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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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Answer:

a) Everyone on the team talks until the entire team agrees on one decision.

Step-by-step explanation:

Option B consists of voting and not everyone would like the outcome. Option C is making an outsider the decision maker, which can't be helpful since he / she won't have as strong opinions as the team itself. Option D is just plain wrong as it defeats the purpose of team work and deciding as one team. So, I believe option A makes the most sense

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Find the surface area of the prism.
Vanyuwa [196]

Answer:

400cm²

Step-by-step explanation:

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Answer with Step-by-step explanation:

We are given that

f(x)=\left\{\begin{matrix}\dfrac{x^2-2x}{x^2-4}&,if\ \ x\neq 2 \\ 1&,if\ \ x=2\end{matrix}\right.

We have to explain that why the function is discontinuous at x=2

We know that if function is continuous at x=a then LHL=RHL=f(a).

f(x)=\frac{x(x-2)}{(x+2)(x-2)}=\frac{x}{x+2}

LHL=Left hand limit when x <2

Substitute x=2-h

where h is small positive value >0

\lim_{h\rightarrow 0}f(x)=\lim_{h\rightarrow 0}\frac{2-h}{2-h+2}

\lim_{h\rightarrow 0}\frac{2-h}{4-h}=\frac{2}{4}=\frac{1}{2}

Right hand limit =RHL when x> 2

Substitute

x=2+h

\lim_{h\rightarrow 0}f(x)=\lim_{h\rightarrow 0}\frac{2+h}{2+h+2}=\lim_{h\rightarrow 0}\frac{2+h}{4+h}

=\frac{2}{4}=\frac{1}{2}

LHL=RHL=\frac{1}{2}

f(2)=1

LHL=RHL\neq f(2)

Hence, function is discontinuous at x=2

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