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SCIE 4101 Review Packet #4 Answers November 2, 2009

Posted by Ms. Miller in UTA.
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Here are the answers for the other review packet I handed out on Thursday, 10/29. The quiz over this packet will be on 11/12.

14. B (1 sq. yard = 9 sq. feet) 15. B (\frac{180-176}{180})
16. B 17. D
12. G 13. B
14. H 1. B
54. H 30. H (\tan 30\textdegree=\frac{2400}{x})
24. G 4?. G (S=\frac{120}{360}\pi 36^2)
39. D (\frac{5280}{15840}=\frac{6600}{6600+x}) 27. D (\overline{AB} \parallel \overline{CD}; y-2=-\frac{4}{3}(x-5))
33. A (\frac{3\pi + 7\pi}{16\pi}) 28. B (angle is bisected)

SCIE 4101 Review Packet #3 Answers October 28, 2009

Posted by Ms. Miller in UTA.
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Assume perimeter is in \mathrm{units}, area is in \mathrm{units^2}, and volume is in \mathrm{units^3} unless stated otherwise.

1a. To find the length, use the Pythagorean Theorem (a^2+b^2=c^2): 5^2+b^2=13^2, which means that the length is 12.

P=2l+2w=2(12)+2(5)=34
A=bh=(12)(5)=60

1b. Using the Pyth. Theorem as before, we have 2a^2=(60\sqrt{2})^2, which works out to a=60.

P=60+60+60\sqrt{2}=120+60\sqrt{2}
A=\frac{1}{2}bh=\frac{1}{2}(60)(60)=1800

2a.

C=2 \pi r=2(16)\pi=32\pi \approx 100.53
A=\pi r^2=256\pi

2b.

C=\pi d=21\pi \approx 65.97
A=\pi(10.5)^2=110.25\pi \approx 346.36

3a. Since the perimeter is 64 and we know two sides are 25, we know that the third side is 14. Half of that is the short leg of a right triangle, so we can use the Pyth. Theorem to solve for x: 7^2+x^2=25^2, which means that x=24.

3b.

(x+4)(x-10)=240
x^2-6x-40=240
x^2-6x-280=0
Factoring
(x-20)(x+14)=0		 Quadratic Formula
x=\frac{6 \pm \sqrt{(-6)^2-4(1)(-280)}}{2(1)}
x=\frac{6 \pm \sqrt{1156}}{2}
x=3 \pm 17
x=20,-14 x=20,-14

(It pays to remember how to factor!)
3c.

\pi r^2=121\pi
r=11\ \Rightarrow \ y=2r=22

3d. There are actually two different ways to work this.

Method #1 Method #2
Divide the trapezoid into a rectangle flanked by two right triangles. This means that the longer base can be broken up into three lengths: 3, 9, and 3, and the right triangles have one leg that is 3 and a hypotenuse of 5, and the height is the other leg.
3^2+h^2=5^2
h=4		 A=\frac{1}{2}h(b_1+b_2)
48=\frac{1}{2}h(9+15)
48=\frac{1}{2}(24)h
48=12h
h=4

4a. S=Ph+2B (P is the perimeter of the base, h is the height of the prism, and B is the area of the base.)

S=[2(20)+2(8)](12)+2(20)(8)=992

Notice that if we selected the other rectangle as our base, we’d still get the same answer:
S=[2(8)+2(12)](20)+2(8)(12)=992

V=Bh

V=[(8)(20)](12)=1920

Again, it doesn’t matter which rectangle we select as our base.
4b. The formula for the surface area of a cylinder is S=Ch+2\pi r^2.

S=(8\pi)(9)+2\pi(4^2)=72\pi+32\pi=104\pi.

The volume of a cylinder is the same as a prism: V=Bh, where B is the area of the base, in this case a circle.

V=\pi(4^2)(9)=144\pi

4c.

