1. ## SAT Math

These are the last two questions in my practice test for the section. I am stuck and have no idea how to get the answer. The answer is given but i do not know how it was gotten. Any help will be greatly appreciated. q.17-412

1) In the xy-coordinate plane, the graph of $\displaystyle x^2=y^2-4$ intersects line $\displaystyle l$ at $\displaystyle (0,p)$ and $\displaystyle (5,t)$. That is the greatest possible slope of line $\displaystyle l$? The answer is 5 but i do not understand that. If the y-values are unknown the rise can be anything, right?

2)Esther drove to work in the morning at an average speed of 45 miles per hour. she returned home in the evening along the same route and averaged 30 miles per hour. If Esther spent a total of one hour commuting to and from work, how many miles did Esther drive to work in the morning? The answer is 18 i got 58. Very confused. Help with be gladly accepted.

Thanks again.

2. 1. Okay, you got the equation $\displaystyle x^2 = y^2 - 4$
If $\displaystyle x = 0$, what is $\displaystyle y = p =$ ? Replace in the equation above.
If $\displaystyle x = 5$, what is $\displaystyle y = t =$ ? Replace in the equation above

As for getting the greatest possible slope, consider this. The equation for slope is:

$\displaystyle m = \frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

Now, I already got the values of x fixed. What's left is the denominator, and to achieve the greatest possible slope, I need to get the highest possible value in the denominator. Suppose that $\displaystyle y_{2} = \pm a$ and $\displaystyle y_{1} = \pm b$, the highest value would be:
$\displaystyle y_{2}-y_{1} = (+a) - (-b) = a + b$

2. Here, they gave you two different speeds along the SAME distance but with different time intervals. Let me emphasize SAME again Establish a ratio between the two speeds.

$\displaystyle v_{1} = 45$
$\displaystyle v_{1} = \frac{d}{t_{1}}$
$\displaystyle 45 = \frac{d}{t_{1}}$

$\displaystyle v_{2} = 30$
$\displaystyle v_{2} = \frac{d}{t_{2}}$
$\displaystyle 30 = \frac{d}{t_{2}}$

Equate these two rates, and then try to get a ratio of $\displaystyle \frac{t_{1}}{t_{2}}$. Remember, if you want to get the distance according to the ratio:

$\displaystyle \frac{t_{1}}{t_{2}}$

then $\displaystyle \frac{t1}{t1+t2}$ should be multiplied with 45 to get your distance.

P.S. Are you sure that slope in first answer is 5?

3. 1. Okay, you got the equation x^2 = y^2 - 4
If x = 0, what is y = p = ? Replace in the equation above.
If x = 5, what is y = t = ? Replace in the equation above

As for getting the greatest possible slope, consider this. The equation for slope is:
m = (y2-y1)/(x2-x1)

Now, I already got the values of x fixed. What's left is the denominator, and to achieve the greatest possible slope, I need to get the highest possible value in the denominator. Suppose that y2 = ħa and y1 = ħb, the highest value would be:
y2 - y1 = (+a) - (-b) = a + b
That's good !

Hmmm, is there something wrong ? I find 1 as the greatest slope and (y2-y1) max is 5.

2. Here, they gave you two different speeds along the SAME distance but with different time intervals. Let me emphasize SAME again Establish a ratio between the two speeds.

V1 = 45
V1 = d/t1
45 = d/t1

V2 = 30
V2 = d/t2
30 = d/t2

Equate these two rates, and then try to get a ratio of t1/t2. Remember, if you want to get the distance according to the ratio:

t1/t2

then t1/(t1+t2) should be multiplied with 45 to get your distance.
Note to the OP : this is because $\displaystyle t_1+t_2=1$

Hey tasty dish ! that was good !

4. That's good to hear. But as for the slope, I found it to be 1.477 (max) and (y2-y1) is 7.385 max.

5. Hmm how did you get that ?

$\displaystyle 0=p^2-4 \implies p=\pm 2$

$\displaystyle t=\pm \dots$

6. Yes, and $\displaystyle t = \pm \sqrt{29}$

$\displaystyle m_{max} = \frac{\sqrt{29} - (-2)}{5 - 0} = 1.477032961$

7. You're correct

Sorry for the confusion !

