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3 years ago

13

by selling a cell phone for 2400rs a shopkeeper makes a profit of 25% what would be his profit percentage if it was 2040rs

1 answer:

xxMikexx [17]3 years ago

7 0

Answer:

6.25%

Step-by-step explanation:

We first begin by solving for the original price of the cell phone.

if 2400 = 125%

x = 100%

By cross multplying, we get

125x = 2400 * 100

x = 240000/125

x = 1920rs

To find the profit percentage on the new sale price, we repeat another set of equations

100% = 1920

x = 2040

1920x = 2040 * 100

x = 204000/1920

x = 106.25%

His new profit percentage is 6.25%

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Use the equation and type the ordered-pairs. y = log 3 x {(1/3, a0), (1, a1), (3, a2), (9, a3), (27, a4), (81, a5)

vagabundo [1.1K]

Answer:

Considering the given equation $y = log_{3}x\\$

And the ordered pairs in the format $(x, y)$

I don't know if it is log of base 3 or 10, but I will assume it is 3.

For $(\frac{1}{3}, a_{0} )$

$x=\frac{1}{3}$

$y=a_{0}$

$y = log_{3}x\\y = log_{3}(\frac{1}{3} )\\y=-\log _3\left(3\right)\\y=-1$

So the ordered pair will be $(\frac{1}{3}, -1 )$

For $(1, a_{1} )$

$x=1$

$y=a_{1}$

$y = log_{3}x\\y = log_{3}1\\y = log_{3}(1)\\Note: \log _a(1)=0\\y = 0$

So the ordered pair will be $(1, 0 )$

For $(3, a_{2} )$

$x=3$

$y=a_{2}$

$y = log_{3}x\\y = log_{3}3\\y = 1$

So the ordered pair will be $(3, 1 )$

For $(9, a_{3} )$

$x=9$

$y=a_{3}$

$y = log_{3}x\\y = log_{3}9\\y=2\log _3\left(3\right)\\y=2$

So the ordered pair will be $(9, 2 )$

For $(27, a_{4} )$

$x=27$

$y=a_{4}$

$y = log_{3}x\\y = log_{3}27\\y=3\log _3\left(3\right)\\y=3$

So the ordered pair will be $(27, 3 )$

For $(81, a_{5} )$

$x=81$

$y=a_{5}$

$y = log_{3}x\\y = log_{3}81\\y=4\log _3\left(3\right)\\y=4$

So the ordered pair will be $(81, 4 )$

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3 years ago

a number has a 4 in the thousands place the number also has a digit whose value is 10 times the value of the 4 on the thousands

maks197457 [2]

Well I can still answer that question without the following and it will be 44,000

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3 years ago

A ball is thrown upward from the top of a 40 foot high building at a speed of 80 feet per second. The ball's height above

tekilochka [14]

it takes 5 seconds to reach a height of 40 ft. and It takes 2.5 seconds to reach maximum height.

Let h(t) represent the height of the ball at time t.

Given that the height is given by:

h(t) = -16t² + 80t + 40

1) For the ball to reach a height 0f 40 ft, h(t) = 40, hence:

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16t² - 80t = 0

t(t - 5) = 0

t = 0 or t = 5

Hence it takes 5 seconds to reach a height of 40 ft.

2) The maximum height is at h'(t) = 0,

h'(t) = -32t + 80

-32t + 80 = 0

32t = 80

t = 2.5

It takes 2.5 seconds to reach maximum height.

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Answer:

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Step-by-step explanation:

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