S=4\pi r^2
S=4\pi (8^2)=256\pi
V=\frac{4}{3}\pi r^3
V=\frac{4}{3}\pi(8^3)=\frac{2048\pi}{3}

4d. To find the surface area of a pyramid, you must have the “slant height” (l). To find the slant height of this pyramid, look at the right triangle created by the height, half the base, and the lateral face. 8^2+6^2=l^2\ \Rightarrow \ l=10

S=\frac{1}{2}(4)(12)(10)+12^2=384
V=\frac{1}{3}(12^2)(8)=384

Remember, even thought the two numbers are the same, one represents surface area (square units) and the other volume (cubic units).

5. An interior angle of a regular polygon is found using the formula \frac{(n-2)(180)}{n}, where n is the number of sides.
a. For a regular hexagon, each interior angle is \frac{4(180)}{6}=120\textdegree.

7x+15=120
7x=105
x=15

b. For a regular pentagon, each interior angle is \frac{3(180)}{5}=108\textdegree.

x^2-3x=54 (angle is half of 108°)
x^2-3x-54=0
(x-9)(x+6)=0
x=9,-6

6. If all the dimensions are doubled, the surface area will increase by a factor of 22, or 4. The volume will increase by a factor of 23, or 8.

S=(150)(4)=600\ \mathrm{cm^2}
V=(225)(8)=1800\ \mathrm{cm^3}

7.

4\pi r^2=180\pi
r^2=\frac{180\pi}{4\pi}=45
r=\sqrt{45}=3\sqrt{5}

V=\frac{4}{3}\pi(\sqrt{45})^3
V=\frac{4}{3}\pi(45\sqrt{45})
V=\frac{4}{3}\pi(135\sqrt{5})
V=180\sqrt{5}\pi\ \mathrm{m^3}

SCIE 4101 Review Packet #3 October 28, 2009

Posted by Ms. Miller in UTA.
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Math Review Packet 3

I will try to post the answers for this on Wednesday.

SCIE 4101 Review Packet #2 Answers October 22, 2009

Posted by Ms. Miller in UTA.
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Transformations
A transformation is a change in the position, size, or shape of a figure. The original shape is called the preimage, and the resulting shape is called the image. There are four types of transformations:

  • Translation (slide) – all of the coordinates move the same distance in the same direction
  • Reflection (flip) – each point and its image are the same distance from the line of reflection
    • Across the x-axis: the x-coordinate stays the same, the y-coordinate changes sign. (x,y) \longrightarrow (x,-y).
    • Across the y-axis: the y-coordinate stays the same, the x-coordinate changes sign. (x,y) \longrightarrow (-x,y)
    • Across the line y=x: the coordinates trade places. (x,y) \longrightarrow (y,x)
  • Rotation (turn) – each point and its image are the same distance from the point of rotation
    • 90° about the origin: (x,y) \longrightarrow (-y,x)
    • 180° about the origin: (x,y) \longrightarrow (-x,-y)
  • Dilation (size) – if centered about the origin, multiply each coordinate by the scale factor

Linear Functions
A function is a rule that is applied to an independent variable. The most common form of this is called the slope-intercept form of a line: y=mx+b, where m is the slope and b is the y-intercept. Another form is the point-slope form: y-y_{1}=m(x-x_{1}).

Slope (also called rate of change) is the ratio of the difference in the y-values to the difference in the x-values.

Direct and Inverse Variation
A related type of function is variation. Direct variation is a linear equation of the form y=kx: x and y both increase or both decrease. Inverse variation is a nonlinear relationship of the form y=\frac{k}{x}: x and y move in opposite directions. In both cases, k is the constant of variation.

Vertical Asymptotes
Vertical asymptotes are boundaries on rational functions at points where the function is undefined. Generally, to find a vertical asymptote, evaluate the denominator of the function to find out where/if it can be equal to zero. Whenever a function evaluates to \frac{0}{0}, you have a “hole”. As the function gets closer and closer to the vertical asymptote, its value becomes more and more extreme. It does not behave this way around a hole.

