Madison C.

Civil Engineering Student

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Geometry

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Question:

For a triangle with vertices A (1,6), B (5,1), and C (1,1): a) find each side length and the perimeter b) find the area c) find the each of the three inside angles

Madison C.

Answer:

The very first step for this problem is to sketch the triangle. Draw an x-y coordinate plane, where the y-axis starts at 0 and goes to at least 6, and the x-axis also starts at 0 and goes to at least 5. Now, plot each of the three points A, B, and C. Connect them to form your triangle. a) With the sketch, it should be simple enough to count the lengths of sides AC and BC, but this can also be done without the sketch, using only the known points. For AC, subtract the y of C from the y of A: 6 - 1 = 5 For BC, subtract the x of C from the x of A: 5 - 1 = 4 Thus we have AC = 5 units and BC = 4 units. Now to find AB, we will have to do some calculations. We can see from the sketch that the triangle is a right triangle and AB is the hypotenuse. So to find AB, we can use the Pythagorean theorem: a^2 + b^2 = c^2 For this equation, c is the length of the hypotenuse and a and b are the lengths of the other two sides. So we have: 5^2 + 4^2 = c^2 Solving for c: 25 + 16 = c^2 41 = c^2 c = sqrt(41) = 6.4 Thus we have AB = 6.4 units. Now to find the perimeter, we add all three side lengths: 5 + 4 + 6.4 = 15.4 Thus our perimeter is 15.4 units. b) The area of a triangle is: A=bh/2 A is area, b is base, and h is height. Because this is a right triangle, our height will be length AC and our base will be length BC. area = AC * BC / 2 = 5 * 4 / 2 = 20 / 2 = 10 Therefore our area is 10 units squared. c) Because we know that triangle ABC is a right triangle, we already know that one angle must be 90 degrees. From the sketch, we see that this angle is and ACB in the bottom left corner of the plane. Therefore, we have angle ACB = 90 degrees To find the other 2 angles, we will have to use sine, cosine, and tangent. This can be a bit tricky, but there's a helpful saying that can get you through it! Remember this. I'm in college and I still use it sometimes, so it can be very very useful!! The saying is: soh cah toa soh: sine = opposite / hypotenuse cah: cosine = adjacent / hypotenuse toa: tangent = opposite / adjacent Let's start with angle CAB. If you've sketched your triangle, it will be the angle at the very top, underneath point A. For every angle we will calculate, the hypotenuse will always be the same: 6.4, the length of AB that we calculated in part b. However, the opposite and adjacent lengths will be different. This is because each angle has a different line opposite to it and adjacent to it than the other angle. The line that is opposite will be the one that is (of course) opposite from the angle, and the line that is adjacent will be the one that is next to (or attached to) the angle. So for this angle, we have: hypotenuse = AB = 6.4 opposite = BC = 4 adjacent = AC = 5 Now because we have all three line lengths we can use whichever method we would like (sin, cos, or tan). Generally, though, it is safest to use the two lengths that you did not have to calculate. Since we calculated the hypotenuse, we should use the opposite and adjacent lengths. From above, you can see that tangent is the only function that does not use the hypotenuse. Letting x be our angle, we can find it like this: tangent = opposite / adjacent tan(x) = BC / AC tan(x) = 4/5 x = arctan(4/5) = 38.66 degrees Note that arctan is the same as tan^-1 (inverse tan). We can find the number with a calculator. If you want to check your answer, you can use the same method with sine and cosine, using the side lengths as seen in soh-cah-toa. So, we have angle CAB = 38.66 degrees. Lastly, we must find angle ABC. Now, we can do this one of two ways: by using sin/cos/tan or by the fact that all 3 angles must add up to 180 degrees. Here, I will use the sin/cos/tan method first, and then we can check the answer using the 180 degrees method. Remember that the hypotenuse will be the same as for angle CAB, but the opposite and adjacent side lengths will be switched. For this angle, we have: hypotenuse = AB = 6.4 opposite = AC = 5 adjacent = BC = 4 Again, we will use tangent because it does not require the calculated hypotenuse. This time, we will let our angle be z. tan(z) = AC / BC tan(z) = 5/4 z = arctan(5/4) = 51.34 degrees Therefore, we have angle ABC = 51.34 degrees. Now we will check it to make sure that the sum of all the angles is equal to 180 degrees. ACB + CAB + ABC 90 + 38.66 + 51.34 = 180 Therefore our calculations were correct. To summarize all of our answers: a) AC = 5 units and BC = 4 units and AB = 6.4 units perimeter is 15.4 units b) area is 10 units square c) ACB = 90 degrees, CAB = 38.66 degrees, ABC = 51.34 degrees

