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AfilCa [17]
3 years ago
7

50 points! Tell whether the orthocenter is inside, on, or outside the triangle. Then find the coordinates of the orthocenter.

Mathematics
1 answer:
Nady [450]3 years ago
6 0

Answer:

  • (1,3) is inside the triangle

Step-by-step explanation:

Orthocenter is the intersection of altitudes.

We'll calculate the slopes of the two sides and their altitudes ad find the intersection.

<h3>Side QR</h3>
  • m = (3 - 5)/(4 - (-1)) = -2/5

<u>Perpendicular slope:</u>

  • -1/m = 5/2

<u>Perpendicular line passes through S(-1, -2):</u>

  • y - (-2) = 5/2(x - (-1)) ⇒ y = 5/2x + 1/2
<h3>Side RS</h3>
  • m = (-2 - 3)/(-1 -4) = -5/-5 = 1

<u>Perpendicular slope:</u>

  • -1/m = -1/1 = -1

<u>Perpendicular line passes through Q(-1, 5):</u>

  • y - 5 = -(x - (-1)) ⇒ y = -x + 4

The intersection of the two lines is the orthocenter.

<u>Solve the system of equations to get the coordinates of the orthocenter:</u>

  • 5/2x + 1/2 = x + 4
  • 5x + 1 = -2x + 8
  • 7x = 7
  • x = 1

<u>Find y-coordinate:</u>

  • y = -1 + 4 = 3

The orthocenter is (1, 3)

If we plot the points, we'll see it is inside the triangle

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3 years ago
When a is divided by 7, the remainder is 5; and when b is divided by 7, the remainder is 4. What is the remainder when a + b is
Makovka662 [10]

Answer:

9

Step-by-step explanation:

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2 years ago
Find a particular solution to the nonhomogeneous differential equation y??+4y?+5y=?10x+e^(?x).
Firdavs [7]

Answer:

A) Particular solution:

2x+\frac{1}{2}e^{-x}-\frac{8}{5}

B) Homogeneous solution:

y_{h}=e^{-2x}(c_{1}cos(x)+c_{2}sin(x))

C) The most general solution is

y=e^{-2x}(c_{1}cos(x)+c_{2}sin(x))+2x+\frac{1}{2}e^{-x}-\frac{8}{5}

Step-by-step explanation:

Given non homogeneous ODE is

y''+4y'+5y=10x+e^{-x}---(1)

To find homogeneous solution:

D^{2}+4D+5=0\\D^{2}+4D+4-4+5=0\\\\(D+2)^{2}=-1\\D+2=\pm iD=-2 \pm i\\y_{h}=e^{-2x}(c_{1}cos(x)+c_{2}sin(x))---(2)

To find particular solution:

y_{p}=Ax+B+Ce^{-x}\\\\y'_{p}=A-Ce^{-x}\\y''_{p}=Ce^{-x}\\

Substituting y_{p},y'_{p},y''_{p} in (1)

y''_{p}+4y'_{p}+5y_{p}=10x+e^{-x}\\Ce^{-x}+4(A-Ce^{-x})+5(Ax+B+Ce^{-x})=10x+e^{-x}\\

Equating the coefficients

5Ax+2Ce^{-x}+4A+5B=10x+e^{-x}\\5A=10\\A=2\\4A+5B=0\\B=-\frac{4A}{5}B=-\frac{8}{5}2C=1\\C=\frac{1}{2}\\So,\\y_{p}=2x+\frac{1}{2}e^{-x}-\frac{8}{5}---(3)\\

The general solution is

y=y_{h}+y_{p}

from (2) ad (3)

y=e^{-2x}(c_{1}cos(x)+c_{2}sin(x))+2x+\frac{1}{2}e^{-x}-\frac{8}{5}

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Step-by-step explanation:hope it helps someone.

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