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Degger [83]
3 years ago
5

Cancelling a dance class after several students hurt themselves is an example of which risk management strategy?

Mathematics
1 answer:
MariettaO [177]3 years ago
3 0
Answer:
Avoid strategy as the threat/risk is directly related to identified threats and these sources of threats are cancelled

Explanation:
Risk management is defined as: 
The identification , analysis, control and possibly elimination of any undesired hazards or risks arising from a certain situation or activity

There are four risk management strategies which are:
1- Avoid: To eliminate the risk by totally avoiding the situation/activity
2- Mitigate: To try to reduce the expected hazards of the situation/activity
3- Accept: To accept the hazards and take your chances
4- Transfer: To outsource the risk (as in case of insurance)

In case of cancelling the dance class, we can note that the risks are directly related to identified threats and the threats are totally cancelled in order to avoid those risks. Therefore, the best strategy describing this situation is "avoid strategy"

Hope this helps :)
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Prove that: (Sec A- cosec A)(1+ tan A+cot A) = tan A× sec A - cot A × cosec A
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Answer:

Step-by-step explanation:

(sec A-cosecA)(1+tanA+cotA)=tanAsecA-cotAcosecA

we take the LHS so here goes,

(sec A-cosecA)(1+tanA+cotA)=tanAsecA-cotAcosecA\\(\frac{1}{cosA} -\frac{1}{sinA})(1+\frac{sinA}{cosA}+\frac{cosA}{sinA})\\\\(\frac{sinA-cosA}{sinAcosA})(\frac{sinAcosA+sin^2A+cos^2A}{sinAcosA})\\

since , sin^2A+cos^2A=1

the identity becomes,

(\frac{sinA-cosA}{sinAcosA})(\frac{1+sinAcosA}{sinAcosA})\\\\(\frac{sinA+sin^2AcosA-cosA-cos^2AsinA}{sin^2Acos^2A})\\\\

now, we know,

sin^2A=1-cos^2A and cos^2A=1-sin^2A

the identity becomes,

(\frac{sinA+(1-cos^2A)cosA-cosA-(1-sin^2A)sinA}{sin^2Acos^2A} )\\\\

(\frac{sinA+cosA-cos^3A-cosA-sinA+sin^3A}{sin^2Acos^2A})

sin A and cos A cancel out it becomes zero

\frac{sin^3A-cos^3A}{sin^2Acos^2A} \\\\

Splitting the denominator the identity becomes

\frac{sin^3A}{sin^2Acos^2A}-\frac{cos^3A}{sin^2Acos^2A}  \\\\\frac{sinA}{cos^2A} - \frac{cosA}{sin^2A} \\\\\frac{sinA}{cosA}(\frac{1}{cosA})-\frac{cosA}{sinA}(\frac{1}{sinA})\\\\

Hence,

tanAsecA-cotAcosecA

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1. -7<-4

2. -4>-7

Step-by-step explanation:

-4 is greater than -7, because -4 is a small negative number, and is closer to 0.

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Graph the function f(x) = (x + 1)(x – 5). Use the drop-down menus to complete the steps needed to graph the function
Effectus [21]

Answer:

0 = (x+1)(x-5)

x= -1 , x= 5

f(x) = x^2 -4x -5

V_x = -\frac{b}{2a}

With a = 1, b=-4 , c=-5

V_x = -\frac{-4}{2*1}= 2

f(2) = 2^2 -4*2 -5 = -9

f(0) = 0^2 -4*0 -5=-5

So then the x intercept would be (0,-5). And finally we can graph the function as we can see in the figure attached.

Step-by-step explanation:

For this case we know the following function:

f(x) = (x+1)(x-5)

We can begin the zeros or the values where the function is 0 like this:

0 = (x+1)(x-5)

And solving for x we got:

x= -1 , x= 5

Now we can rewrite the expression like this:

f(x) = x^2 -4x -5

And we can find the position for the vertex at x with this formula:

V_x = -\frac{b}{2a}

With a = 1, b=-4 , c=-5

And replacing we got:

V_x = -\frac{-4}{2*1}= 2

And then with the coordinate of x for the vertex we can find the coordinate of y replacing the value of x obtained for the vertex.

f(2) = 2^2 -4*2 -5 = -9

Then we can find the intercept using the value of x=0 and replacing into the function we got:

f(0) = 0^2 -4*0 -5=-5

So then the x intercept would be (0,-5). And finally we can graph the function as we can see in the figure attached.

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