English

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Question:

For the following passage a) denote the errors by putting them in brackets [] b) correct the passage. The well-known book the jungle book came into it's own as one of the animated highlights of the Disney empire. It was published in 1894, and its sequel The Second Jungle Book came a year later during the time when Kipling was ensconced at Brattleboro in Vermont. The storie tells of the child Mowgli who is a foundling brought up by wolfs. He learns over time and due to the instructions of varius animal mentors the rules or ‘Laws’ of the jungle. Key figures are the wise black panther, Bagheera, and Baloo the sleepy bear. Both of these friendly beasts contribute to the childs education. We learn about the great enmity between Mowgli and the tiger Shere Khan who killed the boy’s parent's. Like Just So Stories (1902) it portrays the natural world and especially its creatures in a logical anthropomorphized manner, entertaining to adult and childs alike. The simplicity of the concept and the lack of didactic moral overtones have made The Jungle Book a lasting influence on the young. (source: http://www.bibliomania.com/0/-/frameset.html)

Madison C.

Answer:

a) The well-known book [the jungle book] came into [it's] own as one of the animated highlights of the Disney empire. It was published in 1894, and its sequel The Second Jungle Book came a year later during the time when Kipling was ensconced at Brattleboro in Vermont. The [storie] tells of the child Mowgli who is a foundling brought up by [wolfs]. He learns over time and due to the instructions of [varius] animal mentors the rules or ‘Laws’ of the jungle. Key figures are the wise black panther, Bagheera, and Baloo the sleepy bear. Both of these friendly beasts contribute to the [childs] education. We learn about the great enmity between Mowgli and the tiger Shere Khan who killed the boy’s [parent's]. Like Just So Stories (1902) it portrays the natural world and especially its creatures in a logical anthropomorphized manner, entertaining to adult and [childs] alike. The simplicity of the concept and the lack of didactic moral overtones have made The Jungle Book a lasting influence on the young. b) The well-known book The Jungle Book] came into its own as one of the animated highlights of the Disney empire. It was published in 1894, and its sequel The Second Jungle Book came a year later during the time when Kipling was ensconced at Brattleboro in Vermont. The story tells of the child Mowgli who is a foundling brought up by [wolfs]. He learns over time and due to the instructions of various animal mentors the rules or ‘Laws’ of the jungle. Key figures are the wise black panther, Bagheera, and Baloo the sleepy bear. Both of these friendly beasts contribute to the child's education. We learn about the great enmity between Mowgli and the tiger Shere Khan who killed the boy’s parents. Like Just So Stories (1902) it portrays the natural world and especially its creatures in a logical anthropomorphized manner, entertaining to adult and child alike. The simplicity of the concept and the lack of didactic moral overtones have made The Jungle Book a lasting influence on the young.

Algebra

TutorMe

Question:

For the following equation: a) put it in standard form b) find the vertex c) find the y-intercept and critical points. y=2x^2+12x+17

Madison C.

