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3 years ago

HELP !!! | 20 Points.

Mathematics

2 answers:

RUDIKE [14]3 years ago

5 0

Answer:

No. The ladder is too steep.

Step-by-step explanation:

Length of ladder = 17 feet. This is the hypotenuse.

Side a of a right triangle = 16.6

To answer this, you need one of the trigonometric functions.

Opposite = 16.5

hypotenuse = 17

The function you need is the sine function.

Sin(theta) = opposite / hypotenuse

sin(theta) = 16.5 / 17

Sin(theta) = 0.97059

theta = sin-1(0.07059)

theta = 76.07

No the ladder is too steep.

deff fn [24]3 years ago

5 0

Answer:

The ladder will not be safe.

Step-by-step explanation:

The ladder is 17 ft long

The top of the ladder is 16.5 ft above the ground

The ladder makes a right triangle of height = 16.5 ft and hypotenuse = 17 ft

To find the angle the ladder makes with the ground,

Height ÷ hypotenuse = sin angle

i.e $\frac{16.5}{17}$ = sin angle

$sin^{-1}$ angle = 76.1°

So the ladder is not safe since the angle it makes with the ground is greater than 70°.

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How to prove tan z is analytic using cauchy-riemann conditions

Basile [38]

A function $f(z)=f(x+iy)=u(x,y)+i v(x,y)$ is analytic if the C-R conditions are satisfied:

$\begin{cases}u_x=v_y\\u_y=-v_x\end{cases}$

We have

$f(z)=\tan z=\tan(x+iy)$

Recall the angle sum formula for tangent:

$\tan(x+iy)=\dfrac{\tan x+\tan iy}{1-\tan x\tan iy}$

Now recall that

$\tan iy=\dfrac{\sin iy}{\cos iy}=\dfrac{-\frac{e^y-e^{-y}}{2i}}{\frac{e^y+e^{-y}}2}=\dfrac{i\sinh y}{\cosh y}=i\tanh y$

So we have

$\dfrac{\tan x+\tan iy}{1-\tan x\tan iy}=\dfrac{\tan x+i\tanh y}{1-i\tan x\tanh y}$
$=\dfrac{(\tan x+i\tanh y)(1+i\tan x\tanh y)}{(1-i\tan x\tanh y)(1+i\tan x\tanh y)}$
$=\dfrac{\tan x+i\tanh y+i\tan^2x-\tan x\tanh^2y}{1+\tan^2x\tanh^2y}$
$=\dfrac{\tan x(1-\tanh^2y)}{1+\tan^2x\tanh^2y}+i\dfrac{\tanh y(1+\tan^2x)}{1+\tan^2x\tanh^2y}$
$=\underbrace{\dfrac{\tan x\sech^2y}{1+\tan^2x\tanh^2y}}_{u(x,y)}+i\underbrace{\dfrac{\tanh y\sec^2x}{1+\tan^2x\tanh^2y}}_{v(x,y)}$

We could stop here, but taking derivatives may be messy and would be easier to do if we can write this in terms of sines and cosines.

$u(x,y)=\dfrac{\tan x\sech^2y}{1+\tan^2x\tanh^2y}=\dfrac{\frac{\sin x}{\cos x\cosh^2x}}{1+\frac{\sin^2x\sinh^2y}{\cos^2x\cosh^2y}}=\dfrac{\sin x\cos x}{\cos^2x\cosh^2y+\sin^2x\sinh^2y}$
$u(x,y)=\dfrac{\sin2x}{2\cos^2x\cosh^2y+2(1-\cos^2x)(\cosh^2y-1)}=\dfrac{\sin2x}{2\cosh^2y-1+2\cos^2x-1}$
$u(x,y)=\dfrac{\sin2x}{\cos2x+\cosh2y}$

With similar usage of identities, we can find that

$v(x,y)=\dfrac{\sinh2y}{\cos2x+\cosh2y}$

Now we check the C-R conditions.

$u_x=\dfrac{2\cos2x(\cos2x+\cosh2y)-(-2\sin2x)\sin2x}{(\cos2x+\cosh2y)^2}=\dfrac{2+2\cos2x\cosh2y}{(\cos2x+\cosh2y)^2}$
$v_y=\dfrac{2\cosh2y(\cos2x+\cosh2y)-(2\sinh2y)\sinh2y}{(\cos2x+\cosh2y)^2}=\dfrac{2+2\cosh2y\cos2x}{(\cos2x+\cosh2y)^2}$
$\implies u_x=v_y$

Similarly, you can check that $u_y=-v_x$ , hence the C-R conditions are satisfied, and so $\tan z$ is analytic.

