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3 years ago

Please help thank you!

Mathematics

1 answer:

Lena [83]3 years ago

6 0

1 is D, an angle bisector only cuts the original angle in half.

2 is
9x - 18 = 3x
6x - 18 = 0
6x = 18
x = 3

3 is 9(3) - 18 = 27 - 18 = 9

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Simplify (6x2 - 3 + 5x3) - (4x3 - 2x2 - 16

Masteriza [31]

Answer:

x^3 +8x^2 +13

Step-by-step explanation:

(6x^2 - 3 + 5x^3) - (4x^3 - 2x^2 - 16)

Distribute the minus sign

6x^2 - 3 + 5x^3 -4x^3 + 2x^2 + 16

Combine like terms

x^3 +8x^2 +13

8 0

3 years ago

HELP PLEASE I HAVE A COUPLE MORE! Please help and fast would be great! BRAINLEST!!!

Harlamova29_29 [7]

Answer:

b. slope: -5; y-intercept: 7

Step-by-step explanation:

We are given the equation:

y + 5x = 7

To find the slope and y-intercept of the line, it would be helpful to get the equation into slope-intercept form. The slope-intercept form of a line is:

y = mx + b

where m is the slope and b is the y-intercept.

Lets get the given equation into slope-intercept form.

y + 5x = 7

Subtract 5x from both sides.

y = -5x + 7

Now we have the equation in slope-intercept form. By looking at the equation, we can see that the slope is -5 and that the y-intercept is 7.

The correct answer choice would be b.

I hope you find my answer and explanation to be helpful. Happy studying.

3 0

3 years 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:

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