Select all the correct answers.

Mathematics

1 answer:

evablogger [386]3 years ago

6 0

Answer:
The 3rd and 4th option

Explanation:
A function only works when the input (x) only has one output (y).
The first, second, and the last option has the input has two or more outputs.

Fun Fact:
It is okay for the outputs to have more then in input, but not okay for the input to have more then one output.

I hope this help and have a nice day :)

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In order to make the expression below equivalent to 1 2 x 6, which additional operation should be included in the expression? Us

marysya [2.9K]

The additional operation which needed to make the expression equivalent to given expression is $-\dfrac{3}{4} x$ . Thus the option 3 is the correct option.

<h3>What is equivalent expression?</h3>

Equivalent expression are the expression whose result is equal to the original expression, but the way of representation is different.

Given information-

The equation given in the problem is,

$\dfrac{5}{4} x+6$

Let the equation is equation number 1.

This expression is to convert in the expression, which is,

$\dfrac{1}{2}x+6$

Let the above equation is equation number 2.

Let the additional operation which needed to make the expression equivalent to given expression is $f(x)$ .

As the addition of $f(x)$ and equation 1 is equal to the equation 2. Therefore,

$f(x)+\dfrac{5}{4} x+6=\dfrac{1}{2}x+6$

Solve the equation as,

$f(x)=\dfrac{1}{2}x+6-\dfrac{5}{4} x-6\\f(x)=\dfrac{1}{2}x-\dfrac{5}{4} x\\f(x)=\dfrac{2-5}{4} x\\f(x)=\dfrac{-3}{4} x$

Hence, the additional operation which needed to make the expression equivalent to given expression is $-\dfrac{3}{4} x$ . Thus the option 3 is the correct option.

Learn more about the equivalent expression here;

brainly.com/question/2972832

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2 years ago

Evaluate UV if possible.

Mnenie [13.5K]

Answer: B

Step-by-step explanation:

Multiply U and V:

$\begin{pmatrix}-9&1&0\\ \:4&0&2\\ \:0&1&5\end{pmatrix}\begin{pmatrix}2\\ \:-2\\ \:1\end{pmatrix}$

$\mathrm{Multiply \ the \ rows \ of \ the \ first \ matrix \ by \ the \ columns \ of \ the \ second \ matrix:}$

$\begin{aligned}&\left(\begin{array}{ccc}-9 & 1 & 0\end{array}\right)\left(\begin{array}{c}2 \\-2 \\1\end{array}\right)=(-9) \cdot 2+1 \cdot(-2)+0 \cdot 1 \\&\left(\begin{array}{ccc}4 & 0 & 2\end{array}\right)\left(\begin{array}{c}2 \\-2 \\1\end{array}\right)=4 \cdot 2+0 \cdot(-2)+2 \cdot 1 \\&\left(\begin{array}{lll}0 & 1 & 5\end{array}\right)\left(\begin{array}{c}2 \\-2 \\1\end{array}\right)=0 \cdot 2+1 \cdot(-2)+5 \cdot 1\end{aligned}$

$=\begin{pmatrix}\left(-9\right)\cdot \:2+1\cdot \left(-2\right)+0\cdot \:1\\ 4\cdot \:2+0\cdot \left(-2\right)+2\cdot \:1\\ 0\cdot \:2+1\cdot \left(-2\right)+5\cdot \:1\end{pmatrix}$