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SSAT Upper Level Math : SSAT Upper Level Quantitative (Math)

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All SSAT Upper Level Math Resources

6 Diagnostic Tests 311 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #12 : How To Graph Complex Numbers

Add 4+ 7i to its complex conjugate. What is the result?

Possible Answers:

14i

11-3i

11 + 11i

Correct answer:

Explanation:

The complex conjugate of a complex number a + bi is a - bi , so 4+7i has 4-7i as its complex conjugate; the sum of the two numbers is

(4+7i)+ (4-7i) = 4+4 + 7i - 7i = 8

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Example Question #13 : How To Graph Complex Numbers

Multiply the complex conjugate of 3 + 7i by . What is the result?

Possible Answers:

28 + 12i

-12 + 28i

-28+12i

12 + 28 i

12 - 28 i

Correct answer:

28 + 12i

Explanation:

The complex conjugate of a complex number a + bi is a - bi , so the complex conjugate of 3 + 7i is 3 - 7i . Multiply this by :

$(3-7i) \cdot 4i = 3 \cdot 4i - 7 i \cdot 4i$

$= 12i - 7 \cdot 4 \cdot i \cdot i$

$= 12i -28 \cdot (-1)$

= 12i + 28

= 28 + 12i

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Example Question #631 : Ssat Upper Level Quantitative (Math)

Multiply the complex conjugate of 8 by 5 i . What is the result?

Possible Answers:

None of the other responses gives the correct product.

40i

-40i

Correct answer:

40i

Explanation:

The complex conjugate of a complex number a + bi is a - bi . Since 8=8+0i , its complex conjugate is 8-0i = 8 itself. Multiply this by :

$8 \cdot 5i = 40 i$

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Example Question #92 : Graphing

Multiply the complex conjugate of by 3 i . What is the result?

Possible Answers:

-21i

None of the other responses gives the correct product.

21i

Correct answer:

Explanation:

The complex conjugate of a complex number a + bi is a - bi . Since 7i = 0+7i , its complex conjugate is 0-7i = -7i .

Multiply this by :

Recall that by definition i^2=-1 .

$-7i \cdot 3i = -7 \cdot 3 \cdot i \cdot i = -21 (-1) = 21$

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Example Question #21 : How To Graph Complex Numbers

Multiply the following complex numbers:

$\left (5 + i \right )\left (5 + 2i \right )$

Possible Answers:

27 + 5i

23 + 5i

27 - 15i

23 + 15i

23 - 15i

Correct answer:

23 + 15i

Explanation:

FOIL the product out:

$\left (5 + i \right )\left (5 + 2i \right )$

To FOIL multiply the first terms from each binomial together, multiply the outer terms of both terms together, multiply the inner terms from both binomials together, and finally multiply the last terms from each binomial together.

$=5 \cdot 5 + 5 \cdot 2i + 5 \cdot i + i\cdot 2i$

$=25 + 10i + 5 i + 2i^{2}$

Recall that i is an imaginary number and by definition i^2=-1 . Substituting this into the function is as follows.

=25 + 10i + 5 i + 2 (-1)

=25 -2 + 10i + 5 i

=23 + 15i

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Example Question #1 : How To Graph Inverse Variation

Give the equation of the vertical asymptote of the graph of the equation $y = \frac{7}{9x+4}$ .

Possible Answers:

$x= \frac{7}{9}$

$x= - \frac{7}{4}$

x = 9

$x= -\frac{4}{9}$

x=-4

Correct answer:

$x= -\frac{4}{9}$

Explanation:

The vertical asymptote of an inverse variation function is the vertical line of the equation x = a , where is the value for which the expression is not defined. To find , set the denominator to and solve for :

9x+4 = 0

9x= -4

$x= -\frac{4}{9}$ is the equation of the asymptote.

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Example Question #1 : How To Graph Inverse Variation

Give the -intercept of the graph of the equation $y = \frac{5}{2x-7}$ .

Possible Answers:

$\left ( \frac{5}{2}, 0 \right )$

(5,0)

(7,0)

The graph has no -intercept.

