Intercepts

[Susan Smith]

In the first part of the last question, we are simply being asked to find the \(x\)-intercepts. To do so, we let \(y=0\):

[MATH]-0.23x^2+1.87x+1.5=0[/MATH]
And then solve for \(x\). I'd begin by multiplying through by -100, so clear the decimals and get rid of the leading negative sign:

[MATH]23x^2-187x-150=0[/MATH]
Next, apply the quadratic formula...what do you get?

it going to be
X= 187+-radical48769/46
is that right?

 

[Susan Smith]

In the first part of the last question, we are simply being asked to find the \(x\)-intercepts. To do so, we let \(y=0\):

[MATH]-0.23x^2+1.87x+1.5=0[/MATH]
And then solve for \(x\). I'd begin by multiplying through by -100, so clear the decimals and get rid of the leading negative sign:

[MATH]23x^2-187x-150=0[/MATH]
Next, apply the quadratic formula...what do you get?

sorry my post are replying twice

 

[MarkFL]

it going to be
X= 187+-radical48769/46
is that right?

If you mean:

[MATH]x=\frac{187\pm\sqrt{48769}}{46}[/MATH]
then yes, those are the correct roots. Can you use this information to answer the last question?

 

[Susan Smith]

If you mean:

[MATH]x=\frac{187\pm\sqrt{48769}}{46}[/MATH]
then yes, those are the correct roots. Can you use this information to answer the last question?

yes but I do not know how to do it.

 

[MarkFL]

Which of the roots is the missile's final \(x\)-coordinate? What is the missile's initial \(x\)-coordinate?

 

[Susan Smith]

Which of the roots is the missile's final \(x\)-coordinate? What is the missile's initial \(x\)-coordinate?

Sorry mark but i can not understand the last question i have to roots one is positive one is negative

what do i need to do exactly ?

 

[MarkFL]

As I mentioned before, we are assuming that the launch of the missile coincides with \(x=0\), so we can disregard the negative root. So, the final position is at the positive root, and the initial position is at \(x=0\).

Here's a diagram:



To find the total horizontal distance traveled, subtract the initial \(x\)-coordinate from the final one. What do you get?

 

[Susan Smith]

As I mentioned before, we are assuming that the launch of the missile coincides with \(x=0\), so we can disregard the negative root. So, the final position is at the positive root, and the initial position is at \(x=0\).

Here's a diagram:

View attachment 13630

To find the total horizontal distance traveled, subtract the initial \(x\)-coordinate from the final one. What do you get?

is it 9 ?
i am really sorry if i make so many mistakes

 

[MarkFL]

No, it is:

[MATH]\Delta x=\frac{187+\sqrt{48769}}{46}-0=\frac{187+\sqrt{48769}}{46}\approx8.87[/MATH]

 

[Susan Smith]

No, it is:

[MATH]\Delta x=\frac{187+\sqrt{48769}}{46}-0=\frac{187+\sqrt{48769}}{46}\approx8.87[/MATH]

aw i did not see it this way . thank you Mark you are really good at helping and explaining these whole questions were little bit challenging but you made it easy
thanks a lot