Bosco

In \triangle PQR, we have \angle P = 90^\circ, PQ=3, and QR = 5. Find \tan R.   

By Pythagoras  PR² = QR² – PQ²    

                          PR  =  sqrt(25 – 9)  =  4     

                       tan R  =  opposite over adjacent  =  3 / 4      

                       tan R  =  0.7500    

If you draw / label the triangle, the answer becomes clear.    

I'd like to show you, but I don't know how to post a picture.   

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The area of a cross section of a sphere is 64\% of the largest possible cross sectional area of the sphere. If the sphere has radius 1/2, what is the area of the cross section?    

A cross section of a sphere would be a circle, right?  The largest circle would be through the center, so ...     

A_{through center}  =  π r²  =  3.1416 • 0.5²  =  3.1416 • 0.25     

                       =  0.7854    

 0.7854 • 0.64  =  0.5027 unit²    

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Three unit squares and two line segments connecting two pairs of vertices are shown. What is the area of triangle ABC?    

A is at (0,0), B is at (1,0), and C is at (1,1).   

Those three points make a right triangle.    

The base   is (0,0) to (1,0)  =  1 unit    

The height is (1,0) to (1,1)  =  1 unit    

The area of a triangle is (1/2) • base • height  =  (1/2) • 1 • 1  =  (1/2) unit²    

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A and B are two points on a sphere of radius 2. We know the space distance between A and B is 2, What is distance from A to B along the (minor) arc of a great circle?    

Consider the lines from the center of the sphere to A and to B.    

Each of these two lines is the radius of the sphere, namely, 2.    

Since the distance from A to B is 2, an equilateral triangle is formed.    

Every angle of an equilateral triangle is 60^o which is 1/6 of the great circle.    

The circumference of the great circle is π d  =  3.1416 • 4  =  12.5664.    

1/6 • 12.5664  =  2.0944 is the distance from A to B along the curvature of the sphere.    

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A basket contains 7 green marbles and 6 red marbles. Two marbles are taken from the basket at random without replacement (i.e., the first marble is not put back before the second marble is drawn). What is the probability that the first marble is green, and that the second marble is red?    

Prob 1st marble green is   7 / 13    

Prob 2nd marble red is      6 / 12    

Prob both happening is     (7 / 13) x (6 / 12) = 42 / 156  reduces to  7 / 26    

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Simplify (x^4 + x^3 + x^2 + x + 1) + (x^4 - x^3 + x^2 - x + 1).       

stack them for addition    

  x⁴  +  x³  +  x²  +  x  +  1    

  x⁴  –  x³  +  x²  –  x  +  1    

 2x⁴           + 2x²         + 2   =   2(x⁴ + x² + 1)    

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(x+2) / (x)  •  (x+3) / (x+1)  •  (x+4) / (x+2)  =  10    

cancel the (x+2) ' s   

(1) / (x) • (x+3) / (x+1) • (x+4) / (1) = 10    

[ (1) • (x+3) • (x+4) ] / [ (x) • (x+1) • (1) ] = 10    

(x² + 7x + 12) / (x² + x) = 10    

x² + 7x + 12 = 10x² + 10x    

0 = 9x² + 3x – 12              this will factor (it would reduce first, but I'm skipping that step)   

(3x + 4) • (3x – 3) = 0    

                      3x + 4 = 0     this x will not produce integer numerators and denominators, so disregard    

                      3x – 3 = 0    

                            3x = 3    

                              x = 1     therefore (x+2) / (x) = 3 / 1 = 3 is the value of the first fraction   

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what is the answer 65x+17-98=-4    

given that                                            65x + 17 – 98  =  – 4     

Take all the constants to one side,    

by convention the right-hand side       65x  =  – 4 – 17 + 98    

                                                             65x  =  77    

                                                                x  =  77 / 65    

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Find the minimum value of \frac{x^2}{x - 1 + x^3} for x > 1    

I used Desmos to plot y = (x²) / (x – 1 + x³)    

As x increases past (1,1), the value of y decreases asymptotically,    

which means y approaces zero (the x-axis) but it never gets there.    

So there is no minimum value.  We say "it approaches zero".    

