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3 years ago

Ignore the stuff around it ,could someone explain me this?

Mathematics

1 answer:

Ierofanga [76]3 years ago

7 0

Answer:

The lines blue and green are perpendicular

Step-by-step explanation:

we know that

If two lines are perpendicular, then their slopes are opposite reciprocal (the product of their slopes is -1)

The formula to calculate the slope between two points is equal to

$m=\frac{y2-y1}{x2-x1}$

step 1

Find the slope of the blue line

we have the points

(-1,-3) and (0,3)

substitute

$m=\frac{3+3}{0+1}$

$m=\frac{6}{1}=6$

step 2

Find the slope of the red line

we have the points

(3,-3) and (4,2)

substitute

$m=\frac{2+3}{4-3}$

$m=\frac{5}{1}=5$

step 3

Find the slope of the green line

we have the points

(-4,-1) and (2,-2)

substitute

$m=\frac{-2+1}{2+4}$

$m=-\frac{1}{6}$

step 4

Compare the slopes

Blue line ----> $m=6$

Red line ----. $m=5$

Green line ---> $m=-\frac{1}{6}$

The slope of the blue line and the green line are opposite reciprocal ( their product is equal to -1)

therefore

The lines blue and green are perpendicular

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The city of Orlando’s building a dog park in downtown Orlando dog park will be triangular shaped as shown on the coordinate plan

kiruha [24]

Answer:

Perimeter of the dog park = 15.4 yards

Step-by-step explanation:

Coordinates of the vertices of the given triangle are,

P(1, 2), Q(1, 6), R(-4, 2)

Since distance between the two points $(x_1, y_1)$ and $(x_2,y_2)$ is,

d = $\sqrt{(y_2-y_1)^2+(x_2-x_1)^2}$

Length of PQ = $\sqrt{(1-1)^2+(2-6)^2}$

PQ = 4

Length of PR = $\sqrt{(1+4)^2+(2-2)^2}$

PR = 5

Length of QR = $\sqrt{(1+4)^2+(6-2)^2}$

QR = $\sqrt{25+16}$

= $\sqrt{41}$

= 6.4 yards

Therefore, perimeter of the given triangle = PQ + QR + PR

= 4 + 6.4 + 5

= 15.4 yards

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3 years ago

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2 years ago

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GenaCL600 [577]

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1 year ago

What is the side length of a cube that has a volume of 0.008 m

Nataliya [291]

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2 years ago

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Airida [17]

Answer:

89.1° or -1.4°

Step-by-step explanation:

1. Location:

You are on the Mont-Saint-Jean escarpment, near the Belgian town of Waterloo.

The French troops are about 50 m below you and 1.2 km distant.

2. Finding the firing angle

Data:

R = 1200 m

u = 600 m/s

h = -50 m (the height of the target)

a = 9.8 m/s²

We have two conditions.

Horizontal distance

(1) 1200 = 600t cosθ

Vertical distance

(2) -50 = 600t sinθ - 4.9t²

Divide each side of (1) by 600cosθ.

$(3) \, t =\dfrac{2}{\cos \theta}$

Substitute (3) into (2)

$-50 = 600t \sin \theta - 4.9t^{2} = 600 \left( \dfrac{2}{\cos \theta} \right ) \sin \theta - 4.9 \left( \dfrac{2}{\cos \theta} \right )^{2}\\\\(4) \, -50 = 1200 \tan \theta - \dfrac{19.6}{\cos^{2} \theta}$

Recall that

(5) sec²θ = 1/cos²θ = tan²θ + 1

Substitute (5) into (4)

$-50 = 1200 \tan \theta - 19.6 \left(\tan^{2} \theta}+ 1\right )$

Set up a quadratic equation

$\begin{array}{rcl}-50 & = & 1200 \tan \theta - 19.6\tan^{2} \theta -19.6 \\0 & = & 1200 \tan \theta - 19.6\tan^{2} \theta + 30.4\\0 & =&19.6\tan^{2} \theta - 1200 \tan \theta - 30.4\\0 & =&\tan^{2} \theta - 61.224 \tan \theta - 1.551\\\end{array}$

Solve for θ

Use the quadratic formula.

tanθ = 61.249 or -0.025

θ = arctan(61.249) = 89.1° or

θ = arctan(-0.025) = -1.4°

3 0

2 years ago