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Firlakuza [10]
1 year ago
10

First person to help gets brainliest

Mathematics
1 answer:
lions [1.4K]1 year ago
5 0

The solution of the given equation w.r.t m is m = 3 ,i.e., option A

<h3>What are Algebraic equations?</h3>

With one exception, algebraic equations are essentially algebraic expressions.

An = sign is required in all algebraic equations.

No matter what kind of equation it is, all equations use the = sign.

The next topic is algebraic expression, which lacks the operators =,,, >, and.

To put it simply, there is no comparison between two terms in an algebraic statement.

Algebraic expressions frequently use terms like polynomial and square root.

So you now recognize the distinction between the two?

1) An algebraic equation has the symbol =, and 2) an algebraic expression lacks any comparison symbols (such as > and =).

As per the question:

6\frac{1}{9} + 3\frac{1}{3} =28\frac{1}{3}

55/9 + 10/3 = 85/3

(55m+30m)/9 = 85/3

85m/9 = 85/3

85m = (85*9)/3

85m = 85*3

m = (85*3)/85

∴ m = 3

To now more about algebraic equations, visit

brainly.com/question/953809

#SPJ13

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The quadratic function g(x) = a.ca + bx+c has the
Mumz [18]

<em>The value of b is 14 and the value of c is 65</em>

<h2>Explanation:</h2>

The quadratic function is a function of the form:

f(x)=ax^2+bx+c

Here we know that the leading coefficient a=1 so we reduce our equation to:

g(x)=x^2+bx+c

The roots are those values at which g(x)=0

So:

x^2+bx+c=0 \\ \\ First \ root: \\ \\ (-7+4i)^2+b(-7+4i)+c=0 \\ \\ (-7)^2-2(7)(4i)+(4i)^2-7b+4bi+c=0 \\ \\  49-56i+16i^2-7b+4bi+c=0 \\ \\ \\ Simplifying: \\ \\ 49-56i+16(-1)-7b+4bi+c=0 \\ \\ 49-56i-16-7b+4bi+c=0 \\ \\ 33-56i-7b+4bi+c=0 \\ \\ \\

Second \ root: \\ \\ (-7-4i)^2+b(-7-4i)+c=0 \\ \\ (-1)^2(7+4i)^2+b(-7-4i)+c=0 \\ \\ (7)^2+2(7)(4i)+(4i)^2-7b-4bi+c=0 \\ \\  49+56i+16i^2-7b-4bi+c=0 \\ \\ \\ Simplifying: \\ \\ 49+56i+16(-1)-7b-4bi+c=0 \\ \\ 49+56i-16-7b-4bi+c=0 \\ \\ 33+56i-7b-4bi+c=0

So we have:

(1) \ 33-56i-7b+4bi+c=0 \\ \\ (2) \ 33+56i-7b-4bi+c=0 \\ \\ \\ Subtract \ 2 \ from: \\ \\ 33-56i-7b+4bi+c-(33+56i-7b-4bi+c)=0 \\ \\ 33-56i-7b+4bi+c-33-56i+7b+4bi-c=0 \\ \\ \\ Combine \ like \ terms: \\ \\ 33-33-56i-56i-7b+7b+4bi+4bi+c-c=0 \\ \\ -112i+8bi=0 \\ \\ Isolating \ b: \\ \\ b=\frac{112i}{8i} \\ \\ \boxed{b=14}

Finding c from (1):

33-56i-7b+4bi+c=0 \\ \\ \\ Substituting \ b: \\ \\ 33-56i-7(14)+4(14)i+c=0 \\ \\ 33-56i-98+56i+c=0 \\ \\ -65+c=0 \\ \\ \boxed{c=65}

<h2>Learn more:</h2>

Complex conjugate: brainly.com/question/2137496

#LearnWithBrainly

5 0
3 years ago
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