To solve the inequality, the first step is to add \(\displaystyle 7\) to both sides:

 \(\displaystyle 2x - 7 +7 \geq 6+7\)

\(\displaystyle 2x \geq 13\)

The second step is to divide both sides by \(\displaystyle 2\):

\(\displaystyle \frac{2x}{2} \geq \frac{13}{2}\)

\(\displaystyle x \geq 6.5\)

To graph the inequality, you draw a straight number line. Fill in the numbers from \(\displaystyle 6.5\) to infinity. Infinity can be designated by a ray. Be sure to fill in the number \(\displaystyle 6.5\), since the equation indicated greater than OR equal to.

The graph should look like:

If we plug the points \(\displaystyle (3,0)\) and \(\displaystyle (0,3)\) into each equation, we find that these points work only in the equation \(\displaystyle (x-3)^{2}+(y-3)^{2}=9\). This circle has a radius of \(\displaystyle 3\) and is centered at \(\displaystyle (3,3)\).

The key here is to look for the line whose slope is the negative reciprocal of the original slope.

In this case, \(\displaystyle \frac{-1}{3}\) is the negative reciprocal of \(\displaystyle 3\).

Therefore, the equation of the line which is perpendicular to the original equation is:

\(\displaystyle y=\frac{-x}{3}+2\)

The inverse of a function has all the same points as the original function, except the x values and y values are reversed. The same rule applies to polygons such as triangles.

The simplest way to solve this problem is to plug all of the answer choices into the provided equation, and see which one results in a power of \(\displaystyle 96\) \(\displaystyle watts\).  

Alternatively, one could set up the equation,

 \(\displaystyle v{^2}+4v-96=0\) and factor, use the quadratic equation, or graph this on a calculator to find the root. 

If we were to factor we would look for factors of c that when added together give us the value in b when we are in the form,

\(\displaystyle ax^2+bx+c\).

In our case \(\displaystyle a=1, b=4, c=-96\). So we need factors of \(\displaystyle -96\) that when added together give us \(\displaystyle 4\).

Thus the following factoring would solve this problem.

\(\displaystyle (v+12)(v-8)=0\)

Then set each binomial equal to zero and solve for v.

\(\displaystyle v+12=0 \rightarrow v=-12\)

\(\displaystyle v-8=0 \rightarrow v=8\)

Since we can't have a negative power our answer is \(\displaystyle {}8 m/s\).

The \(\displaystyle (+4)\) inside the argument has the effect of shifting the graph \(\displaystyle 4\) units to the left. This can be easily seen by graphing both the original and modified functions on a graphing calculator.  

To answer this question, we need to correctly identify where to plug in our given values and solve for \(\displaystyle a\).

Points on a graph are written in coordinate pairs. These pairs show the \(\displaystyle x\) value first and the \(\displaystyle y\) value second. So, for this data:

\(\displaystyle (1,2a)\) means that \(\displaystyle 1\) is the \(\displaystyle x\) value and \(\displaystyle 2a\) is the \(\displaystyle y\) value.

We must now plug in our \(\displaystyle x\) and \(\displaystyle y\) values into the original equation and solve. Therefore:

\(\displaystyle y=-4x^{2}+6\rightarrow 2a=-4(1)^{2}+6\)

We can now begin to solve for \(\displaystyle a\) by adding up the right side and dividing the entire equation by \(\displaystyle 2\).

\(\displaystyle 2a=-4(1)^{2}+6\rightarrow 2a=-4+6\rightarrow2a=2\)

\(\displaystyle 2a=2\rightarrow \frac{2a}{2}=\frac{2}{2}\rightarrow a=1\)

Therefore, the value of \(\displaystyle a\) is \(\displaystyle 1\).

Complex numbers can be represented on the coordinate plane by mapping the real part to the x-axis and the imaginary part to the y-axis.  For example, the expression \(\displaystyle a+bi\) can be represented graphically by the point \(\displaystyle (a,b)\).

Here, we are given the graph and asked to write the corresponding expression.

\(\displaystyle 3-2i\) not only correctly identifies the x-coordinate with the real part and the y-coordinate with the imaginary part of the complex number, it also includes the necessary \(\displaystyle i\). 

\(\displaystyle 3-2\) correctly identifies the x-coordinate with the real part and the y-coordinate with the imaginary part of the complex number, but fails to include the necessary \(\displaystyle i\).

\(\displaystyle -2+3i\) misidentifies the y-coordinate with the real part and the x-coordinate with the imaginary part of the complex number.

\(\displaystyle -2+3\) misidentifies the y-coordinate with the real part and the x-coordinate with the imaginary part of the complex number.  It also fails to include the necessary \(\displaystyle i\).

Complex numbers can be represented on the coordinate plane by mapping the real part to the x-axis and the imaginary part to the y-axis.  For example, the expression \(\displaystyle a+bi\) can be represented graphically by the point \(\displaystyle (a,b)\).

Here, we are given the complex number \(\displaystyle 4-i\) and asked to graph it.  We will represent the real part, \(\displaystyle 4\), on the x-axis, and the imaginary part, \(\displaystyle -i\), on the y-axis.  Note that the coefficient of \(\displaystyle i\) is \(\displaystyle -1\); this is what we will graph on the y-axis.  The correct coordinates are \(\displaystyle (4,-1)\).