In triangle ABC, M is the midpoint of BC, AB=12, and AC=16. Points E and F are taken on AC and AB respectively, and EF and AM intersect at G. If AE=2AF then what is EG/GF?
\(\text{Let $\angle EAG=\epsilon_1$ } \\ \text{Let $\angle AMC=\epsilon_2$ } \\ \text{Let $\angle AGE=\epsilon_3$ } \\ \text{Let $\angle GAF=\delta_1$ } \\ \text{Let $\angle AMB=\delta_2=180^\circ-\epsilon_2 $ } \\ \text{Let $\angle AGF=\delta_3=180^\circ-\epsilon_3 $ } \\ \text{Let $EG=\mathbf{x}$ } \\ \text{Let $GF=\mathbf{y}$ } \\ \text{Let $AF=f$ } \\ \text{Let $AE=2AF=2f$ } \)
1. sin-rule
\(\begin{array}{|lrcll|} \hline (1) & \dfrac{\sin(\epsilon_1)}{\dfrac{CB}{2}} &=& \dfrac{\sin(\epsilon_2)}{16} \\\\ \hline & \dfrac{\sin(\delta_1)}{\dfrac{CB}{2}} &=& \dfrac{\sin(180^\circ-\epsilon_2)}{12} \\\\ (2) & \dfrac{\sin(\delta_1)}{\dfrac{CB}{2}} &=& \dfrac{\sin(\epsilon_2)}{12} \\\\ \hline & \dfrac{CB}{2}\sin(\epsilon_2) = 16\sin(\epsilon_1)&=& 12\sin(\delta_1) \\ & 16\sin(\epsilon_1)&=& 12\sin(\delta_1) \\\\ & \dfrac{\sin(\epsilon_1)}{\sin(\delta_1)} &=& \dfrac{12}{16} \\\\ (3) & \mathbf{ \dfrac{\sin(\epsilon_1)}{\sin(\delta_1)} } &=& \mathbf{ \dfrac{3}{4} } \\ \hline \end{array} \)
2. sin-rule
\(\begin{array}{|lrcll|} \hline (4) & \dfrac{\sin(\epsilon_1)}{x} &=& \dfrac{\sin(\epsilon_3)}{2f} \\\\ \hline & \dfrac{\sin(\delta_1)}{y} &=& \dfrac{\sin(180^\circ-\epsilon_3)}{f} \\\\ (5) & \dfrac{\sin(\delta_1)}{y} &=& \dfrac{\sin(\epsilon_3)}{f} \\\\ \hline & \dfrac{\sin(\epsilon_3)}{f} = \dfrac{2\sin(\epsilon_1)}{x} &=& \dfrac{\sin(\delta_1)}{y} \\\\ &\dfrac{x}{y} &=& 2\cdot \dfrac{\sin(\epsilon_1)}{\sin(\delta_1)} \quad | \quad \dfrac{\sin(\epsilon_1)}{\sin(\delta_1)} = \dfrac{3}{4} \\\\ &\dfrac{x}{y} &=& 2\cdot \dfrac{3}{4} \\\\ & \mathbf{ \dfrac{x}{y} } &=& \mathbf{ \dfrac{3}{2} } \\ \hline \end{array}\)
\(\mathbf{\dfrac{EG}{GF} = \dfrac32}\)
Thank you, CPhill !
....
CPhill i think you meant S+5>3
bruhhh now i am confused too LOL
3)
(2i)x^2 + x + 3i = 0 factor out the 2i
2i ( x^2 + (1/2i)x + 3/2) = 0 divide both sides by 2i
x^2 + (1/2i)x + 3/2 = 0
x^2 + (1/(2i))x = - 3/2 complete the square on x
Take (1/2) of (1 /(2i)) = (1/(4i))
Square this = (1/(4i)) (1/ (4i)) = 1/(16i^2) = 1/ -16 = -1/16
Add this to both sides
x^2 + (1/(2i))x - 1/16 = -3/2 - 1/16 factor the left.....simplify the right
(x + (1/(4i))^2 = -25/16 take both roots
x + (1/(4i) ) = ±(5/4)i subtract (1/(4i)) from both sides
x = ±(5/4)i - (1/(4i)) multiply top/bottom of the last fraction by i
x = ±(5/4)i - i / (4i^2)
x = ±(5/4)i - (i / (-4) )
x = ±(5/4)i + i/4
x = ±(5/4)i + (1/4)i
x = (5/4)i - (1/4) i = (4/4)i = i
or
x = (-5/4)i - (1/4)i = (-6/4)i = (-3/2)i
So.....the solutions are
x = (-3/2)i and x = i
THX, guest !!!
