Can someone explain why the answers for question 1 and 2 are different than the answers for 3 and 4

1) Parabola is opened up with no x-intercepts so the graph is fully above the x-axis which means all the y-coordinates in our points are positive so that is why the solution is all real numbers because it asked you to solve y>0 for x.

2) Parabola is opened up with no x-intercepts so the graph is fully above the x-axis which means all the y-coordinates in our points are positive so that is why the solution is all real numbers because it asked you to solve y>=0 for x.

3) Parabola is opened up with no x-intercepts so the graph is fully above the x-axis which means all the y-coordinates in our points are positive so that is why the solution is none because we are looking for when y<0 for x.  (y<0 means y is negative)

4) Parabola is opened up with no x-intercepts so the graph is fully above the x-axis which means all the y-coordinates in our points are positive so that is why the solution is none because we are looking for when y<=0 for x.  (y<=0 means negative or 0)

Step-by-step explanation:

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If , we would have a parabola.

Anyways here are few things that might help when solving these:

1) Write ax^2+bx+c in factored form; it can lead to the x-intercepts quickly once you have it

2) The discriminant b^2-4ac:

    A) It tells you if you have two x-intercepts if b^2-4ac is positive

    B)  It tells you if you have one x-intercept if b^2-4ac is zero

    C)  It tells you if you have zero x-intercepts if b^2-4ac negative

3) If a>0, then the parabola opens up.

   If a<0, then the parabola opens down.

4)  You might choose to test before and after the x-intercepts too, using numbers between or before and after.

Let's look at questions labeled (1)-(4).

(1)  x^2-x+2>0

a=1

b=-1

c=2

I'm going to use the discriminant to see how many x-intercepts I have:

b^2-4ac

(-1)^2-4(1)(2)

1-8

-7

Since -7 is negative, then we have no x-intercepts.

The parabola is also opened up.

So if we have no-x-intercepts and the parabola is opened up (a is positive), then the parabola is above the x-axis.

So the y values for x^2-x+2 where y=x^2-x+2 is positive for all real numbers.

The solution to x^2-x+2>0 is therefore all real numbers.

(2)  x^2+5x+7>=0

a=1

b=5

c=7

b^2-4ac

(5)^2-4(1)(7)

25-28

-3

Since -3 is negative, we have no x-intercepts.

The parabola is also opened up because a=1 is positive.

So again x^2+5x+7>=0 has solutions all real numbers.

(3)  x^2-4x+5<0

a=1

b=-4

c=5

b^2-4ac

(-4)^2-4(1)(5)

16-20

-4

Since -4 is negative, we have no x-intercepts.

The parabola is opened up because a=1 is positive.

All the y-coordinates for our points are positive.

So y=x^2-4x+5<0 has no solutions because there are no y's less than 0.

(4)  x^2+6x+10<=0

a=1

b=6

c=10

b^2-4ac

6^2-4(1)(10)

36-40

-4

Since -4 is negative, the parabola has no x-intercepts.

The parabola is opened up because a=1 is positive.

All the y-coordinates on our parabola are positive.

So y=x^2+6x+10<=0 has no solutions because there are no y's less than 0 or equal to 0.

c.   x=2.4cm

step-by-step explanation:

similar triangles are triangles which have the same shape but not the same size. this means their angle measures are equal but their side lengths are not instead they are proportional. to solve, we will set up a proportion and solve.

a proportion is two equal ratios set equal to each other. we form the ratios by finding corresponding sides (sides which match to each other on both triangles by their position). out ratios will be one side of the small triangle over the corresponding side on the big triangle as shown below:

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to solve the proportion, we'll cross multiply by multiplying numerator with denominator across the equal sign.