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4 years ago

What are the solutions of 3x2 + 6x + 6 = 0 –2 ± i –2 ± 2i –1 ± i 1 ± i

1 answer:

4 0

C) Quick way:

3x^2+6x+6=0 ---> x^2+2x+2=0,

sum of solutions is the 'x' coefficient multiplied by -1: (-1+i)+(-1-i)=-2-->2
Product is the independent term: (-1+i)(=1+i) = (-1)^2-i^2 = 1+1=2

Takes longer to write it down!

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2 What is the minimum amount of information you need in order to calculate the slope of a line?

geniusboy [140]

Answer:

y=mx+b

Step-by-step explanation:

The formula to find the slope is y=mx+b

hope this helps

7 0

4 years ago

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3. A company with loud machinery needs to cut its sound intensity to 44% of its original level. By how many decibels would the l

sergeinik [125]

Answer: 3. The correct option is: A) 3.57 decibels.

6. The correct option is: B) $ln(\frac{x^2}{c^5})$ 

Step-by-step explanation:

3. Sound intensity is cut to 44% of its original level. That means, if the original sound intensity was 100, then now sound intensity will be 44.

So, $I_{0}= 100$ and $I= 44$

Using the formula $L= 10log(\frac{I}{I_{0}})$ , we will get.....

$L= 10log(\frac{44}{100})\\ \\ L= 10log(0.44)\\ \\ L= 10(-0.3565...)=-3.565...\approx -3.57$

(Rounded to the nearest hundredth)

So, the loudness would be reduced by 3.57 decibels.

6. Given expression is: $2ln(x) - 5ln(c)$

First applying the property $m*log(x)= log(x^m)$ , we will get.....

$ln(x^2)-ln(c^5)$

Now using the formula, $log(a)-log(b)= log(\frac{a}{b})$ , we will get....

$ln(x^2)-ln(c^5) = ln(\frac{x^2}{c^5})$

Thus, the answer as a single natural logarithm is $ln(\frac{x^2}{c^5})$

8 0

3 years ago

Please help me! I need help on this question!

schepotkina [342]

Answer:

C and A

Step-by-step explanation:

expand the things in parentheses so for C

1/2(x + 16) = 1/2(x) + 1/2(16) = 1/2 x + 8

same process for A

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4 years ago

ILL BRAINLIEST YOU PLEASE HELP ME

Nookie1986 [14]

Answer:

Step-by-step explanation:

7pi2

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3 years ago

Is (3,1) a solution of the equation y=-x+3

kogti [31]

no i don't think so the answer will be (3,0)

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3 years ago

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