New SAT Test Quiz

Choice D is correct. It’s given that at the beginning of the 1st week of the year there was $600 in a savings account and Gabriella deposits $35 in that savings account at the end of each week. Therefore, the amount of money, in dollars, in the savings account at the end of the 4th week of that year is 600 + 4(35), or 740.

Using the slope and the point-slope form from the two points in the table, b is calculated as 33.

Choice A is correct. For the linear function f, it’s given that f(7) = 28. Substituting 7 for a and 28 for f(æ) in the given function yields 28 = 4(7) + b, or 28 = 28 + b. Subtracting 28 from each side of this equation yields 0 = b. Therefore, the value of b is 0.

Choice B is correct. Substituting 72 for f(x) in the given function yields 72 = 8x. Dividing each side of this equation by 8 yields 9 = x. Therefore, f(x) = 72 when the value of x is 9.

The correct answer is 17. It’s given that 2x +3 = 9. Multiplying each side of this equation by 3 yields 3(2x +3) = 3(9), or 6x + 9 = 27. Subtracting 10 from each side of this equation yields 6x + 9 – 10 = 27 – 10, or 6x – 1 = 17. Therefore, the value of 6x – 1 is 17.

Choice C is correct. It’s given that the function f is defined by f(x) = 25x + 30. Substituting 2 for a in this equation yields f(2)=25(2)+30, which is equivalent to f(2)=50+30, or f(2) = 80. Therefore, the value of f(x) is 80 when x = 2.

The slope of line k is -17. The perpendicular slope is the negative reciprocal: 1/17 ≈ 0.176 or 3/17.

Choice D is correct. A line in the xy-plane that passes through the points (x1, y1) and (x2, y2) has slope m, where m = y2 1/1, and can be defined by an equation of the form y – y1 =m(x – x1). One of the lines shown in the graph passes through the points (8, 0) and (3,4). Substituting 8 for 1, 0 for y1, 3 for 2, and 4 for y2 in the equation m =2y1 yields m = 6, orm =. Substituting – for m, 8 for #1 and 0 for y1 in the equation y-y1=m(-1) yields y-0=( -8), which is equivalent to y= + 2 .Adding x to both sides of this equation yields +y = . Multiplying both sides of this equation by -10 yields -8x – 10y= -64. Therefore, an equation of this line is -8x – 10y = -64. Similarly, the other line shown in the graph passes through the points (8, 0) and (3,2). Substituting 8 for #1, 0 for y1, 3 for #2, and 2 for y2 in the equation m = 12 M1 yields m = 2-0, or m = -. Substituting – for m, 8 for 1, and 0 for y1 in the equation y – y/1 =m(x -1) yields y – 0 =- (x – 8), which is equivalent to y=a +. Adding x to both sides of this equation yields x + y = 6. Multiplying both sides of this equation by 10 yields 4x + 10y = 32. Therefore, an equation of this line is 4x + 10y = 32. So, the system of linear equations represented by the lines shown is 4x + 10y = 32 and -8x – 10y =- 64.Choice A is incorrect and may result from conceptual or calculation errors.Choice B is incorrect and may result from conceptual or calculation errors.Choice C is incorrect and may result from conceptual or calculation errors.

The correct answer is 3. Adding the second equation to the first equation in the given system of equations yields (2(8x)-2(8x)) + (4(7y) + 4(7y)) = 12+ 12, or 8(7y) = 24. Dividing both sides of this equation by 8 yields 7y = 3. Substituting 3 for 7y in the first equation, 2(8x) + 4(7y) = 12, yields 2(8x) + 4(3) = 12, or 2(8x) + 12 = 12. Subtracting 12 from both sides of this equation yields 2(8x) = 0. Dividing both sides of this equation by 2 yields 8x = 0. Substituting 0 for 8x and 3 for 7y in the expression 8x + 7y yields 0 + 3, or 3. Therefore, the value of 8x + 7y is 3.

Choice C is correct. Let a equal the number of 120-pound packages, and let b equal the number of 100-pound packages. It’s given that the total weight of the packages can be at most 1,100 pounds: the inequality 120a+100b ≤ 1,100 represents this situation. It’s also given that the helicopter must carry at least 10 packages: the inequality a+ b ≥ 10 represents this situation. Values of a and b that satisfy these two inequalities represent the allowable numbers of 120-pound packages and 100-pound packages the helicopter can transport. To maximize the number of 120-pound packages, a, in the helicopter, the number of 100-pound packages, b, in the helicopter needs to be minimized. Expressing b in terms of a in the second inequality yields b ≥ 10- a, so the minimum value of b is equal to 10- a. Substituting 10- a for b in the first inequality results in 120a+100(10-a) ≤ 1,100. Using the distributive property to rewrite this inequality yields 120a+1,000-100a≤ 1,100, or 20a+ 1,000 ≤ 1,100. Subtracting 1,000 from both sides of this inequality yields 20a ≤ 100. Dividing both sides of this inequality by 20 results in a ≤ 5. This means that the maximum number of 120-pound packages that the helicopter can carry per trip is 5.

Choices A, B, and D are incorrect and may result from incorrectly creating or solving the system of inequalities.