 0/2 for y. When simplified, x would be 3 and y would be 0. This means that the midpoint of the line segment is (3,0)

7 0
3 years ago
Simplify the complex fraction: ((3x-7)/x^2)/(x^2/2)+(2/x)
Archy [21]
Simplify the complex fraction: ((3x-7)/x^2)/(x^2/2)+(2/x)

\dfrac{ \dfrac{3x-7}{x^2} }{ \dfrac{x^2}{2} } + \dfrac{2}{x} =\qquad \qquad x \neq 0\qquad x^4 \neq 0 \\  \\  \\  \dfrac{(3x-7)*2}{x^2*x^2}+ \dfrac{2}{x} =  \\  \\  \\   \dfrac{6x-14}{x^4}+ \dfrac{2}{x} =\\  \\  \\  \dfrac{ 6x-14 }{ x^4 }+ \dfrac{2(x^3)}{x(x^3)}=\\  \\  \\  \boxed{  \dfrac{ 6x-14 +2x^3}{ x^4 } }

6 0
4 years ago
6-y+5y-6y help please with work if you can if not the answer is ok
Goshia [24]
You need to group the like terms (the terms with the same or no variable) together so it is:

5y-6y-y  +6

Then you combine the like term and it becomes
5y-7y +6
-2y+6
which can also be written as 6-2y

Hope this helps :)
5 0
3 years ago
Read 2 more answers
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