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Over [174]
4 years ago
5

Help me with this please!

Mathematics
2 answers:
Mrac [35]4 years ago
6 0

Answer:

Area of triangle = 20√2

Step-by-step explanation:

Formula:

Area of triangle = bh/2

b - base  of triangle

h - height of triangle

Distance formula

d = √(x₂ - x₁)² +(y₂ - y₁)²

From the figure we can see that, the given triangle is right angled triangle.

Base = AC and Height = AB

<u>To find AB and AC</u>

A(1,3), B(4,7) and C(9, -5)

AB =  √(4 - 1)² +(7 - 3)² =  √(3² + 4²) = 5

AC = √(9 - 1)² +(-5 - 3)² =  √(8² + -8²) = 8√2

<u>To find the are of triangle</u>

Area = bh/2 = (5* 8√2)/2 = 20√2

Schach [20]4 years ago
6 0

Answer:

Area =  20√2 is the area of triangle.

Step-by-step explanation:

We have given the triangle in the graph.

We have to find the area of the triangle.

The formula for the area of triangle is :

Area = 1/2(base×height)

In the graph, the base is represented by  AC, and height is represented by AB.

First, we use distance formula to find AC and AB.

Base = AC =√(9-1)² + (-5-3)² = 8√2

Height = AB=√(4-1)² +(7-3)² = 5

Put this values of height and base in the formula we get,

Area = 1/2 (8√2 ×5)

Area =  20√2 is the area of triangle.

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3 years ago
Verify the identity. cot(x - pi/2) = -tan(x)
gizmo_the_mogwai [7]

Answer:

See below.

Step-by-step explanation:

\cot(x-\frac{\pi}{2})=-\tan(x)

Convert the cotangent to cosine over sine:

\frac{\cos(x-\frac{\pi}{2} )}{\sin(x-\frac{\pi}{2})} =-\tan(x)

Use the cofunction identities. The cofunction identities are:

\sin(x)=\cos(\frac{\pi}{2}-x)\\\cos(x)=\sin(\frac{\pi}{2}-x)

To convert this, factor out a negative one from the cosine and sine.

\frac{\cos(-(\frac{\pi}{2}-x ))}{\sin(-(\frac{\pi}{2}-x))} =-\tan(x)

Recall that since cosine is an even function, we can remove the negative. Since sine is an odd function, we can move the negative outside:

\frac{\cos((\frac{\pi}{2}-x ))}{-\sin((\frac{\pi}{2}-x))} =-\tan(x)\\-\frac{\sin(x)}{\cos(x)} =-\tan(x)\\-\tan(x)\stackrel{\checkmark}{=}-\tan(x)

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<h3>Hey .... </h3>

here's your answer in the given attachment

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