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3 years ago

Will mark brainliest if right!

1 answer:

8 0

When x = - 6, y = 3
When x = 1, y =?

-6 : 3
1 :?

$- 6 \div 3 = - 2 \\ 1 \div - 2 = - 0.5$
y = - 0.5

Answer is A.

Hope this helps. - M

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For 25 points: x + 5 = 10 2x + ? = 20 Answer for "?"

Phoenix [80]

Answer:

x= 5

?=10

Step-by-step explanation:

Mark me Brainliest Pls

8 0

3 years ago

A child walks due east on the deck of a ship at 4 miles per hour.

andrey2020 [161]

If you drew the two vectors for the velocity you find a triangle rectangle since the two directions East and North form an angle of 90°. I drew first the vector towards East and second the vector towards North.

Therefore, to get the speed relative to the surface you have to sum the two vectors, which means you can apply the Pythagorean theorem to find the hypotenuse, having the two legs.

h = √(4² + 12²) = √160 = 4√10 = 12.65mph

In order to find the direction, you need to use trigonometry:
- the sine of an angle is given by the ratio between the opposite side divided the hypotenuse;
- the cosine of an angle is given by the ratio between the adjacent side divided by the hypotenuse.

You can use either one, I will show you both:
sinα = O / H
inverting the formula: α = arcsin (O / H) = sin⁻¹ (O / H)
these two forms are equivalent.

Similarly:
cosβ = A / H
β = arccos (A / H) = cos⁻¹ (A / H).

Substituting your numbers:
α = sin⁻¹ (12 / 12.65) = sin⁻¹ (0.9486) = 1.25rad
β = cos⁻¹ (4 / 12.65) = cos⁻¹ (0.3162) = 1.25rad

As you can see, α = β so you can use either formula.

Your answer will then be: the child is moving relative to the surface of the water with a speed of 12.65mph at a direction of 1.25rad from East towards North.

ATTENTION!
If you drew first the vector towards North and then the one towards East, you get an equivalent triangle, which is rotated by 180° respect to the previous one. This way, what you earlier called Opposite is now Adjacent and vice-versa, but the formulas are still valid.
You will get:
α = sin⁻¹ (4 / 12.65) = sin⁻¹ (0.3162) = 0.32rad
β = cos⁻¹ (12 / 12.65) = cos⁻¹ (0.9486) = 0.32rad
And your answer will be: the child is moving relative to the surface of the water with a speed of 12.65mph at a direction of 0.32rad from North towards East.

3 0

3 years ago

El 5% de los focos producidos por una fabrica son defectuosos. Al extraer una muestra de 6 focos. ¿Cual es la probabilidad de qu

Angelina_Jolie [31]

Answer:

3/10 no estoy segura pero creo que es esto

3 0

3 years ago

(-7)(2)(3) 12 0 42 42 0-13

oee [108]

Is one of the answer choices -42, because that would be my answer.

4 0

3 years ago

Read 2 more answers

This problem asks for Taylor polynomials forf(x) = ln(1 +x) centered at= 0. Show Your work in an organized way.(a) Find the 4th,

stich3 [128]

Answer:

a) The 4th degree , 5th degree and sixth degree polynomials

$f^{lV} (x) = \frac{(2(-3))}{(1+x)^4} (1)= \frac{((-1)^3(3!))}{(1+x)^4}$

$f^{V} (x) = \frac{(2(-3)(-4))}{(1+x)^5} =\frac{(-1)^4 (4!)}{(1+x)^5}$

$f^{V1} (x) = \frac{(-120))}{(1+x)^6} (1) = \frac{(-1)^5 5!}{(1+x)^6}$

b)The nth degree Taylor polynomial for f(x) centered at x = 0, in expanded form.

