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3 years ago

Which of the following statements is false?

Mathematics

1 answer:

lisov135 [29]3 years ago

3 0

(1): "c0" was replaced by "c^0".

(2): Dot was discarded near "1.j". 1 more similar replacement(s)

Unauthorized use of the imaginary unit "i" or syntax error in complex arithmetic expression

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If T(x, y) = (x + 5, y + 6) and Pris the image of P, what is the rule for the translation in which P is the image of P'? T(x, y)

xxTIMURxx [149]

Answer:

The translation is;

$T(x,y)=(x-5,y-6)$

Explanation:

Given the translation rule;

$T(x,y)=(x+5,y+6)$

when P' is the image of P.

For the inverse, when P is the image of P', the Translation rule would become;

$\begin{gathered} x=x^{\prime}+5 \\ x^{\prime}=x-5 \\ y=y^{\prime}+6 \\ y^{\prime}=y-6 \\ So,\text{ the translation rule becomes;} \\ T(x,y)=(x-5,y-6) \end{gathered}$

The translation is;

$T(x,y)=(x-5,y-6)$

3 0

1 year ago

jamie has 8/10 of a candy bar leftover. He wants to spilt it into 1/3 pieces . How many 1/3 pieces can he make

Len [333]

I think it might be 2.4 or 2

8 0

4 years ago

Joseph earns $15 for every lawn he mows in the amount of money he earns proportional to the number of long as he mows make a tab

ExtremeBDS [4]

Answer:

15x = y

Step-by-step explanation:

4 0

3 years ago

Solving two-step equations 5x+8=23

andrey2020 [161]

You have to subtract 8 from the left and then 8 from the right. If you do that then you'll have 5x=15.
Then you divide "x" by 5 and 15 by 5.
X=3

7 0

3 years ago

Read 2 more answers

Which of the following graphs shows the solution set for the inequality below? 3|x + 1| < 9

Bas_tet [7]

Step-by-step explanation:

The absolute value function is a well known piecewise function (a function defined by multiple subfunctions) that is described mathematically as

$f(x) \ = \ |x| \ = \ \left\{\left\begin{array}{ccc}x, \ \text{if} \ x \ \geq \ 0 \\ \\ -x, \ \text{if} \ x \ < \ 0\end{array}\right\}$ .

This definition of the absolute function can be explained geometrically to be similar to the straight line $\textbf{\textit{y}} \ = \ \textbf{\textit{x}}$ , however, when the value of x is negative, the range of the function remains positive. In other words, the segment of the line $\textbf{\textit{y}} \ = \ \textbf{\textit{x}}$ where $\textbf{\textit{x}} \ < \ 0$ (shown as the orange dotted line), the segment of the line is reflected across the x-axis.

First, we simplify the expression.

$3\left|x \ + \ 1 \right| \ < \ 9 \\ \\ \\\-\hspace{0.2cm} \left|x \ + \ 1 \right| \ < \ 3$ .

We, now, can simply visualise the straight line, $y \ = \ x \ + \ 1$ , as a line having its y-intercept at the point $(0, \ 1)$ and its x-intercept at the point $(-1, \ 0)$ . Then, imagine that the segment of the line where $x \ < \ 0$ to be reflected along the x-axis, and you get the graph of the absolute function $y \ = \ \left|x \ + \ 1 \right|$ .

Consider the inequality

$\left|x \ + \ 1 \right| \ < \ 3$ ,

this statement can actually be conceptualise as the question

$``\text{For what \textbf{values of \textit{x}} will the absolute function \textbf{be less than 3}}"$ .

Algebraically, we can solve this inequality by breaking the function into two different subfunctions (according to the definition above).

Case 1 (when $x \ \geq \ 0$ )

$x \ + \ 1 \ < \ 3 \\ \\ \\ \-\hspace{0.9cm} x \ < \ 3 \ - \ 1 \\ \\ \\ \-\hspace{0.9cm} x \ < \ 2$

Case 2 (when $x \ < \ 0$ )

$-(x \ + \ 1) \ < \ 3 \\ \\ \\ \-\hspace{0.15cm} -x \ - \ 1 \ < \ 3 \\ \\ \\ \-\hspace{1cm} -x \ < \ 3 \ + \ 1 \\ \\ \\ \-\hspace{1cm} -x \ < \ 4 \\ \\ \\ \-\hspace{1.5cm} x \ > \ -4$

*remember to flip the inequality sign when multiplying or dividing by

negative numbers on both sides of the statement.

Therefore, the values of x that satisfy this inequality lie within the interval

$-4 \ < \ x \ < \ 2$ .

Similarly, on the real number line, the interval is shown below.

The use of open circles (as in the graph) indicates that the interval highlighted on the number line does not include its boundary value (-4 and 2) since the inequality is expressed as "less than", but not "less than or equal to". Contrastingly, close circles (circles that are coloured) show the inclusivity of the boundary values of the inequality.

3 0

3 years ago