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Ann [662]
3 years ago
14

SOLVE 3x²-7x-12=0 correct answer to 2 decimal places

Mathematics
1 answer:
Bas_tet [7]3 years ago
5 0

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hodyreva [135]
K and m are corresponding angles
7 0
2 years ago
Kristy ask Lonnie to think add 3 to it multiply the sum by 2 and the subtract 6 let n stand for lonnie starting number which exp
Kryger [21]

Answer:

2(n+3)-6

n=0

Step-by-step explanation:

2(3+n)-6=

Simplifying

2(3 + n) + -6 = 0

(3 * 2 + n * 2) + -6 = 0

(6 + 2n) + -6 = 0

Reorder the terms:

6 + -6 + 2n = 0

Combine like terms: 6 + -6 = 0

0 + 2n = 0

2n = 0

Solving

2n = 0

Solving for variable 'n'.

Move all terms containing n to the left, all other terms to the right.

Divide each side by '2'.

n = 0

Simplifying

n = 0

5 0
3 years ago
What should be added to both sides of this equation to solve for the variable?<br> 45 = 30 + m
timurjin [86]
Subtract 30 from both sides

4 0
4 years ago
Read 2 more answers
CUMPLEIE
beks73 [17]

the magnitude of an earthquake that is 5,011 times more intense than a standard earthquake is 3.7

<u>Step-by-step explanation:</u>

Here we have , to find  the magnitude of an earthquake that is 5,011 times more intense than a standard earthquake . Let's find out:

We know that , Value of earthquake is given by formula

⇒ M = Log(\frac{I_1}{I_0} ) , Where M is magnitude of earthquake

I_1  is the measured magnitude of intensity of  earthquake .

I_0  is the standard magnitude of intensity of  earthquake .

According to question , I_1 = 5011I_0 , Hence

⇒ M = Log(\frac{I_1}{I_0} )

⇒ M = Log(\frac{5011(I_0)}{I_0} )

⇒ M = Log(5011 )

⇒ M =3.69924    { On rounding off we get }

⇒ M =3.7

Therefore , the magnitude of an earthquake that is 5,011 times more intense than a standard earthquake is 3.7

6 0
3 years ago
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brainly is flooded with spam comments like almost every question i click on i see a spam comment talking about clicking a link.
nika2105 [10]

Answer:

Ikr, it annoys me. I can't even get a good answer to my questions either! ITS SO ANNOYING.

7 0
3 years ago
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