Test (No calculator) - 2 - SAT MCQ

To determine the price per pound of beef when it was equal to the price per pound of chicken, determine the value of x (the number of weeks after July 1) when the two prices were equal. The prices were equal when b = c; that is, when 2.35 + 0.25x = 1.75 + 0.40x. This last equation is equivalent to 0.60 = 0.15x, and so x = 0.60/0.15 = 4. Then to determine b, the price per
pound of beef, substitute 4 for x in b = 2.35 + 0.25x, which gives b = 2.35 + 0.25(4) = 3.35 dollars per pound.
Choice A is incorrect. It results from substituting the value 1, not 4, for x in b = 2.35 + 0.25x. Choice B is incorrect. It results from substituting the value 2, not 4, for x in b = 2.35 + 0.25x. Choice C is incorrect. It results from substituting the value 3, not 4, for x in c= 1.75 + 0.40x.

In the xy-plane, all lines that pass through the origin are of the form y = mx, where m is the slope of the line.
Therefore, the equation of this line is y = 1/7x, or x = 7y. A point with coordinates (a, b) will lie on the line if and only if a = 7b. Of the given choices, only choice D, (14, 2), satisfies this condition: 14 = 7(2).
Choice A is incorrect because the line determined by the origin (0, 0) and (0, 7) is the vertical line with equation x = 0; that is, the y-axis.
The slope of the y-axis is undefined, not 1/7. Therefore, the point (0, 7) does not lie on the line that passes the origin and has slope 1/7.
Choices B and C are incorrect because neither of the ordered pairs has a y-coordinate that is 1/7 the value of the corresponding x-coordinate.

To rewrite  multiply by  ThiS results in the expression  Which
is equivalent to the expression in choice B.
Choices A, C, and D are incorrect and could be the result of common algebraic errors that arise while manipulating a complex fraction.

One approach is to express so that the numerator and denominator are expressed with the same base. Since 2 and 8 are both powers of 2, substituting 2³ for 8 in the numeratorwhich can be rewritten as Since the numerator and denominator of have a common base, this expression can be rewritten as 2^3x-y. It is given that 3x - y = 12, so one can substitute 12 for the exponent, 3x - y, given that the expression is equal to 2¹².
Choice B is incorrect. The expression can be rewritten as , or 2^3x-y. If the value of 2^3x-y is 44, which can be rewritten as 28, then 2^3x-y = 28, which results in 3x - y = 8, not 12. Choice C is incorrect. If the value of  which results in 3x - y = 6, not 12.
Choice D is incorrect because the value of can be determined.

One can find the possible values of a and b in {ax + 2) (bx + 7) by using the given equation a + b = 8 and finding another equation that relates the variables a and b. Since (ax + 2)(hx + 7) = 15x² + ex + 14, one can expand the left side of the equation to obtain abx² + 7ax + 2bx + 14 = 15x² + ex + 14. Since ab is the coefficient of x2 on the left side of the equation and 15 is the coefficient of x² on the right side of the equation, it must be true that ab = 15. Since a + b = 8, it follows that b = 8- a. Thus, ab = 15 can be rewritten as a 8- a) = 15, which in turn can be rewritten as a² - 8a + 15 = 0. Factoring gives (a - 3)(a -5) = 0. Thus, either a = 3 and b = 5, or a = 5 and b = 3.If a = 3 and b = 5, then (ax + 2)(hx + 7) = (3x + 2)(5x + 7) = 15x² + 31x + 14. Thus, one of the possible values of c is 31. If a = 5 and b = 3, then (ax + 2)(hx + 7) = (5x + 2)(3x + 7) = 15x² + 41x + 14. Thus, another possible value for c is 41. Therefore, the two possible values for c are 31 and 41.
Choice A is incorrect; the numbers 3 and 5 are possible values for a and b, but not possible values for c. Choice B is incorrect; if a = 5 and b = 3, then 6 and 35 are the coefficients of x when the expression (5x + 2)(3x + 7) is expanded as 15x² + 35x + 6x + 14. However, when the coefficients of x are 6 and 35, the value of c is 41 and not 6 and 35. Choice C is incorrect; if a = 3 and b = 5, then 10 and 21 are the coefficients of x when the expression (3x + 2)(5x + 7) is expanded as 15x² + 21x + 10x + 14. However, when the coefficients of x are 10 and 21, the value of c is 31 and not 10 and 21.

The correct answer is 2. To solve for t, factor the left side of t² - 4 = 0, giving (t - 2)(t + 2) = 0. Therefore, either t - 2 = 0 or t + 2 = 0.If t - 2 = 0, then t = 2, and if t + 2 = 0, then t = -2. Since it is given that t > 0, the value of t must be 2.
Another way to solve for t is to add 4 to both sides of t² - 4 = 0, giving t² = 4. Then, taking the square root of the left and the right side of the equation gives t = ±√4 = ± 2. Since it is given that t >0, the value of t must be 2.

It is given that ∠AEB and ∠CDB have the same measure. Since ∠ABE and ∠CBD are vertical angles, they have the same measure. Therefore, triangle EAB is similar to triangle DCB because the triangles have two pairs of congruent corresponding angles (angle-angle criterion for similarity of triangles). Since the triangles are similar, the corresponding sides are in the same  proportion; thus  Substituting the given values of 800 for CD, 700 for BD, and 1400 for EB in  Therefore, 

Subtracting the left and right sides of x + y = -9 from the corresponding sides of x + 2y = -25 gives (x + 2y) -{x + y = -25 - (-9)), which is equivalent to y = -16. Substituting -16 for y in x + y = -9 gives x + (-16) = -9, which is equivalent to x = -9- (-16) = 7.

The correct answer is 4/5 or 0.8. By the complementary angle relationship for sine and cosine, sin(x°) = cos(90° - x°). Therefore, cos(90° - x°) = 4/5. Either the fraction 4/5 or its decimal equivalent, 0.8, maybe gridded as the correct answer.
Alternatively, one can construct a right triangle that has an angle of mea sure x° such that sin(x°) = 4/5, as shown in the figure below, where sin(x°) is equal to the ratio of the length of the side opposite the angle measuring x° to the length of the hypotenuse, or 4/5.

Since two of the angles of the triangle are of measure x° and 90°, the third angle must have the measure 180° - 90° - x° = 90° - x°. From the figure, cos(90° - x°), which is equal to the ratio of the length of the side adjacent to the angle measuring 90° - x° to the hypotenuse, is also 4/5.

Since a = 5√2, one can substitute 5√2 for a in 2a = , giving 10√2 = . Squaring each side of 10√2 =  gives which simplifies to  or 200 = 2x. This gives x = 100. To verify, substitute 100 for x and 5√2 for a in the equation 2a = √2x, which yields 2(5√2) =  this is true since