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The real numbers x and y satisfy    
x^2 + y^2 - 8x + 6y + 23 = 0.    
Find the largest possible value of x + y.      

Given                               x² + y² – 8x + 6y + 23  =  0     

Group together               (x² – 8x     )  +  (y² + 6y     )  + 23  =  0    

                                       (x² – 8x     )  +  (y² + 6y     )           =  – 23   

Complete both squares and add the same    

amount on the right as you add on the left.    

                                         (x² – 8x + 16) + (y² + 6y + 9)  =  – 23  + 16  + 9     

                                         (x – 4)²  +  (y + 3)²  =  2    

This draws a circle centered at (+4, –3) with a radius the sqrt(2)     

We don't care about the center, all we care about is the radius.    

By formula, the largest x+y of a circle is its radius times the sqrt(2).    

So, the largest x+y is the sqrt(2) times the sqrt(2)         x + y = sqrt(2) • sqrt(2)    

                                                                                       largest x + y = 2    

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Compute    

(2 + 3i) + (2 - 3i) + (-2 + 3i) + (-2 - 3i).     

given                                                           (2 + 3i) + (2 – 3i) + (–2 + 3i) + (–2 – 3i)    

without the parentheses                               2 + 3i + 2 – 3i – 2 + 3i – 2 – 3i    

rearranged for ease of recognition             +2 +2 –2 –2   +3i +3i –3i –3i    

looks like everything cancels out                 zero    

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Give an example of a quadratic function that has zeroes at x = 2 and x = -2, and that takes the value 0 when x = 2.   

zeros at +2 & –2 gives us (x – 2)(x + 2)    

then multiplied out, the quadratic is x² – 4    

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Alex chooses a number at random from the set {1, 2, 3, 4, 5}.  Winnie also chooses a number at random from the same set.  (They can choose the same number.)  What is the probability that the product of their numbers is at least 4?    

There are 25 combinations possible.    

20 of them multiply to 4 or more:     1,4  1,5    

                                                         2,2  2,3  2,4  2,5    

                                                         3,2  3,3  3,4  3,5     

                                                         4,1  4,2  4,3  4,4  4,5     

                                                         5,1  5,2  5.3  5,4  5,5    

Therefore the probability is 20 / 25 or 80%.    

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Evaluate the infinite geometric series    
0.4 + 0.036 + 0.0000324 + . . .     

0.036 / 0.4              =  0.09

0.0000324 / 0.036  =  0.0009    

Definition, from the internet:  In Maths, Geometric Progression is a type of sequence where each succeeding term is produced by multiplying each preceding term by a fixed number, which is called a common ratio.    

The ratios are not the same, so the series in question is not a geometric series.    

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(x+2)/(x)  •  (x+3)/(x+1)  •  (x+4)/(x+2)  =  10    

cancel the (x+2) ' s   

(1)/(x) • (x+3)/(x+1) • (x+4)/(1) = 10    

[ (1) • (x+3) • (x+4) ] / (x) • (x+1) • (1) ] = 10    

(x² + 7x + 12) / (x² + x) = 10    

x² + 7x + 12 = 10x² + 10x    

0 = 9x² + 3x – 12     this will factor    

(3x + 4) • (3x – 3) = 0    

                      3x + 4 = 0     this x will not be an integer, so disregard    

                      3x – 3 = 0    

                            3x = 3    

                              x = 1     therefore (x+2) / (x) = 3/1 = 3 is the value of the first fraction   

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The quadratic equation $x^2-mx+24 = 10$ has roots $x_1$ and $x_2$. If $x_1$ and $x_2$ are integers, how many different values of $m$ are possible?    

x² – mx + 24 = 10     

Subtract 10 from both sides.    

x² – mx + 14 = 0    

m has to be the sum of two negative integers that multiply to 14.    

These could be –1 & –14 or –2 & –7.    

This makes 2 possible values for m; i.e., m could be –15 or –9.    

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A9 multiply by b9 = 2001 .    

Find the value of A and B? A and B are single digit number.     

The factors of 2001 are 1, 3, 23, 29, 69, 87, 667, and 2001.     

Only two factors end with a 9, so they have to be 29 and 69.      

29 • 69 = 2001     A and B are 2 and 6.    

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