$log(1+x) = x - \frac{x^2}{2} +\frac{x^3}{3} - \frac{x^4}{4} + \frac{x^5}{5} - \frac{x^6}{6}+\\.. (-1)^{n-1}\frac{x^n}{n} +..$

Step-by-step explanation:

Given the polynomial function f(x) = log(1+x) …...(1) centered at x=0

f(x) = log(1+x) ……(1)

using formula $\frac{d}{dx} logx =\frac{1}{x}$

Differentiating Equation(1) with respective to 'x' we get

$f^{l} (x) = \frac{1}{1+x} (\frac{d}{dx}(1+x)$

$f^{l} (x) = \frac{1}{1+x} (1)$ ….(2)

At x= 0

$f^{l} (0) = \frac{1}{1+0} (1)= 1$

using formula $\frac{d}{dx} x^{n-1} =nx^{n-1}$

Again Differentiating Equation(2) with respective to 'x' we get

$f^{l} (x) = \frac{-1}{(1+x)^2} (\frac{d}{dx}((1+x))$

$f^{ll} (x) = \frac{-1}{(1+x)^2} (1)$ ….(3)

At x=0

$f^{ll} (0) = \frac{-1}{(1+0)^2} (1)= -1$

Again Differentiating Equation(3) with respective to 'x' we get

$f^{lll} (x) = \frac{(-1)(-2)}{(1+x)^3} (\frac{d}{dx}((1+x))$

$f^{lll} (x) = \frac{(-1)(-2)}{(1+x)^3} (1)= \frac{(-1)^2 (2)!}{(1+x)^3}$ ….(4)

At x=0

$f^{lll} (0) = \frac{(-1)(-2)}{(1+0)^3} (1)$

$f^{lll} (0) = 2$

Again Differentiating Equation(4) with respective to 'x' we get

$f^{lV} (x) = \frac{(2(-3))}{(1+x)^4} (\frac{d}{dx}((1+x))$

$f^{lV} (x) = \frac{(2(-3))}{(1+x)^4} (1)= \frac{((-1)^3(3!))}{(1+x)^4}$ ....(5)

$f^{lV} (0) = \frac{(2(-3))}{(1+0)^4}$

$f^{lV} (0) = -6$

Again Differentiating Equation(5) with respective to 'x' we get

$f^{V} (x) = \frac{(2(-3)(-4))}{(1+x)^5} (\frac{d}{dx}((1+x))$

$f^{V} (x) = \frac{(2(-3)(-4))}{(1+x)^5} =\frac{(-1)^4 (4!)}{(1+x)^5}$ .....(6)

At x=0

$f^{V} (x) = 24$

Again Differentiating Equation(6) with respective to 'x' we get

$f^{V1} (x) = \frac{(2(-3)(-4)(-5))}{(1+x)^6} (\frac{d}{dx}((1+x))$

$f^{V1} (x) = \frac{(-120))}{(1+x)^6} (1) = \frac{(-1)^5 5!}{(1+x)^6}$

and so on...

The nth term is

$f^{n} (x) = = \frac{(-1)^{n-1} (n-1)!}{(1+x)^n}$

Step :-2

Taylors theorem expansion of f(x) is

$f(x) = f(a) + \frac{x}{1!} f^{l}(x) +\frac{(x-a)^2}{2!}f^{ll}(x)+\frac{(x-a)^3}{3!}f^{lll}(x)+\frac{(x-a)^4}{4!}f^{lV}(x)+\frac{(x-a)^5}{5!}f^{V}(x)+\frac{(x-a)^6}{6!}f^{VI}(x)+...….. \frac{(x-a)^n}{n!}f^{n}(x)$

At x=a =0

$f(x) = f(0) + \frac{x}{1!} f^{l}(0) +\frac{(x)^2}{2!}f^{ll}(0)+\frac{(x)^3}{3!}f^{lll}(0)+\frac{(x)^4}{4!}f^{lV}(0)+\frac{(x)^5}{5!}f^{V}(0)+\frac{(x)^6}{6!}f^{VI}(0)+...….. \frac{(x-0)^n}{n!}f^{n}(0)$

Substitute all values , we get

$f(x) = f(0) + \frac{x}{1!} (1) +\frac{(x)^2}{2!}(-1)+\frac{(x)^3}{3!}(2)+\frac{(x)^4}{4!}(-6)+\frac{(x)^5}{5!}(24)+\frac{(x)^6}{6!}(-120)+...….. \frac{(x-0)^n}{n!}f^{n}(0)$

On simplification we get

$log(1+x) = x - \frac{x^2}{2} +\frac{x^3}{3} - \frac{x^4}{4} + \frac{x^5}{5} - \frac{x^6}{6}+\\.. (-1)^{n-1}\frac{x^n}{n} +..$

4 0

4 years ago