Need help with SAT 2, Math level 2 problems?

Question

Its been a while since i took precal and cal and im rusty on some things. I just took a practice test and here are some problems i need someone to explain how to do. There are 7 or 8 problems here, if you can do any or all of them, it would greatly help.

Would really appreciate the help + 10 points to whoever helps

1) What is the length of a major axis of the ellipse whose equation is 60x^2 + 30y^2 = 150

Answer is 4.47 but i want to know how to solve this. I presume its with a formula i cant quite remember.

2) In how many ways can 10 people be divided into two groups, one with 7 people and the other with 3 people?

Answer is 120, but again, want to know how to solve this. I think its with a permutation/combination formula i cant remember.

3) If the 20th term of an arithmetic sequence is 100 and the 40th term of the sequence is 250, what is the first term of the sequence?

Answer is -42.5, i think it has to do with the artihmatic sequence formula, but cant rmbr how to do it =(

4) If (6.31)^m = (3.02)^n what is the value of m/n?
This has to do with rule of logs/ln and i tried doing it, but i must have messed up somewhere. Answer is .6, not sure how to get this.

5) If f(x) = log(base2)x for x>0, then f(x)^-1 (The inverse of f(x) basically) = ?

Answer is log(basex)2 but i dont know how to do this.

6)If 3x-4y+7 = 0 and 2y-x^2 = 0 for x>=0 , then x = ?

Answer is 2.77, not sure how to do this

7) What is the degree measure of the largest angle of a triangle of length 7,6 and 6?

Answer is 71.37%. This one doesnt seem hard but when i tried to solve it it wouldnt work, so if anyone can do it..

Answer 1

1) Given equation can be rewritten as ,x²/ (5/2) + y² / 5 = 1

here a² = 5/2 and b² = 5 { here b > a }

so length of major axis = 2b ==> 2√(5) = 4.472

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2) select 7 people out of 10, second group of 3 people will automatically be formed.



so, no of ways = 10 C 3 ==> (10 * 9 * 8 * 7 ! )/ (3 ! * 7 !)



=> 120

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3) Given: T(20) = 100 & T(40) = 250



=> 100 = a + 19d & 250 = a + 39d



=> 20d = 150 ==> d = 7.5 therefore, a = 100 - 19 * 7.5



=> a = - 42.5

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4) Given; (6.31)^m = (3.02)^n



=> m (log 6.31) = n (log 3.02)



=> (m/n) = (log 3.02) / (log 6.31)



=> (m/n) = 0.48 / 0.80 => 48/80



=> (m/n) = 3/5

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5) let y = log ( base2 ) x



=> x = 2^y



=> f^(-1)y = 2^y



=> f^(-1)x = 2^x

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6) 2y = (3x + 7) / 2



=> (3x + 7) / 2 - x² = 0



=> 3x + 7 - 2x² = 0 ==> 2x² - 3x - 7 = 0



=> x = [3 ± √(-3)^2 + 56)] / 4



=> x = [3 + √65)] / 4 { x > 0 }



=> x = 2.77

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7) let largest angle be θ ( opposite to side 7 )



sin (θ/2) = 3.5/6 ==> sin (θ/2) = 7/12



=> θ/2 = 35.68 ==> θ = 71.37 degrees

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Answer 2

1)

Put 60x^2 + 30y^2 = 150 in standard form. That means divide each term by 150 so you get 1 on the right. The standard form of ellipse is x^2/a^2 + y^2/b^2 = 1. You have to make your equation look like that.



You should end up with 60x^2/150 +30y^2/150 = 1

This reduces to 2x^2/5 + y^2/5 = 1.

We are not in standard form yet! Standard form has no coefficients in the numerator (top) of the fraction. To get rid of it, you flip the fraction and put in on the bottom. They are equivalent... it's like how 2 = 1/(1/2). So the standard form of this ellipse is x^2/(sq. rt(5/2))^2 + y^2/(sq rt.((5))^2 = 1. You have to take a square root and then square it if you don't have a perfect square like 4, 9, 16, etc.



