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3 years ago

A tree double in weight in three months. How long will it take to be 1600% in weight? (Linear or exponential growth?)

Mathematics

1 answer:

Ray Of Light [21]3 years ago

6 0

Answer: 12 months

Step-by-step explanation:

Given : A tree double in weight in three months.

Since the weight is increasing by growth factor of 2 , therefore its is an exponential growth.

The exponential growth equation is given by :-

$y=Ab^x$ (1)

, where A is the initial values , b is the growth factor and x is the time period.

As per given , b= 2

Since the tree doubles in weight in three months, so time period x = $\dfrac{t}{3}$ , where t= number of months.

Substitute the value of b and x in (1) , we get

$y=A(2)^{\dfrac{t}{3}}$ , where y= weight of tree after t months and A is initial weight of tree.

When it will be 1600% of his initial weight , the weight of tree : y= 1600% of A = $\dfrac{1600}{100}\times A=16A$

At y= 16 A , $16A=A(2)^{\dfrac{t}{3}}$

$\Rightarrow\ 16=2^{\dfrac{t}{3}}$

$\Rightarrow\ 2^{4}=2^{\dfrac{t}{3}}$

$\Rightarrow\ 4=\dfrac{t}{3}\Rightarrow\ t=3\times4=12$

Hence, it will take 12 months to be 1600% in weight.

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Answer:

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Step-by-step explanation:

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3 years ago

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Answer:

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Step-by-step explanation:

To solve this you need to compare the lengths given to you in the question statement.

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Both triangles are similar triangles, so we can see triangle ADE as a larger version of angle AGF.

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3 years ago

For a scale of 1cm: 12km what is the actual length represented by 1.7 cm

madreJ [45]

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3 years ago

Which is the simplified form of the expression ((2 Superscript negative 2 Baseline) (3 Superscript 4 Baseline)) Superscript nega

AlekseyPX

Answer:

The option "StartFraction 1 Over 3 Superscript 8" is correct

That is $\frac{1}{3^8}$ is correct answer

Therefore $[(2^{-2})(3^4)]^{-3}\times [(2^{-3})(3^2)]^2=\frac{1}{3^8}$

Step-by-step explanation:

Given expression is ((2 Superscript negative 2 Baseline) (3 Superscript 4 Baseline)) Superscript negative 3 Baseline times ((2 Superscript negative 3 Baseline) (3 squared)) squared

The given expression can be written as

$[(2^{-2})(3^4)]^{-3}\times [(2^{-3})(3^2)]^2$

To find the simplified form of the given expression :

$[(2^{-2})(3^4)]^{-3}\times [(2^{-3})(3^2)]^2$

$=(2^{-2})^{-3}(3^4)^{-3}\times (2^{-3})^2(3^2)^2$ ( using the property $(ab)^m=a^m.b^m$ )

$=(2^6)(3^{-12})\times (2^{-6})(3^4)$ ( using the property $(a^m)^n=a^{mn}$

$=(2^6)(2^{-6})(3^{-12})(3^4)$ ( combining the like powers )

$=2^{6-6}3^{-12+4}$ ( using the property $a^m.a^n=a^{m+n}$ )

$=2^03^{-8}$

$=\frac{1}{3^8}$ ( using the property $a^{-m}=\frac{1}{a^m}$ )

Therefore $[(2^{-2})(3^4)]^{-3}\times [(2^{-3})(3^2)]^2=\frac{1}{3^8}$

Therefore option "StartFraction 1 Over 3 Superscript 8" is correct

That is $\frac{1}{3^8}$ is correct answer

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3 years ago

Read 2 more answers

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