What is the value of x?

The granola summer buys used to cost $6.00 per pound but it has been marked up %15 A. How much did it cost summer to buy 2.6 pou

Maksim231197 [3]

The granola summer buys used to cost $6.00 per pound but it has been marked up %15 Question: How much did it cost summer to buy 2.6 pounds of granola at the old price.=> previous price of granola = 6.00 dollars per pound=> marked up 15% = 15% /100% = .15Solve:=> 6 * .15 = .9=> 6 + .9 = 6.9 dollars per poundNow, you want to buy 2.6 pounds=> 6.9 * 2.6 = 17.94 dollars <span>
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3 0

2 years ago

1. What is the value of x? Show your work to justify your answer.

Reil [10]

Answer:

x=72 and the exterior angle is 154

Step-by-step explanation:

We will call the unknown angle in the triangle y. Angle y and the angle (2x +10) form a straight line so they make 180 degrees.

y + 2x+10 =180

Solve for y by subtracting 2x+10 from each side.

y + 2x+10 - (2x+10) =180 - (2x+10)

y = 180-2x-10

y = 170-2x

The three angles of a triangle add to 180 degrees

x+ y+ 82 = 180

x+ (170-2x)+82 = 180

Combine like terms

-x +252=180

Subtract 252 from each side

-x+252-252 = 180-252

-x = -72

Multiply each side by -1

-1*-x = -72*-1

x=72

The exterior angle is 2x+10. Substitute x=72 into the equation.

2(72)+10

144+10

154

8 0

3 years ago

Use the Taylor series you just found for sinc(x) to find the Taylor series for f(x) = (integral from 0 to x) of sinc(t)dt based

Marina CMI [18]

In this question (brainly.com/question/12792658) I derived the Taylor series for $\mathrm{sinc}\,x$ about $x=0$ :

$\mathrm{sinc}\,x=\displaystyle\sum_{n=0}^\infty\frac{(-1)^nx^{2n}}{(2n+1)!}$

Then the Taylor series for

$f(x)=\displaystyle\int_0^x\mathrm{sinc}\,t\,\mathrm dt$

is obtained by integrating the series above:

$f(x)=\displaystyle\int\sum_{n=0}^\infty\frac{(-1)^nx^{2n}}{(2n+1)!}\,\mathrm dx=C+\sum_{n=0}^\infty\frac{(-1)^nx^{2n+1}}{(2n+1)^2(2n)!}$

We have $f(0)=0$ , so $C=0$ and so

$f(x)=\displaystyle\sum_{n=0}^\infty\frac{(-1)^nx^{2n+1}}{(2n+1)^2(2n)!}$

which converges by the ratio test if the following limit is less than 1:

$\displaystyle\lim_{n\to\infty}\left|\frac{\frac{(-1)^{n+1}x^{2n+3}}{(2n+3)^2(2n+2)!}}{\frac{(-1)^nx^{2n+1}}{(2n+1)^2(2n)!}}\right|=|x^2|\lim_{n\to\infty}\frac{(2n+1)^2(2n)!}{(2n+3)^2(2n+2)!}$

Like in the linked problem, the limit is 0 so the series for $f(x)$ converges everywhere.

7 0

3 years ago

M/n=p-6 what is n how to solve

madreJ [45]

M/n=p-6
n= m/(p-6)
i swapped the n and p-6

4 0

3 years ago

A square garden has a side of length that is 3 meters shorter than the length of a rectangle garden. Find the perimeter of the r

Ray Of Light [21]

If the square perimeter is 8y, then the square side would be:
square perimeter= 4*side
8y= 4*side
side=2y

The square garden side is 3m shorter than the length of the rectangle garden. The rectangle garden length will be:
rectangle length= square sides +3
rectangle length= 2y+3

rectangle perimeter= 2(rectangle length + rectangle side)
rectangle perimeter= 2(2y+3 + 2y+1)
rectangle perimeter= 2(4y+4)
rectangle perimeter= 8y + 8

5 0

3 years ago