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Oliga [24]
3 years ago
14

Write an algebraic expression that illustrates the multiplicative identity.

Mathematics
1 answer:
Alla [95]3 years ago
4 0
The multiplicative identity shows us that any time you multiply something by the number 1, the result would be what you started with.

Ex: 1(m) or 1 x m.

The reason for thusbis that when you are multiplying you are representing groups of something, so 1 group of anything or m, would just be that anything or m in this case!
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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 <em>x</em>-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 <em>x</em>-intercept at the point (-1, \ 0). Then, imagine that the segment of the line where x \ < \ 0 to be reflected along the <em>x</em>-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 <em>x</em> 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
Prove-: sin5A = 5cos^4 A sinA - 10cos^2 A sin^3 A + sin^5 A
jeyben [28]
A=x\\\\sin5x=5cos^4xsinx-10cos^2xsin^3x+sin^5x\\\\L=sin(3x+2x)=sin3xcos2x+sin2xcos3x=(*)\\-------------------------------\\sin3x=sin(2x+x)=sin2xcosx+sinxcos2x\\=2sinxcosxcosx+sinx(cos^2x-sin^2x)\\=2sinxcos^2x+sinxcos^2x-sin^3x=3sinxcos^2x-sin^3x\\-------------------------------

sin3xcos2x=(3sinxcos^2x-sin^3x)(cos^2x-sin^2x)\\=3sinxcos^4x-3sin^3xcos^2x-sin^3xcos^2x+sin^5x\\=3sinxcos^4x-4sin^3xcos^2x+sin^5x\\-------------------------------

cos3x=cos(2x+x)=cos2xcosx-sin2xsinx\\=cosx(cos^2x-sin^2x)-2sinxcosxsinx\\=cos^3x-sin^2xcosx-2sin^2xcosx=cos^3x-3sin^2xcosx\\-------------------------------

sin2xcos3x=2sinxcosx(cos^3x-3sin^2xcosx)\\=2sinxcos^4x-6sin^3xcos^2x\\-------------------------------

&#10;(*)=3sinxcos^4x-4sin^3xcos^2x+sin^5x+2sinxcos^4x-6sin^3xcos^2x\\\\=5sinxcos^4x-10sin^3xcos^2x+sin^5x=R

=======================================\\\\sin(\alpha+\beta)=sin\alpha cos\beta+sin\beta cos\alpha\\\\cos(\alpha+\beta)=cos\alpha cos\beta-sin\alpha sin\beta\\\\sin2\alpha=2sin\alpha cos\alpha\\\\cos\alpha=cos^2\alpha-sin^2\alpha
5 0
3 years ago
Which conditional statement has a false converse?
exis [7]

Consider the converses:

a) If two planes have no points in common, then they are parallel. (true)

b) If a point lies on the y-axis, then it has x-coordinate 0. (true)

c) If two angles have the same measure, then they are congruent. (true)

d) If a figure has four sides, then it is a square. (FALSE) (A figure with 4 sides may not even be a plane figure.)

8 0
4 years ago
Plz plz help plz plz I’ll give brainlist
swat32

Answer: 10

Step-by-step explanation:

hope this helps!

3 0
3 years ago
Read 2 more answers
Amy ran 3/4 mile. which decimal shows how many miles she ran?
Olin [163]
Out of 100 it is 0.75
6 0
3 years ago
Read 2 more answers
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