Math Exam
Question 1: Algebra
Solve the following equation for xxx: 2(x−3)+4=3x−52(x - 3) + 4 = 3x - 52(x−3)+4=3x−5
Answer:
1. Distribute the 2 on the left side: 2x−6+4=3x−52x - 6 + 4 = 3x - 52x−6+4=3x−5
2. Simplify the left side: 2x−2=3x−52x - 2 = 3x - 52x−2=3x−5
3. Subtract 2x2x2x from both sides: −2=x−5-2 = x - 5−2=x−5
4. Add 5 to both sides: 3=x3 = x3=x
Solution: x=3x = 3x=3
Question 2: Geometry
Find the area of a triangle with a base of 8 units and a height of 5 units.
Answer:
1. Use the formula for the area of a triangle: Area=12×base×height\text{Area} =
\frac{1}{2} \times \text{base} \times \text{height}Area=21×base×height
2. Substitute the given values: Area=12×8×5\text{Area} = \frac{1}{2} \times 8 \times
5Area=21×8×5
3. Calculate the area: Area=12×40=20\text{Area} = \frac{1}{2} \times 40 =
20Area=21×40=20
Solution: The area of the triangle is 202020 square units.
Question 3: Calculus
Find the derivative of the function f(x)=3x3−4x2+2x−7f(x) = 3x^3 - 4x^2 + 2x -
7f(x)=3x3−4x2+2x−7.
Answer:
1. Use the power rule for derivatives: if f(x)=axnf(x) = ax^nf(x)=axn, then
f′(x)=a⋅n⋅xn−1f'(x) = a \cdot n \cdot x^{n-1}f′(x)=a⋅n⋅xn−1.
2. Differentiate each term: ddx[3x3]=3⋅3⋅x3−1=9x2\frac{d}{dx}[3x^3] = 3 \cdot 3 \cdot
x^{3-1} = 9x^2dxd[3x3]=3⋅3⋅x3−1=9x2 ddx[−4x2]=−4⋅2⋅x2−1=−8x\frac{d}{dx}[-4x^2]
= -4 \cdot 2 \cdot x^{2-1} = -8xdxd[−4x2]=−4⋅2⋅x2−1=−8x ddx[2x]=2\frac{d}{dx}[2x] =
2dxd[2x]=2 ddx[−7]=0\frac{d}{dx}[-7] = 0dxd[−7]=0
3. Combine the results: f′(x)=9x2−8x+2f'(x) = 9x^2 - 8x + 2f′(x)=9x2−8x+2
Question 1: Algebra
Solve the following equation for xxx: 2(x−3)+4=3x−52(x - 3) + 4 = 3x - 52(x−3)+4=3x−5
Answer:
1. Distribute the 2 on the left side: 2x−6+4=3x−52x - 6 + 4 = 3x - 52x−6+4=3x−5
2. Simplify the left side: 2x−2=3x−52x - 2 = 3x - 52x−2=3x−5
3. Subtract 2x2x2x from both sides: −2=x−5-2 = x - 5−2=x−5
4. Add 5 to both sides: 3=x3 = x3=x
Solution: x=3x = 3x=3
Question 2: Geometry
Find the area of a triangle with a base of 8 units and a height of 5 units.
Answer:
1. Use the formula for the area of a triangle: Area=12×base×height\text{Area} =
\frac{1}{2} \times \text{base} \times \text{height}Area=21×base×height
2. Substitute the given values: Area=12×8×5\text{Area} = \frac{1}{2} \times 8 \times
5Area=21×8×5
3. Calculate the area: Area=12×40=20\text{Area} = \frac{1}{2} \times 40 =
20Area=21×40=20
Solution: The area of the triangle is 202020 square units.
Question 3: Calculus
Find the derivative of the function f(x)=3x3−4x2+2x−7f(x) = 3x^3 - 4x^2 + 2x -
7f(x)=3x3−4x2+2x−7.
Answer:
1. Use the power rule for derivatives: if f(x)=axnf(x) = ax^nf(x)=axn, then
f′(x)=a⋅n⋅xn−1f'(x) = a \cdot n \cdot x^{n-1}f′(x)=a⋅n⋅xn−1.
2. Differentiate each term: ddx[3x3]=3⋅3⋅x3−1=9x2\frac{d}{dx}[3x^3] = 3 \cdot 3 \cdot
x^{3-1} = 9x^2dxd[3x3]=3⋅3⋅x3−1=9x2 ddx[−4x2]=−4⋅2⋅x2−1=−8x\frac{d}{dx}[-4x^2]
= -4 \cdot 2 \cdot x^{2-1} = -8xdxd[−4x2]=−4⋅2⋅x2−1=−8x ddx[2x]=2\frac{d}{dx}[2x] =
2dxd[2x]=2 ddx[−7]=0\frac{d}{dx}[-7] = 0dxd[−7]=0
3. Combine the results: f′(x)=9x2−8x+2f'(x) = 9x^2 - 8x + 2f′(x)=9x2−8x+2
