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creativ13 [48]
3 years ago
12

Mario says that the expression 4+3n² has four terms: 4, 3, n, and 2. Is he correct? explain.

Mathematics
1 answer:
stich3 [128]3 years ago
8 0

Mario says that the expression 4+3n^2 has four terms: 4, 3, n, and 2. Mario is incorrect

<em><u>Solution:</u></em>

Given that the expression is:

4+3n^2

Given that, Mario says that the above expression has four terms

But Mraio is incorrect

Because the given expression has two terms only

4 is one of the term

3n^2 is another term

So there are totally 2 terms only

A term can be a signed number, a variable, or a constant multiplied by a variable or variables

Here 3 is a constant multiplied by n^2

So, 3n^2 is one term

Each term in an algebraic expression is separated by a + sign or - sign

Thus there are two terms in mario expression

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What is the greatest common factor of 72 and 48?
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2 years ago
Solve the matrix equation for a, b, c, and d. [1 2] [a b] [6 5][3 4] [c d]= [19 8]
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Answer:

The answer is "\bold{\left[\begin{array}{cc}a&b\\c&d\end{array}\right] = \left[\begin{array}{cc}7&-2\\ -\frac{1}{2}&\frac{7}{2}\end{array}\right]}".

Step-by-step explanation:

\bold{\left[\begin{array}{cc}1&2\\3&4\end{array}\right] \left[\begin{array}{cc}a&b\\c&d\end{array}\right] = \left[\begin{array}{cc}6&5\\ 19&8\end{array}\right]}

Solve the L.H.S part:

\left[\begin{array}{cc}1&2\\3&4\end{array}\right] \left[\begin{array}{cc}a&b\\c&d\end{array}\right]\\\\\\\left[\begin{array}{cc}a+2c&b+2d\\3a+4c&3b+4d\end{array}\right]

After calculating the L.H.S part compare the value with R.H.S:

\left[\begin{array}{cc}a+2c&b+2d\\3a+4c&3b+4d\end{array}\right]= \left[\begin{array}{cc}6&5\\ 19&8\end{array}\right]} \\\\

\to a+2c =6....(i)\\\\\to b+2d =5....(ii)\\\\\to 3a+4c =19....(iii)\\\\\to 3b+4d = 8 ....(iv)\\\\

In equation (i) multiply by 3 and subtract by equation (iii):

\to 3a+6c=18\\\to 3a+4c=19\\\\\text{subtract}... \\\\\to 2c = -1\\\\\to  c= - \frac{1}{2}

put the value of c in equation (i):

\to a+ 2 (- \frac{1}{2})=6\\\\\to a- 2 \times \frac{1}{2}=6\\\\\to a- 1=6\\\\\to a =6 +1\\\\\to a = 7\\

In equation (ii) multiply by 3 then subtract by equation (iv):

\to 3b+6d=15\\\to 3b+4d=8\\\\\text{subtract...}\\\\\to 2d = 7\\\\\to d= \frac{7}{2}\\

put the value of d in equation (iv):

\to 3b+4 (\frac{7}{2})=8\\\\\to 3b+4 \times \frac{7}{2}=8\\\\\to 3b+14=8\\\\\to 3b =8-14\\\\\to 3b = -6\\\\\to b= \frac{-6}{3}\\\\\to b= -2

The final answer is "\bold{\left[\begin{array}{cc}a&b\\c&d\end{array}\right] = \left[\begin{array}{cc}7&-2\\ -\frac{1}{2}&\frac{7}{2}\end{array}\right]}".

4 0
3 years ago
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