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Lady bird [3.3K]

3 years ago

5

Find, corrrect to the nearest degree, the three angles of the triangle with the given vertices. A(1,0), B(3,6), C(-1,4)

1 answer:

Karo-lina-s [1.5K]3 years ago

4 0

Answer:

45°, 45°, 90°

Step-by-step explanation:

Find the vectors to represent each sides

AB =<3-1, 6-0>=<2,6>

AC = < -1-1, 4 - 0 > = < -2, 4>

Magnitude of the vectors

AB = √(2²+6²) = 6.32

AC = √ ((-2)² + 4²) = 4.47

cosθ = vector of AB × vector AC / ( Product of the magnitude of AB and AC) = 2 × (-2) + (6×4)/ (6.32×4.47) = 20 / 28.2504

θ = arcos(20 / 28.2504 ) approx = 45°

Magnitude of the vectors

BA =<1-3,0-6>=<-2,-6>

BC =<-1-3,4-6>=<-4,-2>

Magnitude of the vectors equals

BA = √((-2)² + (-6)²) = 6.325

BC = √((-4)² + (-2)²) = 4.4721

cosθ = (-2×-4) + (-6 ×-2) / (6.325 × 4.4721) = 20 / 28.286

θ = arcos (20 / 28.286 ) = 45°

Magnitude of the vectors

CB =<3--1, 6-4>=<4,2>

CA=<1--1,0-4>=<2,-4>

Magnitude of the vector =

CB = √(4² + 2²) = 4.4721

CA = √(2² + (-4)²) = 4.4721

cosθ = (4×2) + (2×-4) / (4.4721×4.4721) = 0

θ = arcos 0 = 90°

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Answer:

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Step-by-step explanation:

Given

See attachment for complete question

Required

Determine the equilibrium solutions

We have:

$\frac{dR}{dt} = 0.09R(1 - 0.00025R) - 0.001RW$

$\frac{dW}{dt} = -0.02W + 0.00004RW$

To solve this, we first equate $\frac{dR}{dt}$ and $\frac{dW}{dt}$ to 0.

So, we have:

$0.09R(1 - 0.00025R) - 0.001RW = 0$

$-0.02W + 0.00004RW = 0$

Factor out R in $0.09R(1 - 0.00025R) - 0.001RW = 0$

$R(0.09(1 - 0.00025R) - 0.001W) = 0$

Split

$R = 0$ or $0.09(1 - 0.00025R) - 0.001W = 0$

$R = 0$ or $0.09 - 2.25 * 10^{-5}R - 0.001W = 0$

Factor out W in $-0.02W + 0.00004RW = 0$

$W(-0.02 + 0.00004R) = 0$

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$W = 0$ or $-0.02 + 0.00004R = 0$

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$-0.02 + 0.00004R = 0$

$0.00004R = 0.02$

Make R the subject

$R = \frac{0.02}{0.00004}$

$R = 500$

When $R = 500$ , we have:

$0.09 - 2.25 * 10^{-5}R - 0.001W = 0$

$0.09 -2.25 * 10^{-5} * 500 - 0.001W = 0$

$0.09 -0.01125 - 0.001W = 0$

$0.07875 - 0.001W = 0$

Collect like terms

$- 0.001W = -0.07875$

Solve for W

$W = \frac{-0.07875}{ - 0.001}$

$W = 78.75$

$W \approx 79$

$(R,W) \to (500,79)$

When $W = 0$ , we have:

$0.09 - 2.25 * 10^{-5}R - 0.001W = 0$

$0.09 - 2.25 * 10^{-5}R - 0.001*0 = 0$

$0.09 - 2.25 * 10^{-5}R = 0$

Collect like terms

$- 2.25 * 10^{-5}R = -0.09$

Solve for R

$R = \frac{-0.09}{- 2.25 * 10^{-5}}$

$R = 4000$

So, we have:

$(R,W) \to (4000,0)$

When $R =0$ , we have:

$-0.02W + 0.00004RW = 0$

$-0.02W + 0.00004W*0 = 0$

$-0.02W + 0 = 0$

$-0.02W = 0$

$W=0$

So, we have:

$(R,W) \to (0,0)$

Hence, the points of equilibrium are:

$(0,0)$ $(4000,0)$ and $(500,79)$

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