MATHEMATICS HIGH SCHOOL

Find an ordered pair (x, y) that is a solution to the equation.
5x - y = 8

Answers

Answer 1
Answer:

(2,2) could be a solution. :)

(0,-8)  could be another.

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Find the area of blah blah blah blah blah

Answers

Answer: hope this helps ♡

The area of blah blah blah blah blah is blah*blah LOL

Use the graph below for this question:graph of parabola going through 0, 3 and 2, negative 5.

What is the average rate of change from x = 0 to x = 2? (1 point)


−4

−8

1

5

Answers

Answer:

It should be the 2nd one

Step-by-step explanation:
3rd one is the correct answer for your queation

Identify the range of the function. A) (2, 8) B) [2, 8] C) (−5, 9) D) [−5, 9]

Answers

Answer:

The answer is "Option A"

Step-by-step explanation:

The domain is the collection of the value, which belongs to the separate variable (horizontal axis). So, to find a region with a graph, it must search for the function, which starts and end. And at all these levels we are searching at x-values.  

Its starting point is (2,9) and the ending point is (8,3).  Therefore, x= 2 to x=8 is the domain.

How to ask someone out ?

Answers

maybe tell them your feelings first then proceed to ask them out i would ask if they like you back first

Answer:

There are so many ways! Do want me to help?

Step-by-step explanation:

Write the quadratic equation in general form and then choose the value of "b."(2x - 1)(x + 6) = 0

Answers

first of all to transfer (2x - 1)(x + 6) = 0 into a form where we can plug it into the quadratic formula we need to use the FOIL method.

(2x - 1)(x + 6) = 0
 becomes
2x*x+2x*6+-1*x+-1*6
which simplifies to
2x^2 + 12x - x - 6
and then we add like terms to get
2x^2 + 11x - 6

now that it is in the correct form we can identify "a" "b" and "c" by following this form
ax^2 + bx + c
looking back at the equation we got earlier
2x^2 + 11x - 6

a=2,b=11,and c=-6

Answer:

The answer is b=11

Step-by-step explanation:

You are welcome.

Is this correct ? .......................

Answers

To find the midpoint of the two points (x₁ ,y₁) and (x₂, y₂) we need to use the formula:

\sf{Midpoint=((x_1 + x_2)/(2), (y_1+y_2)/(2))}

So to find the midpoint of E (a,a) and F (3a, a), let's plug it in to the formula:

\sf{Midpoint=((a + 3a)/(2), (a+a)/(2))}

Simplify the numerator:

\sf{Midpoint=((4a)/(2), (2a)/(2))}

Simplifying the fractions more:

\sf{Midpoint=(2a, a)}


So the midpoint of EF is (2a,a).

Your answer of (a,a) would be wrong.
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