# geometry2 Homework Help

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• geometry2
Since A'C' is parallel to AC, then the triangles BA'C' and BAC are similar. We'll use the proportionality of the lengths of the sides to determine the unknowns: A'C'/AC = BA'/BA = BC'/BC Let BC' =...

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• geometry2
To evaluate the area of the triangle ABC, we must find out first the x and y intercepts of the graph of the function f(x) = -2x + 5. To calculate x intercept, we'll have to cancel y = 0. But y =...

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• geometry2
Let d = diagonal, l = length, w = widthSince we know that the diagonal of the rectangle splits the rectangle into two right angle triangles, being the hypotenuse of both right angle triangles.To...

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• geometry2
We must solve the system formed by the equations of the given lines to verify if it has any solutions. (x-1)/2 - y=5 We'll multiply by 2: x - 1 - 2y = 10 We'll add 1: x - 2y = 11 (1) x + (1-y)/4=6...

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• geometry2
To calculate the sum a +3b + 5c, we'll have to calculate first the vectors 3b and 5c: 3b = 3(2i - 3j) 5c = 5(- i +j) Computing the sum of 3 vectors, yields: a +3b + 5c = a + 3(2i - 3j) + 5(- i +j)...

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• geometry2
Since the given points lie on the graph of f(x), their coordinates verify the expression of f(x). f(1) = 3 and f(3) =7 We'll put x = 1 in the expression of condition from enunciation: a*f(1+2) +...

• geometry2
We'll get the abscisa and the ordinate of a point, tracing perpendiculars from the given point to x and y axis. We'll form a right angle triangle, whose hypotenuse is the distance from origin to...

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• geometry2
First, we'll write the equation of the line that passes AB. (xB - xA)/(x-xA) = (yB-yA)/(y-yA) We'll substitute the coordinates for A and B: (b-3)/(x-3) = (-4-2)/(y-2) (b-3)/(x-3) = -6/(y-2) We'll...

• geometry2
We'll write the perimeter of the triangle as the sum of the lengths of the sides: x + y + 10 = 24 (1) Since it is a right angle triangle, we'll use the Pythagorean theorem to calculate the...

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• geometry2
To verify if the pair (-1,-3) is the solution of the system, we have to substitute x and y by the values -1 and -3 to check if they cancel the equations. We'll substitute (-1,-3) in the first...

• geometry2
For the given quadrilateral ABCD: Length of AB = sqrt(8^2 + 1^2) = sqrt 65, slope of AB = (1/8) Length of BC = sqrt(4^2 + 7^2) = sqrt 65, slope of BC = -7/4 Length of CD = sqrt(8^2 + 1^2) = sqrt...

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• geometry2
Let the length and width of the rectangle be L and W. The perimeter of a rectangle is 7 times its width. => 2L + 2W = 7W => 2L = 5W => L = (5/2)W Area = 40 => L*W = 40 => (5/2)W*W =...

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• geometry2
It is given that the circle is tangent to the y axis and has a radius of three units. This implies that the x-coordinate is 3 or -3. The center lies in the third quadrant. So x = -3 The center...

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• geometry2
We'll have to use the law of cosines to determine the length of the side "a". a^2 = b^2 + c^2 - 2b*c*cos (b,c) (1) We know that the side "a" is facing to the angle A, then the side "b" is facing to...

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• geometry2
Since the lengths of the 3 sides are given, we'll use the law of cosines to determine the measure of the largest angle. The largest angle is the opposite angle to the largest length of triangle....

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• geometry2
We know that an isosceles triangle has 2 equal angles. Since the given angle is an obtuse angle, then the other 2 angles are equal and they are of x degrees. We know that the sum of all angles of a...

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• geometry2
We'll consider the three angles of triangle as: A,B,C. By definition, the sum of angles of a triangle is 180 degrees. A+B+C = 180 degrees We'll consider A as being the obtuse angle of the triangle...

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• geometry2
The distance between any two points (x1 , y1) and (x2 , y2) is given as sqrt ((x1 - x2)^2 + (y1 - y2)^2) Here the points are ( 2x+3, 8) and (2x, 4) The distance between the two is sqrt [(2x + 3 -...

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• geometry2
Area and perimeter have different units, so I'll consider only their numeric values. Let the side of the square be S. The perimeter is 4S and the area is S^2 As the area is 60 more than the...

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• geometry2
Area and perimeter have different units, so I'll consider only their numeric values. Let the side of the square be S. The perimeter is 4S and the area is S^2 As the area is 60 more than the...

Asked by jane1992 on via web

• geometry2
For the triangle ABC AB = 6, B=pi/4, C=pi/6. As the angles of a triangle have a sum of pi. A = 7*pi/12 Use the property sin A/a = sin B/b = sin C/c c = 6, C = pi/6, B = pi/4 and A = 7*pi/12 =>...

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• geometry2
The center of the circle is (1,1). The point (3, 5) lies on the circle. The distance from (3,5) to (1,1) is the radius of the circle. This is equal to sqrt[(3 - 1)^2 + (5 - 1)^2] = sqrt [ 4 + 16] =...

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• geometry2
For the equation of the tangent we need the slope of the tangent and one point it passes through. The slope of a tangent to any curve at a point is the value of the first derivative at that point....

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• geometry2
The equation of a line between points (x1, y1) and (x2, y2) is: (y - y1)/(x - x1) = (y2 - y1)/(x2 - x1) Substituting the values given, the equation of the line is: (y + 5)/(x - 7) = (1 + 5)/(3 -...

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• geometry2
The general equation of a circle with center (a,b) and radius r is: (x - a)^2 + (y - b)^2 = r^2 Here the center is (0,0) and the radius is 3. Substituting the values x^2 + y^2 = 9 The equation of...

