How is finding the area of a trapezoid different from finding the area of a parallelogram and a triangle?

Mathematics

1 answer:

trasher [3.6K]3 years ago

8 0

There are different formulas for each different shape.

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Simplify the expression cos x cot x+ sin x please select the best answer from the choices provided a.0, b.csc x, c. Tan x, d sec

expeople1 [14]

Answer:

$cot x = \frac{cos x}{sin x}$

$cos x \frac{cos x}{sin x} + sin x$

$\frac{cos^2 x}{sin x} +sin x$

$sin^2 x + cos^2 x =1$

Solving for $cos^2 x$ we got $cos^2 x =1 -sin^2 x$ and replacing this we got:

$\frac{1-sin^2 x}{sin x} +sin x$

$\frac{1}{sin x} -\frac{sin^2 x}{sin x} +sin x$

$csc x -sin x + sin x = csc x$

And then the best option for this case would be:

b.csc x

Step-by-step explanation:

For this case we have the following expression given:

$cos x cot x + sin x$

We know from math properties that the definition for cot is $cot x = \frac{cos x}{sin x}$

If we use this definition we got:

$cos x \frac{cos x}{sin x} + sin x$

$\frac{cos^2 x}{sin x} +sin x$

Now we can use the following identity:

$sin^2 x + cos^2 x =1$

Solving for $cos^2 x$ we got $cos^2 x =1 -sin^2 x$ and replacing this we got:

$\frac{1-sin^2 x}{sin x} +sin x$

$\frac{1}{sin x} -\frac{sin^2 x}{sin x} +sin x$

$csc x -sin x + sin x = csc x$

And then the best option for this case would be:

b.csc x

4 0

4 years ago

Rectangle R has varying length l and width w but a constant perimeter of 4 ft. A. Express the area A as a function of l. What do

ivanzaharov [21]

Given:
l = length of the rectangle
w = width of the rectangle
P = 4 ft, constant perimeter

Because the given perimeter is constant,
2(w + l) = 4
w + l = 2
w = 2 - l (1)

Part A.
The area is
A = w*l
= (2 - l)*l
A = 2l - l²
This is a quadratic function or a parabola.

Part B.
Write the parabola in standard form.
A = -[l² - 2l]
= -[ (l -1)² - 1]
= -(l -1)² + 1
This is a parabola with vertex at (1, 1). Because the leading coefficient is negative the curve is downward, as shown below.

The maximum value occurs at the vertex, so the maximum value of A = 1.
From equation (1), obtain
w = 2 - l = 2 - 1 = 1.
The maximum value of the area occurs when w=1 and l=1 (a square).

Answer:
The area is maximum when l=1 and w=1.
The geometric argument is based on the vertex of the parabola denoting maximum area.