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4 years ago

Which statements are true for verifying the solution set of

Mathematics

2 answers:

Feliz [49]4 years ago

8 0

The 3rd one . shows how to substitute "x" . The most frequent Way

forsale [732]4 years ago

4 0

Answer:

its 1 and 4

Step-by-step explanation:

You might be interested in

Write the standard form of an equation with p=3 sqrt2, theta= 135°

Arlecino [84]

Answer:

x - y + 6 = 0

Step-by-step explanation:

In normal form of a straight line, the equation is given by

$x\cos \theta + y\sin \theta = p$

where p is the perpendicular distance of the line from the origin and $\theta$ is the angle between the perpendicular line and the positive direction of the x-axis.

Here, in our case $p = 3\sqrt{2}$ and $\theta = 135$ Degree,

Therefore, the normal form of the straight line equation is

$x \cos 135 + y \sin 135 = 3\sqrt{2}$

⇒ $x \cos (180 - 45) + y \sin (180 - 45) = 3\sqrt{2}$

⇒ $- x \cos 45 + y \sin 45 = 3\sqrt{2}$ {Since, Cos (180 - Ф) = - Cos Ф and Sin (180 - Ф) = Sin Ф}

⇒ $- \frac{x}{\sqrt{2}} + \frac{y}{\sqrt{2}} = 3\sqrt{2}$

⇒ - x + y = 3√2 × √2 = 6

⇒ x - y + 6 = 0

So, the standard form of the equation is x - y + 6 = 0. (Answer)

7 0

4 years ago

How can i look at an equation and determine if it is the same line, intersecting or parallel?

BaLLatris [955]

Answer: It is simple, if you look at the slope of the 2, if they have the same slope then it means it definitely is not intersecting, and if you look at which points that the equation begins at you can tell if it starts at the same points with the same slope then it is same line, if it starts at different points with same slope then it is parallel lines

7 0

3 years ago

53 is the difference of Gail’s age and 18

lys-0071 [83]

53-x+18

Gail’s age would be x (or any variable of our choice).

5 0

3 years ago

If you simplify 2(4x-13) what would be the answer?

lina2011 [118]

Answer:

8x-26

Step-by-step explanation:

Apply the distributive law: 2·4x-2·13
Multiply 2·x and 2·13: 8x-26

4 0

4 years ago

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