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4 years ago

Which answer is equal to the quotient in the expression below?

1 answer:

7 0

FIRST section: x^2-3x+2 = x^2-x-2x+2=(x^2-x) -2x+2 = (x^2-x) -2(x-a million) = x(x-a million) -2(x-a million) = (x-2)(x-a million) 2nd section: x^-4 = x^2- 2^2 = (x-2)(x+2) So now your equation looks like this: FIRST section / 2nd section or (x-2)(x-a million) / (x-2)(x+2) and this comes out at (x-a million) / (x+2), so the respond is B.

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This is 15 points plz help ( no links ) EXPLAIN YOUR ANSWER plz

goldfiish [28.3K]

A.
EXPLANATION:
The mode is the peak of the data, in this data set, 5/8
The median is the center of data, in the data set, 1/2
5/8>1/2
Mode>Median

6 0

3 years ago

You travel from point A to point B in a car moving at a constant speed of 70 km/h. Then you travel the same distance from point

tankabanditka [31]

Answer:

Yes, the average speed for the entire trip from A to C is equal to $80\frac{km}{h}$

Step-by-step explanation:

The average speed of an object is defined as the distance traveled divided by the time elapsed. Velocity is a vector quantity, and average velocity can be defined as the displacement divided by the time. For the special case of straight line motion in the x direction, the average velocity takes the form:

$V_a_v_e_r_a_g_e=\frac{x_2-x_1}{t_2-t_1}=\frac{Δx}{Δt}$

If the beginning and ending velocities for the motion are known, and the acceleration is constant, the average velocity can also be expressed as:

$V_a_v_e_r_a_g_e=\frac{V_1+V_2}{2}$

We Know that:

$V_1=70\frac{km}{h} \\\\V_2=90\frac{km}{h}$

Replacing the values:

$V_a_v_e_r_a_g_e=\frac{70+90}{2} =\frac{160}{2}=80\frac{km}{h}$

5 0

3 years ago

Solve: 2a-3=9 Solve: -3(b+2)= -9

Softa [21]

The Solution are : a=3.
b=1.

3 0

3 years ago

Read 2 more answers

You ask a friend to think of a number feom two to twelve. What is the probability that his numberwill ne 4?

antiseptic1488 [7]

The probability that his number will be 4 is one in eleven, or 1/11.

4 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