Discriminants
Discriminants are an indicator of whether a quadratic function will have imaginary roots or zeros. If you’ll recall, for a quadratic equation of the form ax^x+bx+c=0, the quadratic formula is x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}. The discriminant is the middle part: b^2-4ac.

  • If this is less than zero, then the equation will have imaginary roots, and the graph does not cross the x-axis.
  • If it is equal to zero, the equation will have one solution, which is where the graph touches the x-axis.
  • If it is greater than zero, the equation will have two real roots, where the graph crosses the x-axis.

Answers

1. U'(-2,1), T'(2,-4),A'(3,1); U'(2,-1), T'(-2,4), A'(-3,-1)
2. H'(5,2), A'(8,-1), N'(6,-11), D'(2,-3)
3. M'(-3,2\frac{1}{2}), A'(4,-2\frac{1}{2}), V'(-3,-2\frac{1}{2})
4. M'(-5,-6), A'(5,8), V'(5,-6)
5. 5n-2 6. y=4x-3
7. {0,1,4,5} 8. 6350
9. \frac{x^2-ax}{2} 10. y\geqslant 2x-3
13. x=4; solutions are 0 and 8
14a. direct; y=3x 14b. inverse; y=\frac{2.4}{x}
15. k=96 16a. vertical asymptote at x=-2 and a hole at x=-3
16b. vertical asymptote at x=-4 and a hole at x=-4 16c. vertical asymptote at x=3 and a hole at x=-2
17a. D=61; two real roots 17b. D=-31; two imaginary roots
17c. D=0; one real root 17d. D=64; two real roots

SCIE 4101 Math Review Packet #2 October 19, 2009

Posted by Ms. Miller in UTA.
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Here is Math Review Packet 2(#16 is updated!). I do not plan to give a quiz over it until 10/29, but you may want to look over it before class.

SCIE 4101 FYI October 19, 2009

Posted by Ms. Miller in UTA.
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The problems from the Preparation Manual that correspond to the first review packet are Competencies 001-003 (or problems 1-6).

This Thursday, we will be looking at Domain II, which is Competencies 004-007 (or problems 7-13). If you have any questions about any of these problems, be sure to ask me about them in class.

SCIE 4101 Math Review Packet #1 October 16, 2009

Posted by Ms. Miller in UTA.
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It was good to finally meet with y’all last night. I hope my introduction made some sort of sense to you.

Math Review Packet 1

Answers

1. N, Z, Q, R 2. Q, R 3. R 4. N, Z, Q, R 5. Q, R
6. R 7. Z, Q, R 8. N, Z, Q, R 9. C 10. R
11. Example: \frac{3}{7} or 2.5 12. Example: -15 13. 12
14. Example: \pi or \sqrt{5} 15. Example: \sqrt{-1} 16. Example: 20
17. Yes 18. No; 5-6 19. Yes
20. No; \frac{-3}{4} 21. Yes 22. No; \frac{20}{4}
23. R is always closed except for even fractional exponents (square roots, fourth roots, etc.).
Group Properties Addition in the Reals Multiplication in the Reals
Commutative p+q=q+p ab=ba
Associative p+(q+r)=(p+q)+r a(bc)=(ab)c
Identity There exists a number 0 such that for all p, p+0=0+p=p There exists a number 1 such that for all a, a \cdot 1=1 \cdot a=a
Inverse For all p, there exists a number -p such that p+-p=0. For a\neq 0, there exists a number \frac{1}{a} such that a \cdot \frac{1}{a}=1.
25. \frac{3}{4} 26. 7 27. 2
28. \frac{5}{6} 29. \frac{5}{2} 30. -\frac{13}{36}
31. 8.7654 \times 10^7 32. 5.477 \times 10^{-4} 33. 9.7 \times 10^13
34. 54,487,000 35. 0.000000000684 36. 0.000165585
37. -7 38. 24
39. 2^{64}=18,446,744,073,709,551,616\ \mathrm{gold}\ \mathrm{pieces}
40. Since there are at most 366 days in a year, if there are 370 people, at least two of them must share a birthday.
41. 38=34+3+1