The major axis is the larger one. So the equation is telling you to go sq rt.(5/2) right and left from the origin (x direction is horizontal), and sq rt. (5) up and down (the y direction is vertical). Those are the boundaries of the ellipse. The major axis is twice the length of sq rt.(5), because otherwise you'd just have half the length! This value on your calculator, 2(sq rt. (5)) = 4.47214.





2) and 3)

There are formulas for these in any standard algebra or pre-calculus text. I don't have any way of writing out the notation for them. You could probably find them on Wikipedia.



4)



(6.31)^m = (3.02)^n



Take the logarithm of both sides.



ln(6.31)^m = ln(3.02)^n



Use the property of logs that says you can bring the power down to the front.



mln(6.31) = nln(3.02)



divide both sides by ln(6.31) to get m by itself.



m = nln(3.02)/ln(6.31)



now divide both sides by n to get n on the other side



m/n = ln(3.02)/ln(6.31)



Go on your calculator and do m/n = ln(3.02)/ln(6.31). You should get .5999 = .6





5) f(x) = log(base2)x



To find the inverse, change f(x) to y. Then switch x and y. Then solve for y. This new function is your inverse.



y = log(base2)x

x = log(base2)y



To get rid of a logarithm and get y by itself, all you have to do is create a base 2 on both sides to cancel out the log.



2^x = 2^(log(base2)y)



I can do this because it's a property of logarithms. It's less common, so you may not have seen it.



2^x = y (everything else that was on the right side cancels out)



f^-1(x) = 2^x



I'm not sure why you have log(basex)2 as the answer. Why would the variable become a base? The inverse of a log function is always an exponential, and vice-versa.







6)

If 3x-4y+7 = 0 and 2y-x^2 = 0



This is a system of equations in 2 variables. It can be solved with the substitution, elimination or graphing methods. I'll do substitution. That means solve one of your equations for x or y (it doesn't matter which equation or which variable you choose) and then put it into the OTHER equation.



I will solve 2y-x^2 = 0 for y.



2y = x^2 (add x^2 to both sides)



divide both sides by 2 to get y by itself



y = x^2/2 <----- This is what you plug into the other equation



3x-4y+7 = 0



3x-4(x^2/2)+7 = 0



3x-2x^2+7 = 0 The 4/2 becomes a 2



-2x^2 + 3x + 7 = 0



This is a QUADRATIC equation and can be solved by using the quadratic formula. I can't type that here. It's an easy formula that you plug stuff into. Search for it on google, then put in a = -2, b = 3 and c = 7.





7) A triangle with 2 equal sides is isoceles. That means you can split it down the middle and get 2 right triangles. The sides of the new triangle are 7/2 (horizontal), 6 diagonal and vertical unknown. Use inverse cosine of (7/2)/6 To get the angle of the side whose length is 6. I get 54.3147 degrees. The isoceles triangle we started with has 3 angles whose sum is 180 degrees, 2 of which are 54.3147 degrees. So 180 - 54.3147 - 54.3147 = 71.3707 degrees, which is opposite the largest side.

Answer 3

Since this is an SAT Math Level 2 problem derivatives should not be required to find the solution. To find "How many more hours of daylight does the day with max sunlight have than May 1," all you need to understand is that sin(x) has a maximum value of 1. The day with max sunlight will occur when sin(2*pi*t/365) = 1, giving the max sunlight to be 35/3 + 7/3 = 14 hours Evaluating your equation for sunlight when t = 41, May 1 will have about 13.18 hours of sunlight. The difference is about 0.82 hours of sunlight. Even though it is unnecessary for this problem, finding the actual max sunlight day can be done by solving for t when d = 14, of by the use of calculus. Common min/max problems on the SAT Math Level 2 involve sin and cos, which both have min values of -1 and max values of 1, and also polynomial functions with only even powered variables or variable expressions, which have a min/max when the variable or variable expression equals 0. For example, f(x) = (x-2)^4 + 4 will have a min value of 4 when x = 2.