Mott Community College Math 120 Intermediate Algebra Test ✓ Solved

Mott Community College math 120 intermediate Algebra Test

Solve each of the following: x   ½   −½   xx

Solve each of the following and write your answer in interval notation: 1437 ÷¼ xx ≤ y

Solve each of the following: 1513 y 54310 ÷ y

Solve each of the following and write your answer in interval notation: 8232 ≤ x 2092 ≥ x

Find the value of a if the slope of the line through (a, -2) and (4, -6) is -3. Find the slope of a line parallel to the line that passes through (-7, 1) and (-4, -3). Find the slope of a line perpendicular to 3x – 5y = -4.

Find the equation of the line that passes through (-4, -2) with slope m = 3/2. Write your answer in slope-intercept form.

Find the equation of the line perpendicular to 13 yx that passes through the point (2, 0). Write your answer in slope-intercept form.

Graph each of the following: 1 23 yx xy 3³

Let 82(x) = 2^x and g(x) = 1^x. Find each of the following: g[f(2)] , f[g(2)], f(g(2)).

Let 12(x) = 2^x and 13(x) = x^3. Find each of the following: g[f(2)] , f[g(2)].

The volume of a gas varies directly with its temperature and inversely with the pressure. If the volume of a certain gas is 30 cubic feet at a temperature of 300 K and a pressure of 20 pounds per square inch, what is the volume of the same gas at 340 K when the pressure is 30 pounds per square inch? Write the formula for the general equation if y varies jointly with the square of x and z and inversely with the square root of w.

Sample Paper For Above instruction

The investigation of algebraic principles often involves solving various types of equations, understanding functions, and applying formulas to real-world scenarios. This paper demonstrates responses to such problems, encompassing algebraic manipulations, graphing, and application of variation concepts, to exemplify mastery of intermediate algebra topics as presented in the Mott Community College Math 120 Midterm.

1. Solving algebraic expressions

For the first problem, the expression appears to involve simplifying or solving for an equation involving x, such as x − ½ = xx. Assuming the task is to solve this equation, we proceed by rewriting the equation and solving for x. If the original problem was meant to be (x - ½) = xx, then:

x - ½ = x^2

Rearranging:

x^2 - x + ½ = 0

This quadratic can be solved via the quadratic formula where a=1, b=-1, c=½:

x = [1 ± √(1^2 - 4·1·½)] / 2

Discriminant: 1 - 2 = -1, which is negative, indicating no real solutions. Hence, the solution set in real numbers is empty.

2. Solving inequalities and expressing in interval notation

Given the inequality 1437 ÷ ¼ xx ≤ y, interpreting correctly, we find the solution set by isolating y and analyzing the inequality. Assuming 1437 divided by (¼ xx) ≤ y, and that xx is x squared:

1437 / (¼ x^2) ≤ y

which simplifies to:

1437 * 4 / x^2 ≤ y

or

Approximate y ≥ (5734) / x^2

Since x ≠ 0, the domain is all real x ≠ 0. The solution set can be expressed as an interval where y satisfies y ≥ 5734 / x^2, with x in (-∞, 0) ∪ (0, ∞).

3. Solving for y and x in given contexts

Given 1513 y, which likely indicates an equation such as y = 1513, and 54310 ÷ ¼ x. Without clearer details, assuming this refers to solving for x in 54310 / (¼) = x, then:

x = 54310 * 4 = 217,240

4. More inequalities and interval notation

Given inequalities 8232 ≤ x and 2092 ≥ x, the solution in interval form is:

x ∈ [8232, 2092], which is invalid because 8232 > 2092, so the solution set is empty unless inequalities are meant differently. Alternatively, if the strict inequalities are reversed, the solution set is x ∈ [2092, 8232].

5. Slope calculations and line equations

To find the value of a given the points (a, -2) and (4, -6) with slope -3:

slope = (−6 - (-2)) / (4 - a) = (-6 + 2) / (4 - a) = -4 / (4 - a) = -3

Solving:

-4 = -3(4 - a)

-4 = -12 + 3a

3a = 8

a = 8/3

Next, the slope of a line parallel to the line through (-7, 1) and (-4, -3):

slope = (-3 - 1)/(-4 + 7) = (-4)/3 = -4/3

Perpendicular slope to the line 3x – 5y = -4:

Rewrite in slope form:

3x – 5y = -4 → y = (3/5)x + 4/5

Slope of this line is 3/5, so the perpendicular slope is -5/3.

6. Equation of lines given a point and slope

Line through (-4, -2) with slope m=3/2:

Using point-slope form:

y - (-2) = (3/2)(x - (-4))

y + 2 = (3/2)(x + 4)

Expanding:

y + 2 = (3/2)x + 6

y = (3/2)x + 4

Equation in slope-intercept form is y = (3/2)x + 4.

Equation perpendicular to 13 yx (assuming 13 y = x or similar), passing through (2, 0):

Identify slope of the original line, then find the negative reciprocal for perpendicular:

Suppose the line is y = (1/13)x, then slope is 1/13, so perpendicular slope is -13.

Equation passing through (2, 0):

y – 0 = -13(x – 2)

y = -13x + 26

7. Graphs of functions

Graphs for y = 1/23 x, y = xy, and y = 3³ involve plotting linear, exponential, and polynomial functions respectively. Visualization ensures understanding of their behaviors, intersections, and asymptotic tendencies.

8. Function composition: exponential functions

Let f(x) = 2^x and g(x) = 1^x:

g[f(2)] = g(2^2) = g(4) = 1^4 = 1

f[g(2)] = 2^{g(2)} = 2^{1^2} = 2^1 = 2

f(g(2)) = 2^1 = 2

Similarly, for 12(x) = 2^x and 13(x) = x^3:

g[f(2)] = f(2) = 2^2 = 4; then g(4) = 4^3 = 64

f[g(2)] = 2^{g(2)} = 2^{2^2} = 2^4 = 16

9. Gas law variation formula

The volume (V) varies directly with temperature (T) and inversely with pressure (P):

V = k (T) / P

Given that V₁ = 30 at T₁ = 300 K and P₁ = 20 psi, find velocity at T₂=340 K and P₂=30 psi:

k = V₁ P₁ / T₁ = 30 * 20 / 300 = 2

V₂ = k T₂ / P₂ = 2 * 340 / 30 ≈ 22.67 cubic feet

10. General variation formula

If y varies jointly with x^2 and z, and inversely with √w, then:

y = k x^2 z / √w

where k is the variation constant. This formula encapsulates the relationships described, applicable in physical sciences and engineering contexts.

References

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