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2 years ago

We used these functions to represent the two sides of the equation:

Mathematics

1 answer:

katrin [286]2 years ago

4 0

The solution to the system of equation graphed is (-1, -3)

<h3>Solution to system of equation</h3>

Given the system of equations graphed on the plane expressed as:

f(x) = 3x

g(x) = 4x + 1

Equate

f(x) = g(x)

3x = 4x + 1

3x - 4x = 1

-x = 1

x = -1

Determine the value of y

y = 3x

y = 3(-1)

y = -3

Hence the solution to the system of equation graphed is (-1, -3)

Learn more on system of equation here: brainly.com/question/847634

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Determine the constant that must be added to the given expression to complete the square and express the resulting trinomial as

erma4kov [3.2K]

Constant:9
Expression: (x-3)^2

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3 years ago

The product of 2√ 3 and 3√12 in simplified form is .

borishaifa [10]

Answer:

Step-by-step explanation:

Rewrite this as

(2)(3)(√3)(√12), or

6 * √36, or

6 * 6 = 36

5 0

2 years ago

Nvm I got it now lol

Usimov [2.4K]

Answer:

nice bro good for you :)

Step-by-step explanation:

3 0

2 years ago

Find the domain for each expression. (1/x)+5<br> 1/x<br> X+5/x+5<br> X^2/x<br> 7/(x-2)

wel

Using it's concept, it is found that the domain for the expressions is, respectively, given by:

$x \neq 0, x \neq 0, (-\infty, \infty), (-\infty, \infty), x \neq 2$

<h3>What is the domain of a function?</h3>

It is the <u>set that contains all possible input values</u>.

In a fraction, the denominator cannot be zero, hence:

The domain of the first two expressions is of $x \neq 0$ .

The domain of the last expression is of $x \neq 2$ .

The third expression can be simplified, as:

(x + 5)/(x + 5) = 1.

The same is true for the fourth, as:

x²/x = 1.

Neither has any restriction, hence their domain is all real numbers, represented by $(-\infty, \infty)$ .

More can be learned about the domain of a function at brainly.com/question/25897115

6 0

2 years ago

Expres each of the folowing as a single matrix.

Elden [556K]

Answer:

$\left[\begin{array}{cc}6&-3\\10&-1\end{array}\right]+\left[\begin{array}{cc}-2&8\\3&-12\end{array}\right]=\left[\begin{array}{cc}4&5\\13&-13\end{array}\right]$

Step-by-step explanation:

If you have two matrices:

$A=\left[\begin{array}{cc}a&b\\c&d\end{array}\right]\\and\\B=\left[\begin{array}{cc}e&f\\g&h\end{array}\right]\\\\\\A+B=\left[\begin{array}{cc}a+e&b+f\\c+g&d+h\end{array}\right]$

We have:

$A=\left[\begin{array}{cc}6&-3\\10&-1\end{array}\right]\\and\\B=\left[\begin{array}{cc}-2&8\\3&-12\end{array}\right]$

And we need to express as a single matrix:

$A+B=\left[\begin{array}{cc}6&-3\\10&-1\end{array}\right]+\left[\begin{array}{cc}-2&8\\3&-12\end{array}\right]\\\\\\A+B=\left[\begin{array}{cc}6+(-2)&-3+8\\10+3&-1+(-12)\end{array}\right]\\\\\\A+B=\left[\begin{array}{cc}6-2&5\\13&-1-12\end{array}\right]\\\\\\A+B=\left[\begin{array}{cc}4&5\\13&-13\end{array}\right]$

The answer is:

$\left[\begin{array}{cc}6&-3\\10&-1\end{array}\right]+\left[\begin{array}{cc}-2&8\\3&-12\end{array}\right]=\left[\begin{array}{cc}4&5\\13&-13\end{array}\right]$

It is expressed as a single matrix.

8 0

3 years ago