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• geometry2
To verify if the given lines are parallel, perpendicular or neither, we'll have to put them first, in the point slope form: y = mx + n We'll start with the first equation: 4x+5y=198 We'll keep 5y...

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• geometry2
To determine the length of the side BC, we'll use the law of sines. Since BC is the opposite to the angle A, we'll get: BC/sin A = AC/ sin B We'll determine the measure of the angle B, based on...

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• geometry2
Since BM is the opposite side of the angle A = pi/2, that means that it is hypotenuse of the right angle triangle ABM. To determine BM, we need to calculate the missing length AC. We'll apply...

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• geometry2
We have to find vector v if u*v=15, w*v=17 and u=5i+2j, w=i-j Let v = ai + bj u*v = 15 = 5a + 2b ...(1) w*v = 17 = a - b ...(2) (1) + 2*(2) => 5a + 2b + 2a - 2b = 34 + 15 => 7a = 49 => a =...

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• geometry2
The power of M with respect to circle C is: p(M) = d^2 - r^2 d - is the distance from M to the center of the circle C. We'll calculate d^2 = 8^2 + 3^2 d^2 = 64 + 9 d^2 = 73 p(M) = 73 - 25 p(M) = 48...

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• geometry2
Let the linear function be f(c) = ax + b. y = ax + b is in the slope intercept form with the slope being a. We have f(2) = -6 and f(-2) = 4 2a + b = -6 ...(1) -2a + b = 4 ...(a) (1) - (2) => 4a...

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• geometry2
Let x be one side of the square. The area of the square is: A = x^2 The perimeter of the square is: P = 4x Now, we'll write mathematically the condition given by enunciation: x^2 = 4x - 4 (area is...

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• geometry2
The curve defined by the equation x^2 + y^2 - 16 = 0 can be written as x^2 + y^2 - 16 = 0 => (x - 0)^2 + (y - 0)^2 = 4^2 This is the equation of a circle with center at (0,0) and a radius of 4....

Asked by leeaeel on via web

• geometry2
To find the slope, we need two points from the line. If x = 0, 2y = 12 and y = 6. This means that one of our points is (0,6) If y = 0, 4x = 12 and x = 3. This means that another point is (3,0)....

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• geometry2
To find the slope of a line, we use the formula m = (y1 - y2)/(x1 - x2). In this case, your p value is x and your q value is y. We know that, for any number, q1 - q2 is going to equal -4. This is...

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• geometry2
It is given that the hypotenuse of a right triangle is 26 cm long and one leg is 14 cm longer than the other. Let the length of the shorter leg be x , the other leg is x + 14 AS it is a right...

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• geometry2
We have the points A(2,-2) B(1,1) C(1,4) and D(x,5) and we need to find x if AB and CD are parallel. The slope of AB is (1+2)/(1-2) = 3/-1 = -3 The slope of CD is (5-4)/(x - 1) = 1/(x - 1) For...

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• geometry2
Let the smaller sides of the triangle have a length a and b. The length of the hypotenuse is sqrt( a^2 + b^2) The area of the triangle is (1/2)*a*b = 6 => ab = 12 => a = 12/b The perimeter is...

• geometry2
We know that the slope of the perpendicular bisector is the opposite reciprocal of the slope containing the two given points. Therefore, the slope of the line containing the two given points is...

• geometry2
The equation of a circle is (x - a)^2 + (y - b)^2 = r^2 As we know three points through which the circle passes, we can create three equations. (-2 , 4) (-2 - a)^2 + (4 - b)^2 = r^2 ...(1) (3 ,...

Asked by sirserie on via web

• geometry2
The general equation of a circle has three variables, if the center is (a, b) and r is the radius: (x - a)^2 + (y - b)^2 = r^2. You have provided two points through which the circle passes (0,5)...

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• geometry2
The perpendicular bisector of the segment with endpoints (k,0) and (4,6) has a slope -3. This gives the slope of the line segment with end points (k,0) and (4,6) as 1/3. We get this as the product...

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• geometry2
The slope of two perpendicular lines m1 and m2 are related as m1* m2 = -1. The slope of the line through (1,3) and ( 2,6) is : m = (6 - 3)/(2 - 1) => m = 3 A line perpendicular to this line has...

Asked by sirserie on via web

• geometry2
The mid point of a line joining the points (x1, y1) and ( x2, y2) is given by [(x1 + x2) / 2 , (y1 + y2)/2] For the side, BC the coordinates of B are (7 , 3) and those of C are ( 5, 7). The mid...

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• geometry2
We have the function f(x) = x^2. The points given to us are (t , f(t)) and (t+h , f(t+h)) or (t , t^2) and ((t + h) , (t + h)^2) The gradient between these points is =>[ (t + h)^2 - t^2] / [ t +...

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• geometry2
We have to find the straight line passing through the points ( -3,2) and ( 5,8) The equation of a line passing through the points ( x1, y1) and ( x2, y2) is given by ( y - y1) = [ ( y2 - y1)/(x2 -...

Asked by kamused on via web

• geometry2
Firts, we'll determine the coordinates of the vertex of the parabola. V(-b/2a ; -delta/4a) deta = b^2 - 4ac a,b,c, are the coefficients of the quadratic. a = 1 , b = -2(m-1) , c = m-1 xV =...

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• geometry2
Two lines are parallel if their slopes are equal or the ratios of correspondent coefficients are equal. We'll form the ratios of the correspondent coefficients: (t-3)/1 = 2/(t+1) We are applying...

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• geometry2
Given the lines: f(x)= 2x - 1 g(x)= -4x + 1 In order to find a point on both lines, then we need to determine where the lines intersects. To find intersection points we need to determine x values...

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