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9966 [12]
3 years ago
9

Mr. Brown owned a house, which he rented for $60 a month. The house was assessed at $9000. In 1975 the rate of taxation was incr

eased from $25 to $28 per $1000 assessed valuation. By what amount should the monthly rent have been raised to absorb the increase in that year's taxes?
Mathematics
1 answer:
Alexandra [31]3 years ago
5 0
We first calculate the percentage increase on the tax

Old value = 25×9 = 225 ⇒ The value 9 represents the 9 lots of thousands of the house's value
New value = 28×9 = 252

Increase in tax = 252 - 225 =27  
Percentage increase = (27÷225) ×100 = 12%

The amount of yearly rent would be then increased by 12%

Monthly rent = $60
Yearly rent = 60×12 = $720
Increase by 12% = 720×1.12 = 806.4 ⇒ The value 1.12 is the multiplier, obtained from 100%+12%=112%=1.12

The monthly rent is 806.4÷12 = $67.20 which is an increase of $7.20 per month
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The value of x when y = 2 is x = 36

Step-by-step explanation:

The equation that we have in this problem is:

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In order to find the value of x when y is equal to

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We need to re-arrange the equation making x the subject.

We proceed as follows:

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2) Now we multiply both  sides by 4:

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The height of a stuntperson jumping off a building that is 20 m high is modeled by the equation h = 20 -57, where t is the time
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A stuntman jumping off a 20-m-high building is modeled by the equation h = 20 – 5t2, where t is the time in seconds. A high-speed camera is ready to film him between 15 m and 10 m above the ground. For which interval of time should the camera film him?

Answer:

1\leq t\geq \sqrt{2}

Step-by-step explanation:

Given:

A stuntman jumping off a 20-m-high building is modeled by the equation

h =20-5t^{2}-----------(1)

A high-speed camera is ready to making film between 15 m and 10 m above the ground

when the stuntman is 15m above the ground.

height h = 15m  

Put height value in equation 1

15 =20-5t^{2}

5t^{2} =20-15

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t^{2} =1

t =\pm1

We know that the time is always positive, therefore t=1

when the stuntman is 10m above the ground.

height h = 10m  

Put height value in equation 1

10 =20-5t^{2}

5t^{2} =20-10

5t^{2} =10

t^{2} =\frac{10}{5}

t^{2} =2

t=\pm\sqrt{2}

t=\sqrt{2}

Therefore ,time interval of camera film him is 1\leq t\geq \sqrt{2}

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