Sorry i took so long but i have the explanation to the first problem and i also do not know how to equate the two question from the second problem.

1) The reason the greatest slope is 5 is modeled if you graph $\displaystyle x=y^2-4$ which in terms of $\displaystyle y$ would be $\displaystyle y=\sqrt{(x+4)}$ and $\displaystyle y=-\sqrt{(x+4)}$. This is a horizontal parabola that open to the right with a vertex at $\displaystyle (0,4)$. when $\displaystyle x=0$ then $\displaystyle y$ is $\displaystyle 2$ AND $\displaystyle -2$ since it is a parabola. when $\displaystyle x=5$ then $\displaystyle y$ is $\displaystyle 3$ AND $\displaystyle -3$. So if a line intersects at $\displaystyle (0,p)$ and $\displaystyle (5,t)$, the greatest slope would be 5 because if the two points $\displaystyle (5,3)$ and $\displaystyle (0,-2)$ are chosen, then $\displaystyle m = \frac{3-(-2)}{5 - 0}$, which is 5, the same goes for the other two points.

I hope this explanation is correct and full, but i would still appreciate help for the second problem. How would i equate $\displaystyle 45 = \frac{d}{t_{1}}$ and $\displaystyle 30 = \frac{d}{t_{2}}$ also, how do i get the ratio of $\displaystyle \frac{t_{1}}{t_{2}}$? Thanks a lot again everybody!

9. Originally Posted by OnMyWayToBeAMathProffesor
How would i equate $\displaystyle 45 = \frac{d}{t_{1}}$ and $\displaystyle 30 = \frac{d}{t_{2}}$ also, how do i get the ratio of $\displaystyle \frac{t_{1}}{t_{2}}$? Thanks a lot again everybody!
Since both equations have the same value $\displaystyle d$, we should solve each equation for $\displaystyle d$. This would then enable us to set the two expressions equal to each other.

$\displaystyle 45=\frac{d}{t_1}\implies 45t_1=d$

$\displaystyle 30=\frac{d}{t_2}\implies 30t_2=d$

Set the equations equal to each other:

$\displaystyle 45t_1=30t_2$

Now you can find $\displaystyle \frac{t_1}{t_2}$.

Does this clarify things?

--Chris

so then if $\displaystyle 45t_1=30t_2$ then $\displaystyle \frac{45t_1}{30}=t_2$ and $\displaystyle \frac{30t_2}{45}=t_1$. Thus $\displaystyle \frac{t_1}{t_2}=\frac{\frac{30t_2}{45}}{\frac{45t_ 1}{30}}$. From there it would turn into $\displaystyle \frac{30t_2}{45} * \frac{30}{45t_1}$. what do i do form there? thats where i get stuck, thanks again for all your help.

11. Nonononoo!!!

$\displaystyle 45t_1=30t_2$

Simply cross multiply and move the 45 and t_2:

$\displaystyle \frac{t_{1}}{t_{2}} = \frac{30}{45}$

Now, this is your ratio: $\displaystyle t_{1}:t_{2} = 30:45 = 2:3$

$\displaystyle \frac{t_{1}}{t_{1}+t_{2}} \cdot v_{1}$

ONWTBAMP: You need to revise ratios.

12. Originally Posted by OnMyWayToBeAMathProffesor
so then if $\displaystyle 45t_1=30t_2$ then $\displaystyle \frac{45t_1}{30}=t_2$ and $\displaystyle \frac{30t_2}{45}=t_1$. Thus $\displaystyle \frac{t_1}{t_2}=\frac{\frac{30t_2}{45}}{\frac{45t_ 1}{30}}$. From there it would turn into $\displaystyle \frac{30t_2}{45} * \frac{30}{45t_1}$. what do i do form there? thats where i get stuck, thanks again for all your help.
If $\displaystyle 45t_1=30t_2$, then $\displaystyle 45\frac{t_1}{t_2}=30\implies \frac{t_1}{t_2}=\frac{30}{45}\implies\color{red}\b oxed{\frac{t_1}{t_2}=\frac{2}{3}}$

I hope this clarifies things!

--Chris

13. thanks a lot i get it now