How many real solutions does the equation 2( x-1 )^2 +5=3 have?

You are paid $82.50 for 7 and a half hours of work what is your rate of pay

kodGreya [7K]

$11 an hour, because 82.5/7.5=11

4 0

3 years ago

Aul bought 10 reams of paper at the store for a total of $84 the tax on thepurchase was $4 what was the cost of each ream of pap

yKpoI14uk [10]

84+4/10
hope this helps

4 0

3 years ago

(a) Use the power series expansions for ex, sin x, cos x, and geometric series to find the first three nonzero terms in the powe

Fofino [41]

Answer:

a) $\mathbf{4 + \dfrac{x}{1!}- \dfrac{2x^2}{2!} ...}$

b) See Below for proper explanation

Step-by-step explanation:

a) The objective here is to Use the power series expansions for ex, sin x, cos x, and geometric series to find the first three nonzero terms in the power series expansion of the given function.

The function is $e^x + 3 \ cos \ x$

The expansion is of $e^x$ is $e^x = 1 + \dfrac{x}{1!}+ \dfrac{x^2}{2!}+ \dfrac{x^3}{3!} + ...$

The expansion of cos x is $cos \ x = 1 - \dfrac{x^2}{2!}+ \dfrac{x^4}{4!}- \dfrac{x^6}{6!}+ ...$

Therefore; $e^x + 3 \ cos \ x = 1 + \dfrac{x}{1!}+ \dfrac{x^2}{2!}+ \dfrac{x^3}{3!} + ... 3[1 - \dfrac{x^2}{2!}+ \dfrac{x^4}{4!}- \dfrac{x^6}{6!}+ ...]$

$e^x + 3 \ cos \ x = 4 + \dfrac{x}{1!}- \dfrac{2x^2}{2!} + \dfrac{x^3}{3!}+ ...$

Thus, the first three terms of the above series are:

$\mathbf{4 + \dfrac{x}{1!}- \dfrac{2x^2}{2!} ...}$

b)

The series for $e^x + 3 \ cos \ x$ is $\sum \limits^{\infty}_{x=0} \dfrac{x^x}{n!} + 3 \sum \limits^{\infty}_{x=0} ( -1 )^x \dfrac{x^{2x}}{(2n)!}$

let consider the series; $\sum \limits^{\infty}_{x=0} \dfrac{x^x}{n!}$

$|\frac{a_x+1}{a_x}| = | \frac{x^{n+1}}{(n+1)!} * \frac{n!}{x^x}| = |\frac{x}{(n+1)}| \to 0 \ as \ n \to \infty$

Thus it converges for all value of x

Let also consider the series $\sum \limits^{\infty}_{x=0}(-1)^x\dfrac{x^{2n}}{(2n)!}$

It also converges for all values of x

7 0

3 years ago

What is the answer to 8y-8=37-7y

Brut [27]

Y = 3

Just simplify the both sides, then add 7y to both then add 8 to both sides and divide both sides by 15

8 0

3 years ago

Find the perimeter of the following shape, rounded to the nearest tenth:

Mademuasel [1]

Answer:

B, 19.1

Step-by-step explanation:

Imagine that each side is the hypotenuse of a triangle. You want to use the Pythagorean theorem on each of these triangles to find the length of the side.

Let's start with AB:

One side of the triangle is the distance between the points in the x direction only. A is at the x-coordinate of -3 while B is at the x-coordinate of 2. The side is, thus, 5 units.

The other side of the triangle is the distance between the points in the y direction only. A is at the y-coordinate of 5, while B is at 6. The distance is 1 unit.

So we have a triangle with the base of 5 and the height of 1. The Pythagorean theorem says:

Base² + Height² = Hypotenuse²

Or

a² + b² = c²

where a, b are the sides, and c is the hypotenuse.

Put our values into this, the base is 5 and the height is 1:

5² + 1² = c²

25 + 1 = 26 = c²

c = √26 ≈ 5.1

Repeat this for the lengths BC, CD, and DA.

BC

x-difference = 2

y-difference = 4

4² + 2² = c²

20 = c²

c = √20 ≈ 4.47

CD

x-difference = 5

y-difference = 1

(since this is the same as AB we can use that value ≈ 5.1 )

DA

x-difference = 2

y-difference = 4

(since this is the same as BC, we can use that value ≈ 4.47)

Add all these together:

5.1 + 5.1 + 4.47 + 4.47 ≈ 19.14

(Note: this is not exact, but it's good enough for this purpose)

The closest option to this is B, 19.1 - thus, B is the answer.

5 0

3 years ago