Additional Notes on Probability September 10, 2009

Posted by Ms. Miller in Algebra I.
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I felt we rushed through this, so here are a couple of things I wanted to add:

Combinations

  • OR – when you see this word, you should ADD the individual probabilities.
  • AND – when you see this word, you should MULTIPLY the individual probabilities.
  • NOT – when you see this word, you can either ADD all the other probabilities or you can take the probability of the event’s happening and SUBTRACT that from 1. Example: The probability of NOT rolling a 2 on a die is 1-\frac{1}{6}=\frac{5}{6}.

Whenever you are combining two or more events together, you have to be aware of whether you are resetting the circumstances of the trial if that is possible. For example, if you have two spinners, one with numbers and one with colors, you don’t have to worry about replacing or resetting anything. On the M&M trials we did, however, we really should have talked about replacement. For example: If you had 30 M&Ms, of which 6 were red and 4 were green, and I asked you the theoretical probability of drawing a red M&M and then drawing a green M&M, you would multiply the probabilities (because of the AND), but the number in the denominator would change because you would no longer have 30 M&Ms when you made the second draw: \frac{6}{30} * \frac{4}{29}=\frac{24}{870}=\frac{1}{145}

A standard deck has 52 cards divided into 4 different sets, called “suits”: hearts and diamonds, which are red, and spades and clubs, which are black. Each suit has 13 cards numbered from 1 (ace) to 10, and the the face cards Jack, Queen, and King. So, for example, the probability of drawing a red card would be \frac{2\ \mathrm{suits}}{4\ \mathrm{suits}}=\frac{1}{2}. The probability of drawing a 7 would be \frac{4\ \mathrm{7s}}{52\ \mathrm{cards}}=\frac{1}{13}.

Tests! September 10, 2009

Posted by Ms. Miller in Algebra I, Geometry.
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Don’t forget — tests are Friday and Monday!

Factoring Tricks August 29, 2009

Posted by Ms. Miller in Algebra I, Geometry.
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It’s important, if you are trying to find the LCM of some numbers (such as trying to find a common denominator), that you be able to tell immediately what factors go into a number. My 6th grade teacher, Mrs. Dana, taught me the following tricks, which I want to share with you:

“Casting Out Nines”
This is the name of a process of adding up the digits of a number; it will come in handy as we try to find factors. When we talk about “adding up the digits of a number”, what you should do is:

  • Cross out any 9s that are in the number.
  • If there are any pairs of digits that add up to 9, cross them out as well.
  • Add up the remaining digits. If the sum is greater than 9, continue to add the digits together until you have a one-digit answer. This is the number we are interested in.

Example: What is the sum of the digits of the number 96,216?
We cross off the 9, and then add 6+2+1+6=15. We then find the sum of 15, which is 1+5=6. So the sum of the digits is 6.

(This method also has uses in checking arithmetic, but we’re not concerned about that right now. If you’re interested in reading further about it, check out Wikipedia’s entry on the subject.)

Now, let’s look at how we know whether certain numbers are factors:

Factor Divisible By If …
1 Goes into everything
2 Number ends in 0, 2, 4, 6, or 8
3 The sum of the digits is 3, 6, or 9
4 The last two digits are divisible by 4. Ex. 87,665,9\underline{24}
5 Number ends in 0 or 5
6 Number is divisible by 2 and 3 (Even number whose digits sum to 3, 6, or 9)
7 No rule exists.
8 The last three digits are divisible by 8. Ex. 2783738273\underline{064}
9 The sum of the digits is 9.
10 Number ends in 0
12 Number is divisible by 3 and by 4.
25 Last two digits are divisible by 25 (00, 25, 50, 75).