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1 year ago

Need help (problem and directions in the image):

Mathematics

1 answer:

Alex1 year ago

6 0

Triangle ABC is congruent to triangle CDE using the side - angle - side congruence theorem. The sink hole is 52.2 ft and the Perimeter of ABC is 172.2 ft

<h3>What are congruent triangles?</h3>

Two triangles are said to be congruent if they have the same shape and their corresponding sides are congruent.

In the image shown:

AC = CE, BC = ED and they have the same angle (opposite angles are congruent).

Hence:

Triangle ABC is congruent to triangle CDE using the side - angle - side congruence theorem.

AB = DE = 52.2 ft

Perimeter of ABC = AB + BC + AC = 50 + 70 + 52.2 = 172.2 ft

Find out more on congruent triangles at: brainly.com/question/2938476

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The area of room is 812 square feet, cost of carpet to cover the room is $6739.6 and the remaining square feet of carpet are covered by furniture is 203 square feet.

<h3>What is the area of trapezoid?</h3>

The area of the trapezoid is the space occupied by it on the two dimensional plane. The area of this figure is half of the product of height and sum of sides.

$A=\dfrac{(a+b)h}{2}$

Here, (a and b) are the base sides of the trapezoid and (h) is the height.

A room is shaped like a trapezoid which has the height of 28 feet and the length of sides as 23 ft and 35 ft. Thus, the area of it is,

$A=\dfrac{(a+b)h}{2}\\A=\dfrac{(23+35)28}{2}\\A=812\rm\; ft^2$

Carpet costs $8.30 per square foot. Thus, the cost of carpet with area 812 ft² is,

C=812×$8.30

C=$6739.6

Furniture covers 25% of the floor. Thus, the remaining square feet of carpet are covered by furniture is,

$A=\dfrac{25}{100}\times812\\A=203\rm\; ft^2$

Thus, the area of room is 812 square feet, cost of carpet to cover the room is $6739.6 and the remaining square feet of carpet are covered by furniture is 203 square feet.

Learn more about the area of trapezoid here;

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1 year ago

Use the definition of Taylor series to find the Taylor series, centered at c, for the function. f(x) = sin x, c = 3π/4

anyanavicka [17]

Answer:

$\sin(x) = \sum\limit^{\infty}_{n = 0} \frac{1}{\sqrt 2}\frac{(-1)^{n(n+1)/2}}{n!}(x - \frac{3\pi}{4})^n$

Step-by-step explanation:

Given

$f(x) = \sin x\\$

$c = \frac{3\pi}{4}$

Required

Find the Taylor series

The Taylor series of a function is defines as:

$f(x) = f(c) + f'(c)(x -c) + \frac{f"(c)}{2!}(x-c)^2 + \frac{f"'(c)}{3!}(x-c)^3 + ........ + \frac{f*n(c)}{n!}(x-c)^n$

We have:

$c = \frac{3\pi}{4}$

$f(x) = \sin x\\$

$f(c) = \sin(c)$

$f(c) = \sin(\frac{3\pi}{4})$

This gives:

$f(c) = \frac{1}{\sqrt 2}$

We have:

$f(c) = \sin(\frac{3\pi}{4})$

Differentiate

$f'(c) = \cos(\frac{3\pi}{4})$

This gives:

$f'(c) = -\frac{1}{\sqrt 2}$

We have:

$f'(c) = \cos(\frac{3\pi}{4})$

Differentiate

$f"(c) = -\sin(\frac{3\pi}{4})$

This gives:

$f"(c) = -\frac{1}{\sqrt 2}$

We have:

$f"(c) = -\sin(\frac{3\pi}{4})$

Differentiate

$f"'(c) = -\cos(\frac{3\pi}{4})$

This gives:

$f"'(c) = - * -\frac{1}{\sqrt 2}$

$f"'(c) = \frac{1}{\sqrt 2}$

So, we have:

$f(c) = \frac{1}{\sqrt 2}$

$f'(c) = -\frac{1}{\sqrt 2}$

$f"(c) = -\frac{1}{\sqrt 2}$

$f"'(c) = \frac{1}{\sqrt 2}$

$f(x) = f(c) + f'(c)(x -c) + \frac{f"(c)}{2!}(x-c)^2 + \frac{f"'(c)}{3!}(x-c)^3 + ........ + \frac{f*n(c)}{n!}(x-c)^n$

becomes

$f(x) = \frac{1}{\sqrt 2} - \frac{1}{\sqrt 2}(x - \frac{3\pi}{4}) -\frac{1/\sqrt 2}{2!}(x - \frac{3\pi}{4})^2 +\frac{1/\sqrt 2}{3!}(x - \frac{3\pi}{4})^3 + ... +\frac{f^n(c)}{n!}(x - \frac{3\pi}{4})^n$

Rewrite as:

$f(x) = \frac{1}{\sqrt 2} + \frac{(-1)}{\sqrt 2}(x - \frac{3\pi}{4}) +\frac{(-1)/\sqrt 2}{2!}(x - \frac{3\pi}{4})^2 +\frac{(-1)^2/\sqrt 2}{3!}(x - \frac{3\pi}{4})^3 + ... +\frac{f^n(c)}{n!}(x - \frac{3\pi}{4})^n$

Generally, the expression becomes

$f(x) = \sum\limit^{\infty}_{n = 0} \frac{1}{\sqrt 2}\frac{(-1)^{n(n+1)/2}}{n!}(x - \frac{3\pi}{4})^n$

Hence:

$\sin(x) = \sum\limit^{\infty}_{n = 0} \frac{1}{\sqrt 2}\frac{(-1)^{n(n+1)/2}}{n!}(x - \frac{3\pi}{4})^n$

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2 years ago

Which of the following is characteristic of an epic hero?

jenyasd209 [6]

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3 years ago

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