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Comment on Working with Formulas
You can solve this problem by
V(old cone) = 1/3(∏(r)(r)h) = K
V(transformed) = 1/3(∏(r/2)(r/2)(2h) = 1/3(r)(r)(h)(1/2)
substituting K for 1/3(∏(r)(r)h, we have K(1/2) or K/2.
Perfect solution!
Perfect solution!
Hello Brent,
I can't understand why these two questions have more than 2 values of x, I solved the 2 equations but you mentioned there are more than 2 values:
http://gre.myprepclub.com/forum/what-is-the-sum-of-all-solutions-to-the-equation-3114.html
http://gre.myprepclub.com/forum/topic3144.html
Question #1 link: http:/
Question #1 link: http://gre.myprepclub.com/forum/what-is-the-sum-of-all-solutions-to-the-...
I'm assuming that you found 2 solutions to the equation x^(2x² + 4x – 6) = x^(x² + 8x + 6) by equating the exponents to get the equation 2x² + 4x – 6 = x² + 8x + 6
This a good start. However, we need to also heed the advice that starts at 2:55 of the following video: https://www.greenlighttestprep.com/module/gre-powers-and-roots/video/1047
In that video, we examine some other possible conditions in which we can't assume that two exponents are equal.
For example, we need to consider what happens when the base equals 0.
Take this equation: 0^x = 0^y
Must it be the case that x = y? No.
When the base is 0, then x and y can have any values, and the equation 0^x = 0^y still holds true.
Similarly, we need to consider what happens when the base equals 1.
For example, 1^x = 1^y
Must it be the case that x = y? No.
When the base is 1, then x and y can have any values, and the equation 1^x = 1^y still holds true
So, in addition to equating the exponents, we must consider other cases(see linked video).
-----------------------------------
Question #2 link: http://gre.myprepclub.com/forum/topic3144.html
The same conditions apply here.
In this case, x = 0 is a solution to the equation (x²)^(x² - 2x + 1) = x^(3x² + x + 8)
To see why, replace x with 0 to get: (0²)^(0² - 2(0) + 1) = 0^(3(0²) + 0 + 8)
Simplify: 0^1 = 0^8
Evaluate: 0 = 0
Since x = 0 is one of the solutions, the PRODUCT of all solutions will be 0.
-------------------------------------
Does that help?
Cheers,
Brent
Yes it helps alot, thanks for
Hi Brent,
Just an FYI that the 160-170 question link does not seem to be working from me. Based on the URL text, it may be this question: the-pressure-of-a-fixed-mass-of-gas-increased-from-40-pounds-2157
Thanks for the heads up!
Thanks for the heads up!
I've fixed the link.
Hi Brent,
As for the last 160-170 question - why do we want to minimize the value of 16(t-3)^2? Thanks!
An object thrown directly upward is at a height of h feet after t seconds, where h = −16(t−3)² + 150. At what height, in feet, is the object 2 seconds after it reaches its maximum height?
A. 6
B. 86
C. 134
D. 150
E. 214
The formula h = -16 (t - 3)² + 150 allows us to determine the height of the object at any time.
For what value of t is -16(t-3)² + 150 MAXIMIZED(in other words, the object is at its maximum height)?
It might be easier to answer this question if we rewrite the formula as h = 150 - 16(t-3)²
To MAXIMIZE the value of h, we need to MINIMIZE the value of 16(t-3)² and this means minimizing the value of (t-3)²
As you can see,(t-3)² is minimized when t = 3.
We want to know the height 2 seconds AFTER the object's height is maximized, so we want to know that height at 5 seconds (3+2)
At t = 5, the height = 150 - 16(5 - 3)²
= 150 - 16(2)²
= 150 - 64
= 86
Answer: B
Our goal is to find the
Our goal is to find the height of the object 2 seconds AFTER it reaches its MAXIMUM height.
If we rewrite the formula as h = 150 - 16(t-3)², we can see that word starting with 150 and subtracting 16(t-3)²
So, the smaller we can make 16(t-3)², the greater the object's height will be.
Notice that the smallest value of 16(t-3)² occurs when t = 3.
So, when t = 3, the height is maximized.
Another way to approach this question is to test some values...
Height = 150 - 16(t-3)²
When t = 1, the height = 150 - 16(1-3)² = 150 - 16(-2)² = 150 - 64 = 86 feet.
When t = 2, the height = 150 - 16(2-3)² = 150 - 16(-1)² = 150 - 16 = 134 feet.
When t = 3, the height = 150 - 16(3-3)² = 150 - 16(0)² = 150 - 0 = 150 feet.
When t = 4, the height = 150 - 16(4-3)² = 150 - 16(1)² = 150 - 16 = 134 feet.
When t = 5, the height = 150 - 16(5-3)² = 150 - 16(2)² = 150 - 64 = 86 feet.
As we can see the height is maximized when t = 3
Does that help?