Ask question

Ask question

All categories

3 years ago

11

In December, the average number of daily shoppers at North Point Mall in 2007 was 1600. This number shrank to 1000 daily shopper

s by December 2014. What is the exact percent decrease in the average number of daily shoppers? Show your work.

1 answer:

kykrilka [37]3 years ago

5 0

Answer:

you trying to cheat

Step-by-step explanation:

the answer is 62.5% percent.

You might be interested in

What is the value of y?<br><br> a. 30<br> b. 26<br> c. 28<br> d. 24

Verizon [17]

Answer:

option-b

Step-by-step explanation:

we are given three parallel lines

so, corresponding angles must be equal

we get

$126=5y-4$

now, we can solve for y

Add both sides by 4

$126+4=5y-4+4$

$130=5y$

Divide both sides by 5

and we get

$y=26$ ...........Answer

8 0

4 years ago

Write a comparison sentence to represent this equation 6x7=42.

nasty-shy [4]

Six times seven is fourty-two

3 0

3 years ago

Read 2 more answers

<img src="https://tex.z-dn.net/?f=HELP%20%5C%3A%20%20ME%20%5C%3A%20%20PLEASE%20" id="TexFormula1" title="HELP \: ME \: PLEASE

Olin [163]

Answer:

22°

Step-by-step explanation:

as we know from the angle sum property of the triangle- the sum of all angles of a triangle is 180°

So now in (d)

79°+79°+x=180°

158°+x=180°

x=180°-158° (transversal property)

x=22°

Hope it helps

8 0

3 years ago

Read 2 more answers

Please help with geometry problems

sertanlavr [38]

Answer:

put your best gees

Step-by-step explanation:

6 0

2 years ago

<img src="https://tex.z-dn.net/?f=%20%20%5Cdisplaystyle%20%5Cint%20%5Climits_%7B0%7D%5E%7B%20%5Cfrac%7B%20%5Cpi%7D%7B2%7D%20%7D%

murzikaleks [220]

Let $x = \arcsin(y)$ , so that

$\sin(x) = y$

$\tan(x)=\dfrac y{\sqrt{1-y^2}}$

$dx = \dfrac{dy}{\sqrt{1-y^2}}$

Then the integral transforms to

$\displaystyle \int_{x=0}^{x=\frac\pi2} \tan(x) \ln(\sin(x)) \, dx = \int_{y=\sin(0)}^{y=\sin\left(\frac\pi2\right)} \frac{y}{\sqrt{1-y^2}} \ln(y) \frac{dy}{\sqrt{1-y^2}}$

$\displaystyle \int_{x=0}^{x=\frac\pi2} \tan(x) \ln(\sin(x)) \, dx = \int_0^1 \frac{y}{1-y^2} \ln(y) \, dy$

Integrate by parts, taking

$u = \ln(y) \implies du = \dfrac{dy}y$

$dv = \dfrac{y}{1-y^2} \, dy \implies v = -\dfrac12 \ln|1-y^2|$

For 0 < y < 1, we have |1 - y²| = 1 - y², so

$\displaystyle \int_0^1 \frac{y}{1-y^2} \ln(y) \, dy = uv \bigg|_{y\to0^+}^{y\to1^-} + \frac12 \int_0^1 \frac{\ln(1-y^2)}{y} \, dy$

It's easy to show that uv approaches 0 as y approaches either 0 or 1, so we just have

$\displaystyle \int_0^1 \frac{y}{1-y^2} \ln(y) \, dy = \frac12 \int_0^1 \frac{\ln(1-y^2)}{y} \, dy$

Recall the Taylor series for ln(1 + y),

$\displaystyle \ln(1+y) = \sum_{n=1}^\infty \frac{(-1)^{n+1}}n y^n$

Replacing y with -y² gives the Taylor series

$\displaystyle \ln(1-y^2) = \sum_{n=1}^\infty \frac{(-1)^{n+1}}n (-y^2)^n = - \sum_{n=1}^\infty \frac1n y^{2n}$

and replacing ln(1 - y²) in the integral with its series representation gives

$\displaystyle -\frac12 \int_0^1 \frac1y \sum_{n=1}^\infty \frac{y^{2n}}n \, dy = -\frac12 \int_0^1 \sum_{n=1}^\infty \frac{y^{2n-1}}n \, dy$

Interchanging the integral and sum (see Fubini's theorem) gives

$\displaystyle -\frac12 \sum_{n=1}^\infty \frac1n \int_0^1 y^{2n-1} \, dy$

Compute the integral:

$\displaystyle -\frac12 \sum_{n=1}^\infty \frac1n \int_0^1 y^{2n-1} \, dy = -\frac12 \sum_{n=1}^\infty \frac{y^{2n}}{2n^2} \bigg|_0^1 = -\frac14 \sum_{n=1}^\infty \frac1{n^2}$

and we recognize the famous sum (see Basel's problem),

$\displaystyle \sum_{n=1}^\infty \frac1{n^2} = \frac{\pi^2}6$

So, the value of our integral is

$\displaystyle \int_0^{\frac\pi2} \tan(x) \ln(\sin(x)) \, dx = \boxed{-\frac{\pi^2}{24}}$

6 0

3 years ago