Math Help:Coordinate Geometry??
1. Since the point is on the y-axis, the x-coordiante is obviously 0. The distance formula: D = sqrt[(x2-x1)^2 + (y2-y1)^2] 10 = sqrt[(8-0)^2 + (2-y)^2] 10 = sqrt[8^2 + (2-y)^2] 10 = sqrt[64 + (2-y)^2] Square both sides 100 = 64 + (2-y)^2 100 - 64 = (2-y)^2 36 = (2-y)^2 (2-y)^2 = 36 Get the square root of both sides sqrt((2-y)^2) = +-sqrt(36) 2-y = +- 6 Two answers: 2 - y = 6 2 - 6 = y y = -4 And 2-y = -6 2 + 6 = y y = 8 So your points are (0, -4) and (0, 8) 2. Since the point is on the x-axis, the y-coordiante is obviously 0. Use the distance formula sqrt[(-4-x)^2 + (7-y)^2] = sqrt[(5-x)^2 + (4-y)^2] sqrt[(-4-x)^2 + (7-0)^2] = sqrt[(5-x)^2 + (4-0)^2] sqrt[(-4-x)^2 + 7^2] = sqrt[(5-x)^2 + 4^2] sqrt[(-4-x)^2 + 49] = sqrt[(5-x)^2 + 16] Square both sides (-4-x)^2 + 49 = (5-x)^2 + 16 (-4-x)^2 - (5-x)^2 = 16 - 49 16 + x^2 + 8x - (25 + x^2 -10x) = -33 16 + x^2 + 8x - 25 - x^2 +10x = -33 18x - 9 = -33 18x = -33 + 9 18x = -24 x = -24/18 x = -12/9 So your answer is (-12/9, 0) If you have given me the correct coordiantes for the two points, then your answer is incorrect. I double checked it and my answer is correct Here is the distance to (37/9, 0) from (-4, 7) = sqrt[(x2-x1)^2 + (y2-y1)^2] = sqrt[(-4-37/9)^2 + (7-0)^2] = sqrt[65.79 + 49] = sqrt[114.79] = 10.714 Here is the distance to (37/9, 0) from (5, 4) = sqrt[(x2-x1)^2 + (y2-y1)^2] = sqrt[(5-37/9)^2 + (4-0)^2] = sqrt[0.79 + 16] = sqrt[16.79] = 4.10 As you can see, they are not the same, but the distance to the point I gave you is the same Here is the distance to (-12/9, 0) from (-4, 7) = sqrt[(x2-x1)^2 + (y2-y1)^2] = sqrt[(-4--12/9)^2 + (7-0)^2] = sqrt[7.11 + 49] = sqrt[56.11] = 7.49 Here is the distance to (-12/9, 0) from (5, 4) = sqrt[(5--12/9)^2 + (4-0)^2] = sqrt[40.11 + 16] = sqrt[56.11] = 7.49 I hope this helped Kia
How would I find the coordinates where the line [math]x-y=8[/math] will intersect the y-axis?
As the line intersect at y-axis so x=0 in this casex-y=8 (a)Using [math]x=0[/math]0-y=8-y=8y=-8Using in equ (a)x-(-8)=8x+8=8x=8–8x=0So coordinates will be (0,-8)
PLEASE HELP! Coordinate geometry AS Maths?
a) Just substitute 2 for x in the equation and solve: y = 2x^3 - 2(2)^2 - 2 + 9 = 7 b) The slope of the function is given by the derivative, y ' = 3x^2 - 4x - 1 Now substitute 2 for x y ' = 3(2)^2 - 4(2) - 1 = 3 The slope = m of the tangent line. Just substitute y and x to get b: 7 = 3(2) + b; Solving, b = 1. So the tangent equation is y = 3x + 1. c) The slope of two perpendicular lines are negative reciprocals of each other. Therefore the slope of the new tangent is -1/3. Substitute this into y ' : -1/3 = 3x^2 - 4x - 1; or, 3x^2 - 4x - 2/3 = 0. Using the quadratic formula, x = (1/(2)(3))(-(-4) ± √[(-4)^2 - (4)(3)(-2/3)] which simplifies to: x = (1/3) (2 ± √6) or x ≈ 1.48 and -0.15. Since we are given the restriction x > 0 then only the former x is correct, or (1/3) (2 + √6) .