6 0

3 years ago

attendance at the montgomery country fair was 2,183 the first day 1,837 the second day and 1,947 the last night what was the ave

butalik [34]

328.5

Mark brainliest please

Hope this helps you

3 0

3 years ago

A supermarket has two customers waiting to pay for their purchases at counter I and one customer waiting to pay at counter II. L

Pachacha [2.7K]

Answer:

b. 0.864

Step-by-step explanation:

Let's start defining the random variables.

Y1 : ''Number of customers who spend more than $50 on groceries at counter 1''

Y2 : ''Number of customers who spend more than $50 on groceries at counter 2''

If X is a binomial random variable, the probability function for X is :

$P(X=x)=(nCx)p^{x}(1-p)^{n-x}$

Where P(X=x) is the probability of the random variable X to assume the value x

nCx is the combinatorial number define as :

$nCx=\frac{n!}{x!(n-x)!}$

n is the number of independent Bernoulli experiments taking place

And p is the success probability.

In counter I :

Y1 ~ Bi (n,p)

Y1 ~ Bi(2,0.2)

$P(Y1=y1)=(2Cy1)(0.2)^{y1}(0.8)^{2-y1}$

With y1 ∈ {0,1,2}

And P( Y1 = y1 ) = 0 with y1 ∉ {0,1,2}

In counter II :

Y2 ~ Bi (n,p)

Y2 ~ Bi (1,0.3)

$P(Y2=y2)=(1Cy2)(0.3)^{y2}(0.7)^{1-y2}$

With y2 ∈ {0,1}

And P( Y2 = y2 ) = 0 with y2 ∉ {0,1}

(1Cy2) with y2 = 0 and y2 = 1 is equal to 1 so the probability function for Y2 is :

$P(Y2=y2)=(0.3)^{y2}(0.7)^{1-y2}$

Y1 and Y2 are independent so the joint probability distribution is the product of the Y1 probability function and the Y2 probability function.

$P(Y1=y1,Y2=y2)=P(Y1=y1).P(Y2=y2)$

$P(Y1=y1,Y2=y2)=(2Cy1)(0.2)^{y1}(0.8)^{2-y1}(0.3)^{y2}(0.7)^{1-y2}$

With y1 ∈ {0,1,2} and y2 ∈ {0,1}

P( Y1 = y1 , Y2 = y2) = 0 when y1 ∉ {0,1,2} or y2 ∉ {0,1}

b. Not more than one of three customers will spend more than $50 can mathematically be expressed as :

$Y1 + Y2 \leq 1$

$Y1 + Y2\leq 1$ when Y1 = 0 and Y2 = 0 , when Y1 = 1 and Y2 = 0 and finally when Y1 = 0 and Y2 = 1

To calculate $P(Y1+Y2\leq 1)$ we must sume all the probabilities that satisfy the equation :

$P(Y1+Y2\leq 1)=P(Y1=0,Y2=0)+P(Y1=1,Y2=0)+P(Y1=0,Y2=1)$

$P(Y1=0,Y2=0)=(2C0)(0.2)^{0}(0.8)^{2-0}(0.3)^{0}(0.7)^{1-0}=(0.8)^{2}(0.7)=0.448$

$P(Y1=1,Y2=0)=(2C1)(0.2)^{1}(0.8)^{2-1}(0.3)^{0}(0.7)^{1-0}=2(0.2)(0.8)(0.7)=0.224$

$P(Y1=0,Y2=1)=(2C0)(0.2)^{0}(0.8)^{2-0}(0.3)^{1}(0.7)^{1-1}=(0.8)^{2}(0.3)=0.192$

$P(Y1+Y2\leq 1)=0.448+0.224+0.192=0.864\\P(Y1+Y2\leq 1)=0.864$

7 0

3 years ago

Which solution to the equation 3/2g+8 g+2/g^2-16 is extraneous? A. g = –4 B. g = –4 and g = 16

ollegr [7]

im assuming the equation is:-

3 / (2g + 8) = (g + 2) / (g^2 - 16)

3(g^2 - 16) = (2g + 8)(g + 2)

3g^2 - 48 = 2g^2 + 4g + 8g + 16

Answer is A x = -4

3g^2 - 2g^2 - 12g - 64 = 0

g^2 - 12g - 64 = 0

( g + 4)(g - 16) = 0

g = - 4, 16.

Test these solutions:-

g = -4

left side = 3/0 which is indeterminate

right side = -2/0 indeterminate

So x = -4 is extraneous.

x = 16:- LHS = 0.075 RHS = 0.075 so x = 16 is a root.

5 0

3 years ago

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