Thanks, Oofrence.....that was an excellent answer !!!!
(2/3)t - 1 < t + 7 ≤ -2t + 15
We can split this up into two inequalities
(2/3)t - 1 < t +7 t + 7 ≤-2t + 15
add 1 to both sides, add 2t to both sides,
subtract t from both sides subtract 7 from both sides
(-1/3)t < 8 3t ≤ 8
Multiply both sides by - 3 divide both sides by 3
Change the direction of the inequality sign
t > - 24 t ≤ 8/3
So ....the answer is that
-24 < t ≤ 8/3
There must be 2 math books on either side, and there are 3 to choose from, so there are 3 possible books for one end, and 2 for the other. Therefore there are 6 possible ways to arrange the books.
For every possible way to arrange the two math books at either end, there is a number of possibilities to sort the 1 math textbook and 5 history textbooks. Then to find the possible ways of arranging you would do 6! = 720
720 x 6 = 4320 ways to arrange the books
https://web2.0calc.com/questions/more-questions-halp has the same problem
Here is a shorter way:
5/2 =2.5 - total growth after 30 hours.
2.5^(1/30) =1.03101424786... growth rate per hour.
1.03101424786^25 =2.145935547,,, x 2 =~ 4.29 mg - exponential growth after 25 hours.
Oh i didn't think about the discriminant, thanks!
To have an integer result.....the quantity under the root must be a perfect square
So...we will have 13 values
To see this.....when x = 0 the result will be 12
When x = (144)^3 the result will be 0
So.....we can generate all the integers from 0 - 12 by taking the square root of the quantity under the radical .....so.....13 values of x
The expression: Sqrt(144 - x^(1/3)) is an integer for the following values of x:
x=0, 12,167, 85,184, 250,047, 512,000, 857,375, 1,259,712, 1,685,159, 2,097,152, 2,460,375, 2,744,000, 2,863,288, 2,924,207 = 13 real values of x.
Here's a slightly shorter approach
Note that 4^2 = 16
Which must mean that
log 7 x = 2
So....in exponential form 7^2 = x = 49
We need to solve this
243 (2/3)^x = 30 divide both sides by 243
(2/3)^x = 30/243
(2/3)^x = 10/81 tak the log of both sides
log(2/3)^x = log (10/81) and we can write
x log(2/3) = log (10/81) divide both sides by log(2/3)
x = log (10/81) / log(2/3) ≈ 5.16 = after the 6th bounce
Check:
After the 5th bounce the ball will rise to 243(3/4)^5 = 32 cm
After the 6th bounce the ball will rise to 243(2/3)^6 = 21.3 cm
13, wait you're 10?
that means that hectictar is 01? Me is very confused...
10 pi ≈ 31.4
So we have 31 positive integers + 31 negative integers + 0 = 63 integers
On the first 8 squares we have
2^1 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8 =
2 + 4 + 8 + 16 + 32 + 64 + 128 + 256 =
510 grains (1)
On the 10th square we have 2^10 = 1024 grains (2)
Subtract (2) - (1) for your answer
The only values of x where the domain is not defined is where x^2 - 4 = 0
So
x^2 - 4 = 0
(x + 2)(x - 2) = 0
Setting each factor to 0 and solving for x gives us that x = -2 and x = 2
Look at the graph here : https://www.desmos.com/calculator/ph4j7rnvab
The domain exists on (-infinity, -2) U (-2, 2) U (2, infinity)
Ah thank you so much, I understand now!
Solve for x over the real numbers: 4^(log(x)/log(7)) = 16
Take the logarithm base 4 of both sides: log(x)/log(7) = 2
Multiply both sides by log(7): log(x) = 2 log(7)
2 log(7) = log(7^2) = log(49): log(x) = log(49)
Cancel logarithms by taking exp of both sides: x = 49
2)
Rearrange as
x^2 -2x + y^2 + 4y = -5 complete the square on x, y
x^2 - 2x + 1 + y^2 + 4y + 4 = -5 + 1 + 4
(x - 1)^2 + (y + 2)^2 = 0
This is known as a "degenerate" circle ( circle with radius = 0)
It has a "center" of ( 1, - 2) = (x, y)
I know a couple
y = x^2 + 2x + 7
y = 6x + b
x^2 + 2x + 7 = 6x + b subtract 6x + b from both sides
x^2 - 4x + (7 - b) = 0
If we have only one intersection point......the discriminant will = 0
(-4)^2 - 4(1)(7 - b) = 0 simplify
16 - 28 + 4b = 0
-12 + 4b = 0
4b = 12
b = 3
Look at the graph here : https://www.desmos.com/calculator/2eit022nxs
The intersection point is (2, 15)
Haha, so your 11?