A pair of sneakers is regularly at $52. The store sells it for $78. What is the percent increase?

Could someone answer this pls??

Hitman42 [59]

Answer:

AREA OF SHADED REGION=AREA OF LARGE PARALLELOGRAM-AREA OF SMALL PARALLELOGRAM

=8*12-(6*4)

=96-24

=72unit^2

5 0

2 years ago

Read 2 more answers

Which values of P and Q result in an equation with no solutions? -60x+32=Qx+P

Elena L [17]

Q = -60 and P ≠ 32 will result in an equation with no solutions. (Both conditions must be met.)

_____
For Q = -60 and P = 32, there will be an infinite number of solutions. For any other values of Q and P, the solution is
.. x = (32 -P)/(Q +60)

4 0

3 years ago

a woman walks in and steals 100 from a register. 5 minutes later uses the same 100 to buy merchandise from that same store and b

murzikaleks [220]

She took 100, the store has - 100
she went back and bough
70 bucks in stuff, the store has - 70

yes, she took the $100 bill and gave it, got back $30, that means she only kept $30 from the original bill.

the store however, lost the original $100 bill plus the $70 in merchandise, so they have -$170.

5 0

3 years ago

Find the Fourier series of f on the given interval. f(x) = 1, ?7 < x < 0 1 + x, 0 ? x < 7

Zolol [24]

$f(x)=\begin{cases}1&\text{for }-7$

The Fourier series expansion of $f(x)$ is given by

$\dfrac{a_0}2+\displaystyle\sum_{n\ge1}a_n\cos\frac{n\pi x}7+\sum_{n\ge1}b_n\sin\frac{n\pi x}7$

where we have

$a_0=\displaystyle\frac17\int_{-7}^7f(x)\,\mathrm dx$
$a_0=\displaystyle\frac17\left(\int_{-7}^0\mathrm dx+\int_0^7(1+x)\,\mathrm dx\right)$
$a_0=\dfrac{7+\frac{63}2}7=\dfrac{11}2$

The coefficients of the cosine series are

$a_n=\displaystyle\frac17\int_{-7}^7f(x)\cos\dfrac{n\pi x}7\,\mathrm dx$
$a_n=\displaystyle\frac17\left(\int_{-7}^0\cos\frac{n\pi x}7\,\mathrm dx+\int_0^7(1+x)\cos\frac{n\pi x}7\,\mathrm dx\right)$
$a_n=\dfrac{9\sin n\pi}{n\pi}+\dfrac{7\cos n\pi-7}{n^2\pi^2}$
$a_n=\dfrac{7(-1)^n-7}{n^2\pi^2}$

When $n$ is even, the numerator vanishes, so we consider odd $n$ , i.e. $n=2k-1$ for $k\in\mathbb N$ , leaving us with

$a_n=a_{2k-1}=\dfrac{7(-1)-7}{(2k-1)^2\pi^2}=-\dfrac{14}{(2k-1)^2\pi^2}$

Meanwhile, the coefficients of the sine series are given by

$b_n=\displaystyle\frac17\int_{-7}^7f(x)\sin\dfrac{n\pi x}7\,\mathrm dx$
$b_n=\displaystyle\frac17\left(\int_{-7}^0\sin\dfrac{n\pi x}7\,\mathrm dx+\int_0^7(1+x)\sin\dfrac{n\pi x}7\,\mathrm dx\right)$
$b_n=-\dfrac{7\cos n\pi}{n\pi}+\dfrac{7\sin n\pi}{n^2\pi^2}$
$b_n=\dfrac{7(-1)^{n+1}}{n\pi}$

So the Fourier series expansion for $f(x)$ is

$f(x)\sim\dfrac{11}4-\dfrac{14}{\pi^2}\displaystyle\sum_{n\ge1}\frac1{(2n-1)^2}\cos\frac{(2n-1)\pi x}7+\frac7\pi\sum_{n\ge1}\frac{(-1)^{n+1}}n\sin\frac{n\pi x}7$

3 0

3 years ago

Olga says: If you subtract 8 from

alexandr402 [8]

Answer: 1.5

Step-by-step explanation: Reverse the equation to get the answer.

-13 divided by 2 is -6.5

add 8 and you'll get 1.5

7 0

3 years ago