$\left ( \frac{7}{2}, 0 \right )$

Correct answer:

The graph has no -intercept.

Explanation:

The -intercept of the graph of an equation is the point at which it intersects the -axis. Its -coordinate is 0, so set y = 0 and solve for :

$y = \frac{5}{2x-7}$

$0= \frac{5}{2x-7}$

$0\cdot (2x-7) = \frac{5}{2x-7} \cdot (2x-7)$

0 = 5

This is identically false, so the graph has no -intercept.

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Example Question #3 : How To Graph Inverse Variation

Give the -intercept of the graph of the equation $y = \frac{3}{5x-8}$ .

Possible Answers:

$\left ( 0, -\frac{3}{5}\right )$

The graph has no -intercept.

$\left ( 0, \frac{8}{5}\right )$

$\left ( 0, -\frac{3}{8}\right )$

(0,8)

Correct answer:

$\left ( 0, -\frac{3}{8}\right )$

Explanation:

The -intercept of the graph of an equation is the point at which it intersects the -axis. Its -coordinate is 0, so set x = 0 and solve for :

$y = \frac{3}{5x-8}$

$y = \frac{3}{5 (0)-8}$

$y = \frac{3}{0-8}$

$y =- \frac{3}{8}$

$\left ( 0, -\frac{3}{8}\right )$ is the -intercept.

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Example Question #1 : How To Graph Inverse Variation

Give the slope of the line that passes through the - and -intercepts of the graph of the equation $y = \frac{6}{8x-2}$ .

Possible Answers:

$-\frac{1}{3}$

The line cannot exist as described.

$\frac{4}{3}$

$\frac{3}{4}$

Correct answer:

The line cannot exist as described.

Explanation:

The graph of $y = \frac{6}{8x-2}$ does not have an -intercept. If it did, then it would be the point on the graph with -coordinate 0. If we were to make this substitution, the equation would be

$0= \frac{6}{8x-2}$

and

$0 \cdot (8x-2)= \frac{6}{8x-2} \cdot (8x-2)$

0=6

This is identically false, so the graph has no -intercept. Therefore, the line cannot exist as described.

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Example Question #5 : How To Graph Inverse Variation

Give the -coordinate of a point with a positive -coordinate at which the graphs of the equations $y = \frac{4}{x}$ and x+3y = 4 intersect.

Possible Answers:

$\frac{2}{3}$

$\frac{4}{3}$

The graphs of the equations do not intersect.

Correct answer:

The graphs of the equations do not intersect.

Explanation:

Substitute $\frac{4}{x}$ for in the second equation:

$x+3\cdot \frac{4}{x} = 4$

$x+ \frac{12}{x} = 4$

$x+ \frac{12}{x}- 4 = 4 - 4$

$x- 4 + \frac{12}{x}= 0$

$\left (x- 4 + \frac{12}{x} \right )\cdot x = 0 \cdot x$

$x^{2}- 4 x + 12 = 0$

The discriminant of this quadratic expression is $b^{2} - 4ac$ , where a=1, b= -4, c=12 ; this is

$b^{2} - 4ac = (-4)^{2}-4 \cdot 1 \cdot 12 = 16 -48 = -32$ .

The discriminant being negative, there are no real solutions to this quadratic equation. Consequently, there are no points of intersection of the graphs of the two equations on the coordinate plane.

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All SSAT Upper Level Math Resources

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SSAT Upper Level Math : SSAT Upper Level Quantitative (Math)

Study concepts, example questions & explanations for SSAT Upper Level Math

All SSAT Upper Level Math Resources

Example Questions

Example Question #12 : How To Graph Complex Numbers

Example Question #13 : How To Graph Complex Numbers

Example Question #631 : Ssat Upper Level Quantitative (Math)

Example Question #92 : Graphing

Example Question #21 : How To Graph Complex Numbers

Example Question #1 : How To Graph Inverse Variation

Example Question #1 : How To Graph Inverse Variation

Example Question #3 : How To Graph Inverse Variation

Example Question #1 : How To Graph Inverse Variation

Example Question #5 : How To Graph Inverse Variation

All SSAT Upper Level Math Resources

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