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4 years ago

Math with Exponents

Mathematics

1 answer:

sergey [27]4 years ago

7 0

Example

2^3 4^2
2×2×2=6 4×4=16

All you do is multiply the number by the exponent however many times the exponents is

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Where is the answer to the expression 4 − 8 located on a horizontal number line? 4 units to the left of 4

Vilka [71]

4 - 8 = -4

A negative number is alwats at the left in a horizontal line.

Let's try to rapresent it.

-4 -3 -2 -1 0 +1 +2 +3 +4

So, you count from +4 to -4

Your answer is "8 units to the left of 4"

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Answer:

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3 years ago

An advertising company designs a campaign to introduce a new product to a metropolitan area of population 3 Million people. Let

Advocard [28]

Answer:

$P(t)=3,000,000-3,000,000e^{0.0138t}$

Step-by-step explanation:

Since P(t) increases at a rate proportional to the number of people still unaware of the product, we have

$P'(t)=K(3,000,000-P(t))$

Since no one was aware of the product at the beginning of the campaign and 50% of the people were aware of the product after 50 days of advertising

We have and ordinary differential equation of first order that we can write

$P'(t)+KP(t)= 3,000,000K$

The <em>integrating factor </em>is

$e^{Kt}$

Multiplying both sides of the equation by the integrating factor

$e^{Kt}P'(t)+e^{Kt}KP(t)= e^{Kt}3,000,000*K$

Hence

$(e^{Kt}P(t))'=3,000,000Ke^{Kt}$

Integrating both sides

$e^{Kt}P(t)=3,000,000K \int e^{Kt}dt +C$

$e^{Kt}P(t)=3,000,000K(\frac{e^{Kt}}{K})+C$

$P(t)=3,000,000+Ce^{-Kt}$

But P(0) = 0, so C = -3,000,000

and P(50) = 1,500,000

$e^{-50K}=\frac{1}{2}\Rightarrow K=-\frac{log(0.5)}{50}=0.0138$

And the equation that models the number of people (in millions) who become aware of the product by time t is

$P(t)=3,000,000-3,000,000e^{0.0138t}$

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4 years ago