SAT Math

**OnMyWayToBeAMathProffesor** · July 7th 2008, 05:24 PM

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 $x^2=y^2-4$ intersects line $l$ at $(0,p)$ and $(5,t)$ . That is the greatest possible slope of line $l$ ? 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.

**Chop Suey** · July 7th 2008, 08:08 PM

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:

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

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 $y_{2} = \pm a$ and $y_{1} = \pm b$ , the highest value would be:
$y_{2}-y_{1} = (+a) - (-b) = a + b$

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.

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

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

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

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

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

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

**Moo** · July 8th 2008, 02:08 AM

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 = ±a and y1 = ±b, 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 $t_1+t_2=1$

Hey tasty dish ! that was good !

**Chop Suey** · July 8th 2008, 02:12 AM

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.

**Moo** · July 8th 2008, 02:14 AM

Hmm how did you get that ?

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

$t=\pm \dots$

**Chop Suey** · July 8th 2008, 02:20 AM

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

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

**Moo** · July 8th 2008, 02:23 AM

You're correct

Sorry for the confusion !

**OnMyWayToBeAMathProffesor** · July 19th 2008, 05:44 PM

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 $x=y^2-4$ which in terms of $y$ would be $y=\sqrt{(x+4)}$ and $y=-\sqrt{(x+4)}$ . This is a horizontal parabola that open to the right with a vertex at $(0,4)$ . when $x=0$ then $y$ is $2$ AND $-2$ since it is a parabola. when $x=5$ then $y$ is $3$ AND $-3$ . So if a line intersects at $(0,p)$ and $(5,t)$ , the greatest slope would be 5 because if the two points $(5,3)$ and $(0,-2)$ are chosen, then $<br /> m = \frac{3-(-2)}{5 - 0}$ , 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 $45 = \frac{d}{t_{1}}$ and $30 = \frac{d}{t_{2}}$ also, how do i get the ratio of $\frac{t_{1}}{t_{2}}$ ? Thanks a lot again everybody!

**Chris L T521** · July 19th 2008, 08:58 PM

Originally Posted by OnMyWayToBeAMathProffesor

How would i equate $45 = \frac{d}{t_{1}}$ and $30 = \frac{d}{t_{2}}$ also, how do i get the ratio of $\frac{t_{1}}{t_{2}}$ ? Thanks a lot again everybody!

Since both equations have the same value $d$ , we should solve each equation for $d$ . This would then enable us to set the two expressions equal to each other.

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

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

Set the equations equal to each other:

$45t_1=30t_2$

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

Does this clarify things?

--Chris

**OnMyWayToBeAMathProffesor** · July 20th 2008, 06:34 AM

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

**Chop Suey** · July 20th 2008, 10:42 AM

Nonononoo!!!

$45t_1=30t_2$

Simply cross multiply and move the 45 and t_2:

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

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

To find your distance:

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

ONWTBAMP: You need to revise ratios.

**Chris L T521** · July 20th 2008, 11:11 AM

Originally Posted by OnMyWayToBeAMathProffesor

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

If $45t_1=30t_2$ , then $45\frac{t_1}{t_2}=30\implies \frac{t_1}{t_2}=\frac{30}{45}\implies\color{red}\b oxed{\frac{t_1}{t_2}=\frac{2}{3}}$

I hope this clarifies things!

--Chris

**OnMyWayToBeAMathProffesor** · July 20th 2008, 11:36 AM

thanks a lot i get it now

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