AWILLISMAT140 MODULE 7
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Mathematics
Date
Nov 24, 2024
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docx
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1
Module 7 Assignment
Alisha Willis
MAT140
2
Looking at a rational expression we see it is a quotient of two polynomials, like (3x + 2)/(x - 1). The absolute value represents a number's distance from zero, denoted by |x|, where |-5| equals 5. A linear function, like f(x) = 2x + 3, forms a straight line on a graph, defined by its slope (2) and y-intercept (3). These mathematical concepts find application in diverse fields, facilitating the analysis of relationships and predictions within various contexts.
3
x
2+12
x
=
2
x
+89+1
This equation involves rational expressions on both sides. To solve for x
x
, you can cross-
multiply or try to find a common denominator to eliminate the fractions and then solve the resulting equation. For
93+y+y+13+y3+
y
9+3+
yy
+1
This expression involves fractions with a common denominator of 3+y3+
y
. To simplify, you can combine the numerators over the common denominator:
9+y+13+y3+
y
9+
y
+1
Further simplification yields:
y+103+y3+
yy
+10
This expression represents the simplified form of the given mathematical expression.
∣
3
z
−2
∣
+8=1
3
3z−23
z
−2
To solve for ZZ, we start by isolating the absolute value term:
∣
3z−2
∣
=1−8
∣
3
z
−2
∣
=1−8
Next, simplify the right side
∣
3z−2
∣
=−7
∣
3
z
−2
∣
=−7
Because absolute values are always non-negative, there is no real solution for z
z
in this case. The equation is inconsistent with real numbers.
The inequality ∣
x
∣
>3
∣
x
∣
>3 implies two separate conditions
x>3
x
>3
x<−3
x
<−3
This can be represented on a number line by marking an open circle at x=−3
x
=−3 and x=3
x
=3, indicating that these values are not included in the solution set due to the strict inequality. Then, we can shade the regions to the left of x=−3
x
=−3 and to the right of x=3
x
=3 to denote the solutions. This illustrates that any x
x
outside the interval (−3,3)(−3,3) satisfies the inequality.
The equation of the line with a slope of 4 and passing through the point (5, 1) is f(x)=4x−19
f
(
x
)=4
x
−19.
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4
Circumference (C
C
): C=2πr
C
=2
πr
where r
r
is the radius.
Area (A
A
): A=πr2
A
=
πr
2 where r
r
is the radius.
For a circle with a radius of 4 inches
Circumference: C=2π×4=8π inches
C
=2
π
×4=8
π
inches
Area: A=π×(42)=16π square inches
A
=
π
×(42)=16
π
square inches
So, the circumference is 8π8
π
inches, and the area is 16π16
π
square inches. If we need numerical approximations, we can use π≈3.14
π
≈3.14.
Exploring a circle with a 4-inch radius shows key geometric insights. Its circumference is
calculated using C=2πr
C
=2
πr
, which equals 8 π
π
inches. The area, derived from A=πr2
A
=
πr
2, equals 16 π
π
square inches.
5
References Martin-Gay, E. K. (2020). Algebra Foundations: Prealgebra, introductory algebra, and intermediate algebra
. Pearson Education, Inc.