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hram777 [196]
3 years ago
10

CAN SOMEONE HELP ME FIND THE AREA OF THIS TRIANGLE

Mathematics
2 answers:
Nitella [24]3 years ago
8 0

Answer:

A = 73.1\ m^2

Step-by-step explanation:

We calculate the height of the triangle using the function sin(\theta)

By definition:

sin(\theta) =\frac{h}{hypotenuse}

Where h is the height of the triangle

In this case we have that:

\theta=35\°

hypotenuse=17

Then:

sin(35) =\frac{h}{17}

h=sin(35)*17\\\\\\h =9.75

Then the area of a triangle is calculated as:

A = 0.5 * b * h

Where b is the length of the base of the triangle and h is its height

In this case

b=15

So

A = 0.5 *15*9.75

A = 73.1\ m^2

Lelu [443]3 years ago
4 0

Answer:

Area of triangle = 73.1 m²

Step-by-step explanation:

<u>Points to remember</u>

Area of triangle = bh/2

Where b - base and h - height

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

Let 'h' be the height of triangle

Sin 35 = h/17

h = 17 * Sin 35

 = 17 * 0.5736

 = 9.75 m

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

Here b = 15 m and h = 9.75

Area = bh/2

 = (15 * 9.75)/2

 = 73.125 ≈73.1 m²

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Step-by-step explanation:

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See explanation

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1. Given the expression

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Note that

\sqrt[7]{x^5}=x^{\frac{5}{7}} \\ \\\sqrt[4]{x^2}=x^{\frac{2}{4}}=x^{\frac{1}{2}}

When dividing \sqrt[7]{x^5} by \sqrt[4]{x^2}, we have to subtract powers (we cannot subtract 4 from 7, because then we get another expression), so

\dfrac{5}{7}-\dfrac{2}{4}=\dfrac{5}{7}-\dfrac{1}{2}=\dfrac{5\cdot 2-1\cdot 7}{14}=\dfrac{3}{14}

and the result is x^{\frac{3}{14}}=\sqrt[14]{x^3}

2. Given equation 3\sqrt[4]{(x-2)^3} -4=20

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Divide by 3:

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Rewrite the equation as:

(x-2)^{\frac{3}{4}}=8\\ \\(x-2)^{\frac{3}{4}}=2^3

Hence,

\left((x-2)^{\frac{3}{4}}\right)^{\frac{4}{3}}=(2^3)^{\frac{4}{3}}\\ \\x-2=2^{3\cdot \frac{4}{3}}\\ \\x-2=2^4\\ \\x-2=16\\ \\x-2+2=16+2\\ \\x=18

3 0
3 years ago
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