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3 years ago

What inequality requires more than one step if the answer is x<-3

1 answer:

4 0

Is the question asking to make up an equality that, in essence, is anything but x+3<0?
In that case:

2x+9<0
2x<-9
x<-9/3
x<-3

You might be interested in

You are a personnel director and are interested in predicting the Number of Shares of Company Stock (Y) using the Number of Year

hoa [83]

Answer:

intercept=b0=75

Step-by-step explanation:

The least squares estimate of the intercept b0 can be computed as

b0=ybar-b1*xbar.

ybar=average number of shares of company stock=525.

xbar= average number of years employed=22.5.

slope=b1=20.

Thus,

intercept=b0=ybar-b1*xbar

intercept=b0=525-20*22.5

intercept=b0=525-450

intercept=b0=75.

Thus, the estimate of intercept b0=75.

8 0

3 years ago

Choose ALL answers that describe the quadrilateral VWXYVWXY if VW = 55VW=55, WX = 67WX=67, XY = 45XY=45, YV = 33YV=33, \text{m}\

frozen [14]

The quadrilateral VWXY is a trapezoid.

<h3>What is a quadrilateral?</h3>

A quadrilateral is a polygon with four sides and four angles. Types of quadrilaterals are parallelogram, rectangle, rhombus, square and trapezoid.

A Trapezoid is a quadrilateral with a pair of opposite parallel sides.

Given a quadrilateral VWXY, WX = 67, XY = 45, YV = 33, VW = 55, and all the angles are different, hence:

The quadrilateral VWXY is a trapezoid.

Find out more on quadrilateral at: brainly.com/question/23935806

5 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

PLEASE HELP

ICE Princess25 [194]

C. 20 / 4

20 represents how much time practicing in running has reduced her run time, and dividing by 4 weeks show the amount of seconds lowered per week (5 seconds).

3 0

4 years ago

Ok I need help is this correct? I tried but I gave up.

Bas_tet [7]

X+y=12
8x+10y=102

8x+8y=96
8x+10y=102 _
-2y= -6
y=3

x+y=12
x=12- 3
x=9

or
8x+10y=102
8x+30=102
8x=102-30
8x=72
x=9

4 0

3 years ago