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1 year ago

All the corners of the shape are right angles. The perimeter of the shape is 28 m. Work out the area of ABCE.

Mathematics

1 answer:

Amiraneli [1.4K]1 year ago

4 0

Answer:

13 m²

Step-by-step explanation:

To find the area of the shaded region, subtract the sum of the triangles from the area of the shape.

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3 years ago

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What is the length of the hypotenuse of the triangle below

Hatshy [7]

One way to solve this is to use Pythagorean theorem: the square of one leg of triangle plus square of other leg of the triangle equals c the hypotenuse (longest side of triangle). You might see this as the formula a^2 + b^2 = c^2, where a and b are the legs and c is the hypotenuse.
In this case, the legs are 3√2 and the hypotenuse is h.
Using the formula:
(3√2)² + (3√2)² = h²
18 + 18 = h²
h = 6

The other way to do this is with trigonometric angles.
Remember cosine is adjacent over hypotenuse.
cos(45°) = (3√2) / h
h = (3√2) / cos(45°)
h = 6

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4 years ago

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Can guys please help me, please solve E

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Answer:

Step-by-step explanation:

h(x)=12/x

h(a)= 12/a ( a is a variable number just replace x with a)

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3 years ago

What is the length of the curve with parametric equations x = t - cos(t), y = 1 - sin(t) from t = 0 to t = π? (5 points)

zzz [600]

Answer:

B) 4√2

General Formulas and Concepts:

Calculus

Differentiation

Derivatives
Derivative Notation

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Parametric Differentiation

Integration

Integrals
Definite Integrals
Integration Constant C

Arc Length Formula [Parametric]: $\displaystyle AL = \int\limits^b_a {\sqrt{[x'(t)]^2 + [y(t)]^2}} \, dx$

Step-by-step explanation:

Step 1: Define

Identify

$\displaystyle \left \{ {{x = t - cos(t)} \atop {y = 1 - sin(t)}} \right.$

Interval [0, π]

Step 2: Find Arc Length

[Parametrics] Differentiate [Basic Power Rule, Trig Differentiation]: $\displaystyle \left \{ {{x' = 1 + sin(t)} \atop {y' = -cos(t)}} \right.$
Substitute in variables [Arc Length Formula - Parametric]: $\displaystyle AL = \int\limits^{\pi}_0 {\sqrt{[1 + sin(t)]^2 + [-cos(t)]^2}} \, dx$
[Integrand] Simplify: $\displaystyle AL = \int\limits^{\pi}_0 {\sqrt{2[sin(x) + 1]} \, dx$
[Integral] Evaluate: $\displaystyle AL = \int\limits^{\pi}_0 {\sqrt{2[sin(x) + 1]} \, dx = 4\sqrt{2}$