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SSAT Upper Level Math : Geometry

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Example Questions

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 #411 : Geometry

Multiply the following complex numbers:

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

Possible Answers:

23 + 15i

27 + 5i

27 - 15i

23 + 5i

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 #4 : 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:

The line cannot exist as described.

$-\frac{1}{3}$

$\frac{3}{4}$

$\frac{4}{3}$

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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Example Question #6 : How To Graph Inverse Variation

The graphs of the equations $y = \frac{3}{x}$ and 4x-2y = -10 intersect in two points; one has a positive -coordinate. and one has a negative -coordinate. Give the -coordinate of the point of intersection that has a positive -coordinate.

Possible Answers:

Correct answer:

Explanation:

Substitute $\frac{3}{x}$ for in the second equation:

4x-2y = -10

$4x-2 \left ( \frac{3}{x} \right )= -10$

$4x- \frac{6}{x} = -10$

$4x- \frac{6}{x} + 10 = -10+ 10$

$4x- \frac{6}{x} + 10 = 0$

$\left (4x- \frac{6}{x} + 10 \right ) \cdot \frac{x}{2} = 0 \cdot \frac{x}{2}$

$2x^{2}-3+5x = 0$

$2x^{2}+5x -3= 0$

This quadratic equation can be solved using the method; the integers with product $2 \cdot (-3) = -6$ and sum 5 are and 6, so continue as follows:

$2x^{2}-x+6x -3= 0$

x (2x -1) +3(2x -1)= 0

(x+3 )(2x -1)= 0

Either x+3 = 0 , in which case x = -3 , or

2x-1 = 0 , in which case

2x = 1

x = 0.5

The desired -coordinate is paired with the positive -coordinate, so we substitute 0.5 for in the first equation:

$y = \frac{3}{x} = \frac{3}{0.5} = 6$

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Example Question #5 : How To Graph Inverse Variation

Give the -coordinate of the point at which the graphs of the equations $y = \frac{2}{x}$ and $y = -\frac{3}{x+1}$ intersect.

Possible Answers:

The graphs of the equations do not intersect.

Correct answer:

Explanation:

Using the substitution method, set the values of equal to each other.

$\frac{2}{x} = -\frac{3}{x+1}$

Multiply both sides by $\frac{x(x+1)}{1}$ :

$\frac{2}{x} \cdot \frac{x(x+1)}{1}= -\frac{3}{x+1}\cdot \frac{x(x+1)}{1}$

2(x+1)= -3x

2x+2= -3x

2x+2 + 3x -2 = -3x + 3x -2

5x = -2

$5x \div 5 = -2 \div 5$

x = -0.4

Substitute in either equation:

$y = \frac{2}{x} = \frac{2}{-0.4} = -5$

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Example Question #6 : How To Graph Inverse Variation

A line with slope 4 shares its -intercept with that of the graph of the equation $y = \frac{3}{2x-7}$ . Which of the following is the equation of that line?

Possible Answers:

This line does not exist, since the graph of $y = \frac{3}{2x-7}$ has no -intercept.

8x-2y =- 7

28x-7y=2

8x-2y=-3

28x-7y=3

Correct answer:

28x-7y=3

Explanation:

The -intercept of the graph of $y = \frac{3}{2x-7}$ —the point at which it crosses the -axis—is the point at which x = 0 , so substitute accordingly and solve for :

$y = \frac{3}{2 \cdot 0-7}= \frac{3}{ 0-7}=- \frac{3}{ 7}$

The -intercept of this graph, and that of the line, is $\left (0, - \frac{3}{ 7} \right )$ . Since the slope is 4, the slope-intercept form of the equation of the line is

$y = 4x-\frac{3}{7}$

To put it in standard form:

$y-y + \frac{3}{7} = 4x-\frac{3}{7} -y + \frac{3}{7}$

$\frac{3}{7} = 4x -y$

$\frac{3}{7} \cdot 7 = (4x -y) \cdot 7$

28x-7y=3

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SSAT Upper Level Math : Geometry

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

All SSAT Upper Level Math Resources

Example Questions

Example Question #92 : Graphing

Example Question #411 : Geometry

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 #4 : How To Graph Inverse Variation

Example Question #5 : How To Graph Inverse Variation

Example Question #6 : How To Graph Inverse Variation

Example Question #5 : How To Graph Inverse Variation

Example Question #6 : How To Graph Inverse Variation

All SSAT Upper Level Math Resources

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