Algebra Practice Test 3

To multiply, we can use the distributive property.

First, we can multiply \(2a^2b\) by each term in \(3a^3b+4c\):

\(2a^2b\times 3a^3b=6a^{2+3}b^{1+1}=6a^5b^2\)

\(2a^2b\times 4c=8a^2bc\)

Then, we can multiply \(-3c^3\) by each term in \(3a^3b+4c\):

\(-3c^3\times 3a^3b=-9a^3bc^3\)

\(-3c^3\times 4c=-12c^{3+1}=-12c^4\)

Finally, we can combine all four products:

\(6a^5b^2 + 8a^2bc – 9a^3bc^3- 12c^4\)

First, we need to find the \(y\)-intercept. The graph crosses the \(y\)-axis at \((0,2)\). In slope-intercept form (\(y=mx+b\)), this tells us that \(b=2\).

Then, we need to compute the slope. Start by choosing two clear points on the line:

• \((0,2)\) (the \(y\)-intercept)
• \((2,0)\) (where it crosses the \(x\)-axis)

Then:

\(m=\dfrac{\Delta y}{\Delta x}=\dfrac{0-2}{2-0}\) \(=\dfrac{-2}{2}=-1\)

Now, with slope \(m=-1\) and intercept \(b=2\):

\(y=-1\cdot x+2\) \(\implies y=-x+2\)

Only Choice A matches both the slope and intercept of the graphed line.

Evaluate as follows:

\(2f(x)-3=2(2x^2 + 7)-3\) \(=4x^2 + 14 – 3\) \(= 4x^2 + 11\)

From the slope and intercept you know the equation of the line is \(y=4x+1\).

To see which choice lies on the line, just substitute its \(x\)-coordinate into \(4x+1\) and check the \(y\)-value.

• For \((2,9)\): \(4\cdot 2 + 1 =9\)
• For \((0,-1)\): \(4\cdot 0 + 1 =1\), not -1
• For \((0,0)\): \(4\cdot 0 + 1 =1\), not 0
• For \((4,1)\): \(4\cdot 4 + 1 =17\), not 1
• For \((1,4)\): \(4\cdot 1 + 1 =5\), not 4

Thus, only \((2,9)\) satisfies the equation.

In the graph above, the blue line is \(y=4x+1\), and the five candidate points are marked. As you can see, only \((2,9)\) lies exactly on the line.

First, we need to plug in \(a=32\):

\(-\dfrac{1}{2}\cdot 32=-16\)

The height function is now \(y(t)=-16\:t^2+30,000\).

Then, we need to evaluate at \(t=15\text{s}\). Since \(15^2=225\), the drop distance is as follows:

\(16 \times 225=3,600 \text{ ft}\)

Subtracting from the starting height gives us the following:

\(y(15)=30,000-3,600=26,400 \text{ ft}\)

Rearranging each equation so that it shows \(y\) as a function of the variable \(x\), we have:

\(y=-\tfrac{a}{3}x+\tfrac{18}{3}\)

\(y=-\tfrac{15}{a}x+\tfrac{24}{a}\)

The lines will be parallel if the slopes are equal. Therefore \(-\tfrac{a}{3}=-\tfrac{15}{a}\), which yields \(a^2=45\). Taking the square root yields \(a=\sqrt{45}\).

Since this value is unique, all the other answers are incorrect.

This is a typical plot of an inverse variation, in which the product of the dependent and independent variables, \(x\) and \(y\), is always equal to the same value.

In this case, the product is always equal to 1, so the plot occupies the first and third quadrants of the coordinate plane. As \(x\) increases and approaches infinity, \(y\) decreases and approaches zero, maintaining the constant product.

It isn’t necessary to use the circle circumference formula to solve the problem. Instead, note that 50 kph corresponds to 50,000 meters per hour.

We are given the car tire’s revolutions per minute and the answer must be represented as meters; therefore, the speed must be converted to meters per minute. This corresponds to a speed of \(\tfrac{50,000}{60}\) meters per minute, as there are 60 minutes in an hour.

In any given minute, the car travels \(\tfrac{50,000}{60}\) meters per minute, and each tire rotates 500 times around, or 500 times its circumference. This corresponds to \(\tfrac{50,000}{60\times 500}=\tfrac{10}{6}\) meters per revolution, which is the circumference of the tire.

Compute the product using the FOIL method. As a result, \((a+b)(a-b)=a^2-ab+ba-b^2\). Since \(ab\) is equal to \(ba\), the middle terms cancel out each other, which leaves \(a^2-b^2\).