For question #1, the answer can only be 28 right? You can bring f(x) = (x-a)(x-b) into standard quadratic form which is f(x) = x^2 + (-a-b)x + ab. Then from here you can say that [-b] / [2a] = 14 and substitute b and a values from f(x). The result is [-(-a-b)] / [2(1)] = 14 Solving for a+b gets you 28. I'm struggling to see how 27, 28, and 29 all could be correct answers to question #1. Thanks
"Then from here you can say that [-b] / [2a] = 14" Why are you assuming that -b/2a = 14, i.e. that vertex is at (14, k)? There is nothing provided in the problem which tells us that h must be equal 14.
@@jwmathtutoring Thanks for responding! I'm assuming that the vertex is (14,k) because the x coordinates they gave us (11, 14, and 17) suggest that 14 is the vertex. This is because 11 and 17 are equidistant to 14 (they are both 3 units away from 14 in either direction) and the only way this is possible in a parabola is if 14 was the vertex. Or at least that was my thought process. Thanks.
@@jwmathtutoring Actually nevermind, I realized that my assumption that 14 is h was wrong and that it doesn't have to be the case. Thanks for pointing that out, I played around with Desmos and this graph and realized my mistake. Thanks for all your help!
Wouldn't the acute angles in question 4 simply be 45 degrees each since the right triangles that are formed are isosceles? The value of k here doesn't matter right?
In the original triangle, yes. But not the new one, because you're adding the value of k to the point (3,2) not to (3, 6). If the new point was (3, 6+k), then yes it would remain an isosceles triangle, but it isn't.
@@rhearajashekhar2914 For that, I believe you could graph it and then move the slider for b until you have no x-ints. Basically when there are no x-ints, there are no real solutions and the expression is not factorable.
You'd have to do some trial & error with the factors of 1140. Edit: Actually it does appear to be 68 and the way you can confirm is to graph the function in Desmos with the b in place, and then do a slider for b. Move the slider until it touches/crosses the x-axis (i.e. has real solutions, meaning it is factorable) and that doesn't happen until b = 68.
If you're asking about Question #3, it's because n can be any value. Using n = 1 just makes it easier to understand because the exponent doesn't have a fraction in it.
For question #1, the answer can only be 28 right?
You can bring f(x) = (x-a)(x-b) into standard quadratic form which is f(x) = x^2 + (-a-b)x + ab.
Then from here you can say that [-b] / [2a] = 14 and substitute b and a values from f(x).
The result is [-(-a-b)] / [2(1)] = 14
Solving for a+b gets you 28.
I'm struggling to see how 27, 28, and 29 all could be correct answers to question #1.
Thanks
"Then from here you can say that [-b] / [2a] = 14"
Why are you assuming that -b/2a = 14, i.e. that vertex is at (14, k)? There is nothing provided in the problem which tells us that h must be equal 14.
@@jwmathtutoring Thanks for responding!
I'm assuming that the vertex is (14,k) because the x coordinates they gave us (11, 14, and 17) suggest that 14 is the vertex.
This is because 11 and 17 are equidistant to 14 (they are both 3 units away from 14 in either direction) and the only way this is possible in a parabola is if 14 was the vertex.
Or at least that was my thought process.
Thanks.
@@jwmathtutoring
Actually nevermind, I realized that my assumption that 14 is h was wrong and that it doesn't have to be the case.
Thanks for pointing that out, I played around with Desmos and this graph and realized my mistake.
Thanks for all your help!
Wouldn't the acute angles in question 4 simply be 45 degrees each since the right triangles that are formed are isosceles? The value of k here doesn't matter right?
In the original triangle, yes. But not the new one, because you're adding the value of k to the point (3,2) not to (3, 6). If the new point was (3, 6+k), then yes it would remain an isosceles triangle, but it isn't.
Can i find any question similiar to these on the internet?The one in the questuon bank are easier.
Does College Board allow for these types of wakthroughs of Digital SAT Past Papers?
how would you factor #5 if it asked for the smallest value of b?
@@rhearajashekhar2914 For that, I believe you could graph it and then move the slider for b until you have no x-ints. Basically when there are no x-ints, there are no real solutions and the expression is not factorable.
For question 8 are we just expected to know the formula for the surface area of a cylinder??
@@Jason-uq5lz It has typically been provided on the test for questions which require it but can't guarantee it always will be.
Please i dont understand number1, how you got the y values
Hi, I'm not sure what you mean. All the values used in the problem are x values (the 12, 13 and 15, 16). Can you explain further?
For no. 5, what if the question was minimum value of b?
You'd have to do some trial & error with the factors of 1140.
Edit: Actually it does appear to be 68 and the way you can confirm is to graph the function in Desmos with the b in place, and then do a slider for b. Move the slider until it touches/crosses the x-axis (i.e. has real solutions, meaning it is factorable) and that doesn't happen until b = 68.
@@jwmathtutoring what about doing the same thing but in the opposite direction? Like (38z + 30)(z + 1)
I edited my comment. Did you see the new verson?
Why desmos solution sub n=1
If you're asking about Question #3, it's because n can be any value. Using n = 1 just makes it easier to understand because the exponent doesn't have a fraction in it.
@@jwmathtutoring thank you
Messi is the goat