What is the x-coordinate of any point in the yz-plane? I will appreciate a quick answer?
zero
Math help!? math helpers pleasE?!?
1. Use the quadratic equation x2 + 3x � 10 = 0. Determine the discriminant. (Points: 5) ?28 ?19 37 49 2. Use the quadratic equation x2 + 3x � 10 = 0. Where does the vertex of the parabola y = x2 + 3x ? 10 lie? (Points: 5) below the x-axis above the x-axis on the x-axis on the x-axis and to the left of the y-axis 3. Use the quadratic equation x2 + 3x � 10 = 0. Determine the number of real roots of y = x2 + 3x ? 10. (Points: 5) one real root two real roots no real roots no solution to the equation 4. Use the quadratic equation x2 + 3x � 10 = 0. Find the x-intercepts of y = x2 + 3x ? 10, if they exist. (Points: 5) the parabola does not cross the x?axis ?3 and ?6 ?5 and 2 1 and 5
Math? Help? Please?10 points?
Quadrilateral A B C D has vertices at A 6,3, B 3,9, C 6,6 and D 12,3. It is reflected across the x-axis. What are the new coordinates? Explain. Gino designed a garden for his rectangular backyard. The design he drew on paper is 7 inches wide. The width of his backyard is 42 feet. If his drawing is to scale and is 8 inches long, how many feet long is his backyard? Write a proportion using words. Then, use the proportion to calculate the yard length. Show your work. Darian is buying an entertainment center for his living room wall. The tag on his first choice shows that the furniture is 54 inches wide and 90 inches high. Darian has a space for it that is 5 feet wide and 8 feet high. Will his first choice fit? Explain. Just a little note. PLEASE just answer my questions I really do not need a lecture on this. And before you answer keep this in mind I DO NOT HAVE ANYBODY TO HELP ME!!! I am completely and totally self taught. I don't have books, I just have a computer screen that doesn't explain as well as its supposed to. Yes I do have a teacher but, shes god knows where and doesn't like to answer my calls.
Grade 10 Math help please?
Let the other point B be denoted by the coordinates( x,y ). The x-coordinate of the midpoint is half the sum of the x-coordinates of the endpoints: i.e. 10 = ( 12 + x ) / 2 10*2 = 12 + x 20 = 12 + x x = 20 - 12 x = 8 The y-coordinates of the midpoint is equal to half the sum of the y-coordinate of the midpoint i.e. - 4 = ( 5 + y ) / 2 - 4*2 = 5 + y -8 = 5 + y y = - 8 - 5 y = - 13 Therefore the coordinates of the endpoint B are ( 8, -13 )
Math Question Help PLEASE!!!?
Use the quadratic equation: x^2 +3x - 10 =0 1.Determine the discriminant A. -28 B. -19 C. 37 D. 49 2. Where does the vertex of the parabola lie? A. below the x-axis B. above the x-axis C. on the x-axis D. on the x-axis and to the left of the y-axis 3. Determine the number of real roots. A. 1 real root B. 2 real roots C. no real roots D. no solution to the equation 4. Find the x-intercepts, if they exist. A. the parabola does not cross the x-axis B. -3, -6 C. -5, 2 D. 1,5 PLease please Help me with these. tHANKS IN ADVANCED =)
Coordinating my math.................?
On 1, there are a couple ways. You could complete the square for x: x^2 + 4x = 6y - 10 (x + 2)^2 = 6y - 10 + 4 = 6y - 6 = 6(y - 1) so the vertex is (-2, 1) making the axis be x = -2 Or, solve for y and use the fact that if y = ax^2 + bx + c, the axis of symmetry (as well as the x coordinate of the vertex) is x = -b / (2a) x^2 + 4x + 10 = 6y 1/6 x^2 + 4/6 x + 10/6 = y so x = -(4/6) / (2 • 1/6) = -4/6 over 2/6 = -4/2 = -2 then plug in to get y 4 - 8 + 10 = 6y 6 = 6y 1 = y Sorry, I don't get what you are asking for on #2.