Answer:

a) This is a quadratic function, which we know because it has the form y=ax^2+bx+c. The standard form of a quadratic function is y=a(x-h)^2+k. We can change the given function to its standard form by completing the square. Follow these steps: y=2x^2+12x+17 The coefficient on x^2 must be 1, so we need to factor the 2 out of both x-terms. y=2(x^2+6x)+17 Now we will complete the square. To do this, we have to find a number that we can add to (x^2+6x) that will make it equal to some square in the from of (x+a)^2. We must remember, though, that a basic rule of algebra is "whatever you do to one side of the equation, you must do to the other." So, when we add this number to the right side of the equation, we must add it to the left side as well. It is important to note that this new number will be inside a set of parentheses that is multiplied by 2. This means that the number will also have to be multiplied by 2 when we add it to the other side. This is the "complete the square" equation (the blanks represent where we will add the new number): y+2(__)=2(x^2+6x+__)+17 Finding the number we will add is simple. We use the term 6x. The coefficient of this term is 6. Divide it by 2, and square the answer. This will be our new number 6/2=3 3^2=9 Putting 9 in the blanks, we now have: y+2(9)=2(x^2+6x+9)+17 Now we simplify. First, factor x^2+6x+9. We know that we square 3 to get 9 and complete the square (above). Therefore, we can know that x^2+6x+9=(x+3)^2. Second, we will do 2(9) and move the result to the right side of the equation. 2(9)=18, so subtract 18 from both sides to get y by itself. x^2+6x+9=(x+3)^2 2(9)=18 y+18=2(x+3)^2+17 y+18-18=2(x+3)^2+17-18 y=2(x+3)^2-1 We now have the standard form: y=2(x+3)^2-1 b) We can find the vertex and y-intercept very easily from the standard form of the equation. We know the general standard form of quadratic functions: y=a(x-h)^2+k We have: y=2(x+3)^2-1 From those, we can see that: a = 2 h = -3 k = -1 For quadratic equations in the standard form, it is known that the vertex is at the point (h,k). Therefore, our vertex is (-3,-1). c) To find the y-intercept and critical points, we will use the original form of the equation, y=2x^2+12x+17. For the y-intercept, set x=0 and solve for y. Remember that anything multiplied by 0 equals 0. y=2(0)^2+12(0)+17 y=0+0+17 y=17 Therefore, our y-intercept is at the point (0,17). This is because we used x=0 and found y=17. For the critical points, we will have to find the roots of the equation. Because this equation cannot be factored easily, we will use the quadratic equation: 1: x = [ -b + sqrt( b^2 - 4ac )] / [2a] 2: x = [ -b - sqrt( b^2 - 4ac )] / [2a] Because our original equation is in the form of y=ax^2+bx+c, we can know: a = 2 b = 12 c = 17 Now we plug these numbers into our two quadratic equations to find the two critical points. 1: x = [ -12 + sqrt( 12^2 - 4(2)(17) )] / [2(2)] x = { -12 + sqrt( 144 - 136)] / [4] x = [ -12 + sqrt( 8 )] / [4] sqrt(8) = sqrt(4) * sqrt(2) = 2 * sqrt(2) -12/4= -3 2*sqrt(2)/4 = .5 * sqrt(2) x = -3 + .5 * sqrt(2) 2: x = [ -12 - sqrt( 12^2 - 4(2)(17) )] / [2(2)] x = { -12 - sqrt( 144 - 136)] / [4] x = [ -12 - sqrt( 8 )] / [4] sqrt(8) = sqrt(4) * sqrt(2) = 2 * sqrt(2) -12/4= -3 2*sqrt(2)/4 = .5 * sqrt(2) x = -3 - .5 * sqrt(2) Because critical points are the points at which the line crosses the x axis, they will be at y=0. Therefore, our two critical points are: ( [-3 + .5 * sqrt(2)] , 0 ) ( [-3 - .5 * sqrt(2)] , 0 ) Summarizing our answers, we have: a) y=2(x+3)^2-1 b) vertex is (-3,-1). c) y-intercept (0,17) critical points are ( [-3 + .5 * sqrt(2)] , 0 ) and ( [-3 - .5 * sqrt(2)] , 0 )

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