HI Mike…Thanks for all your help ! Here’s another question

HI Mike…Thanks for all your help ! Here’s another question:

When a buffet restaurant charges $12.00 per meal, the number of meals it sells per day is 400. For each $0.50 increase to the price per meal, the number of meals sold per day decreases by 10. What is the price per meal that results in the greatest sales, in dollars, from meals each day?

A) $16.00
B) $20.00
C) $24.00
D) $28.00

QAS March 2021, Section 3, # 20.

Hi Mike… QAS March 2021, Section 3, # 20. Since “no solution,” can you consider each side a separate parallel line & use corresponding coefficients to confirm that k must equal 1/2? Or, what’s the best way to solve?

1/2x + 5 = kx + 7

In the given equation, k is a constant. The equation has no solution. What is the value of k?

March 21 QAS Sec 3 #11

Hi Mike …can you solve & explain? (from QAS March 2021)
Section 3; #11.
y = (x-1)(x+1)(x+2)
The graph in the xy plane of the equation above contains the point (a,b). If -1 < or = a < or = 1, which of the following is NOT a possible value of b?
A) -2
B) -1
C) 0
D) 1

Hi, Mike. How would you say the two math sections of the SAT are different (besides the obvious calculator use and the timing/number of questions)? I’m thinking in terms of content covered, types of questions, etc. Thanks!

Hi, Mike. How would you say the two math sections of the SAT are different (besides the obvious calculator use and the timing/number of questions)? I’m thinking in terms of content covered, types of questions, etc. Thanks!

Hi! Love your book so far. I just started reviewing…

Hi! Love your book so far. I just started reviewing. I looked at number 10 in the first section on page 29 and I was wondering how would you pick the correct answer if you chose to substitute 8 for d? Because both answer choices a and d work. Would you just try all the answer choices and choose different variable values and try again or is there a different way to approach it?

What am I not seeing?

4x + y = 7
2x – 7y = 1

If I multiply the second equation by 2, I can stack them and subtract:

4x + y = 7
4x – 14y = 2

So, 15y = 5, —> y = 3

Then: 2x – 7(3) = 1 –> 2x – 21 = 3 —> 2x = 24 –> x = 12
But: 4x + 3 = 7 –> 4x = 4 –> x = 1

What am I not seeing? The answer should be x= 5/3.

In this equation, k is a constant…

Some help please, Mike?

x^2 – 12x +k = 0 In this equation, k is a constant. For which values of k does the equation have only one solution? I know I can set the discriminant to zero and solve for k. But is there another way to solve? Thanks!