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Brrunno [24]
3 years ago
13

Factor over the complex numbers: 36x2 + 9.

Mathematics
1 answer:
ludmilkaskok [199]3 years ago
4 0

36x^2+9=(6x)^2-(-9)=(6x)^2-(-3^2)\\\\=(6x)^2-(3i)^2=(6x-3i)(6x+3i)\\\\\text{Used:}\\a^2-b^2=(a-b)(a+b)\\\\i^2=-1

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Factor mn - 4m - 5n + 20
anastassius [24]
Mn - 4m - 5n + 20

In this, mn and -4m have a common factor; m. So we can take that out to make it:

m(n-4) -5n + 20

Now, -5n + 20 also has a common factor which is 5, because 20/5 gives you 4, and 5/5 equals 1. So we can talk out that -5 (it doesn't need to be negative, but negatives on variable, no gusta). So you'd get

m(n-4)-5(n-4)      OR m(n-4)+5(-n+4)
4 0
3 years ago
Read 2 more answers
In 2000, people charged 1,243 billion on the four most used types of credit cards. In 2005, people charged 1,838 billion on thes
baherus [9]
Alright, so the rate of change is the same thing as the slope, so we can find this by using the slope formula (which I took a picture of because it's super messy to type it out like this)

Now that we know the slope formula, we need to find the x values and the y values to plug into the equation. We'll say that the years are our x values and the money spent (in billions) are our y values. Now that we've set up an axis, we can create points! Our points would be (2000, 1243) and (2005, 1838)

Now we can plug our points into our equations! On the top of our equation, we'll put (1838-1243), and on the bottom we'll put (2005-2000). When we simplify that, we end up with 595/5, which simplifies to 119. Now we know that the rate of change for the credit cards was 119 billion!

4 0
2 years ago
Let A(t) represent the amount of money in a bank account at time t years. The rate at which the account is increasing is proport
nikdorinn [45]

Answer:

(C)A(t)=500e^{0.05t}

Step-by-step explanation:

If A(t) represent the amount of money in a bank account at time t years.

The rate at which the account is increasing is proportional to the amount of money in the account, which is modeled by the differential equation:

\dfrac{dA}{dt}=kA

First, we solve this differential equation

\dfrac{dA}{A}=kdt\\$Taking Integrals\\ln A =kt+C, C a constant of integration\\Taking exponents of both sides\\A(t)=Ce^{kt}\\When t=0, A(t)=\$500\\500=C\\Therefore:\\A(t)=500e^{kt}

At the moment when the amount of money in the account is $1000, the amount is increasing at a rate of $50 per year.

When A(t)=$1000, \dfrac{dA}{dt}=50

Therefore:

50=1000k\\k=50/1000=0.05

Substituting into our result from A(t) above:

A(t)=500e^{0.05t}

3 0
3 years ago
More ASAP questions !!!!
Brrunno [24]
4. Natural
5. Irrational
4 0
3 years ago
Multiply and express your answers in mixed numbers 1/3 x 5
schepotkina [342]

\frac{1}{3}  \times  \frac{5}{1}  =  \frac{5}{3}

^Here's the answer. Any number will have a denominator of 1. Then, multiply as usual and you'll get the answer.

7 0
2 years ago
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