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sweet [91]
2 years ago
7

Solve this:16<x+9 or 18≤x+9​

Mathematics
1 answer:
soldi70 [24.7K]2 years ago
5 0

1) 16 < x + 9

x + 9 > 16

x > 16 - 9

x > 9

2) 18 ≤ x + 9

x + 9 ≥ 18

x ≥ 18 - 9

x ≥ 9

_______

Hope it helps ⚜

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Find the area of a rhombus having perimeter 80 em and one diagonal 24 cm.<br>​
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In a triangle ABC, if &lt;B = 60° and the ratio of two sides is a: c= 2: square root 3+1, then &lt;A=
slavikrds [6]

Your answer is 75°.

To answer this question you need to use both the cosine rule and the sine rule. First, we need to find the length of side b by using the cosine rule, where a = 2 and c = √3 + 1. Then you substitute these into the equation:

b² = a² + c² - 2×a×c×cos(B)

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b² = 8 - 2 = 6

b = √6

Then you use this length in the sine rule, and find the angle:

\frac{sin(A)}{\sqrt{3}+1 } =\frac{sin(60)}{\sqrt{6} }

sin(A) = (√6 + √2)/4

A = 75

I hope this helps! Let me know if you have any questions

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Elasticity of Demand The demand function for a certain brand of backpacks is
Trava [24]

Answer:

See explanation

Step-by-step explanation:

Solution:-

- The demand function of a certain brand is given as price P a function of x quantity of goods ( in hundred ) demanded per month. The relation is:

                           P ( x ) = 50 Ln ( 50 / x ).

- The point price elasticity ( E ) of demand is given by:

                           E = \frac{P}{x}*\frac{dP}{dx}  

- Where, dP / dx : is the rate of change of price ( P ) with each hundred unit of good ( x ) is demanded.

- To determine the " dP / dx " by taking the first derivative of the given relation:

                          P ( x ) = 50 Ln ( 50 / x ).

                          d P ( x ) / dx = [ 50*x / 50 ] * [ -1*50 / x^2 ]

                                              = - 50 / x

- Hence the point price elasticity of demand is given by:

                          E = - ( P / x ) * ( 50 / x )

                          E = -50*P / x^2    

- For an inelastic demand, ! E ! is < 1:

                          ! -50*P / x^2 ! < 1

                          50*P / x^2 < 1

                          P < x^2 / 50

- For an unitary demand, ! E ! is =  1:

                          ! -50*P / x^2 !  = 1

                          50*P / x^2 = 1

                          P = x^2 / 50

- For an inelastic demand, ! E ! is > 1:

                          ! -50*P / x^2 ! > 1

                          50*P / x^2 > 1

                          P > x^2 / 50

2)

If the unit price is increased slightly from $50, will the revenue increase or decrease?

- We see from the calculated demand sensitivity d P / dt:

                          d P ( x ) / dx = - 50 / x

- We see that as P increases the from P = $50, the quantity of goods demanded would be:

                          50 = 50 ln(50/x)

                           1 = Ln ( 50 / x )

                           50/x = e

                           x = 50 / e

Then,

                          d P ( x ) / dx = - 50 / ( 50 / e )

                          d P ( x ) / dx = - e

- We see that if price slightly increases from $ 50 then the quantity demanded would decrease by e (hundreds ) goods.

- The decrease in the quantity demanded is higher than the increase in price. The revenue is given by the product of price P ( x ) and x:

                Revenue R ( x ) = P ( x ) * x

                                = 50*x*ln(50/x)

Then the product of price and quantity goods also decreases; hence, revenue decreases.

8 0
3 years ago
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