Pls help pls help pls help pls help pls help pls help

Please help I will mark Brainly

Alekssandra [29.7K]

Here u go: D. integer, B.Rational and C. real

3 0

3 years ago

ASAP HELP ANSWER ALL THESE AND I WILL GIVE THE BRAINLEST

alexandr1967 [171]

The second question is number 3 for the polygon translation.
Question 3 is 2 units right and four units down.

3 0

3 years ago

Read 2 more answers

Please help question in the picture

tigry1 [53]

There really isn’t a possible answer since there is not enough information given. Not from what I have learned so far at least, but using common sense, and the process of elimination I would say 30

8 0

3 years ago

The length of the rectangle is twice its width. Write and solve a system of linear equations to find the length L and width W of

vazorg [7]

Answer:

The length is equal to 12 and the width is equal to 6.

Step-by-step explanation:

In order to find the values here, we start by setting the width equal to x. Now knowing this, we know that the length is twice that long. Therefore, the length would be equal to 2x. Now we can use the perimeter formula to solve the equation.

2L + 2W = P

2(2x) + 2(x) = P

4x + 2x = 36

6x = 36

x = 6

Now with the given value for x, we can tell that the width is 6 and then we multiply it by 2 to get the length value (12).

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

2 years ago