Answer :
Let's go through each of the math problems one by one:
- If an AP has a first term of 6 and a last term of 54 with a common difference of 6, how many terms are there in the AP?
The formula for the n-th term of an arithmetic progression (AP) is:
[tex]a_n = a_1 + (n-1) \cdot d[/tex]
Where:
- [tex]a_n[/tex] is the n-th term
- [tex]a_1[/tex] is the first term
- [tex]d[/tex] is the common difference
Given:
- [tex]a_1 = 6[/tex]
- [tex]a_n = 54[/tex]
- [tex]d = 6[/tex]
Plug these into the formula:
[tex]54 = 6 + (n-1) \cdot 6[/tex]
[tex]48 = (n-1) \cdot 6[/tex]
[tex]n-1 = \frac{48}{6}[/tex]
[tex]n-1 = 8[/tex]
[tex]n = 9[/tex]
Answer: B. 9
- Where do the lines [tex]y = 2x + 3[/tex] and [tex]y = -x + 6[/tex] intersect?
To find the intersection, set the equations equal:
[tex]2x + 3 = -x + 6[/tex]
Combine like terms:
[tex]2x + x = 6 - 3[/tex]
[tex]3x = 3[/tex]
[tex]x = 1[/tex]
Plug [tex]x = 1[/tex] back into one of the equations:
[tex]y = 2(1) + 3 = 5[/tex]
The intersection is at [tex](1, 5)[/tex].
Answer: A. (1, 5)
- Simplify [tex]\log_5 40[/tex]:
Express 40 as factors of 5 and 2:
[tex]40 = 5^1 \cdot 2^3[/tex]
Using logarithm properties:
[tex]\log_5 40 = \log_5 (5^1 \cdot 2^3)[/tex]
[tex]= \log_5 5^1 + \log_5 2^3[/tex]
[tex]= 1 + 3 \log_5 2[/tex]
Answer: C. 1 + 3 \log_5 2
- If [tex]f(x) = 3x + 2[/tex], what is [tex]f^{-1}(x)[/tex]?
To find the inverse, switch [tex]x[/tex] and [tex]f(x)[/tex]:
- [tex]y = 3x + 2[/tex]
- Switch [tex]x[/tex] and [tex]y[/tex]: [tex]x = 3y + 2[/tex]
- Solve for [tex]y[/tex]:
[tex]x - 2 = 3y[/tex]
[tex]y = \frac{x - 2}{3}[/tex]
Answer: A. \frac{x-2}{3}
- Find the coefficient of the 4th term of the binomial expression [tex](x - 3y)^4[/tex]:
Use the binomial theorem:
[tex](x - 3y)^4 = \sum_{k=0}^{4} \binom{4}{k} x^{4-k} (-3y)^k[/tex]
The 4th term corresponds to [tex]k = 3[/tex]:
[tex]\binom{4}{3} x^{1} (-3y)^3[/tex]
[tex]\binom{4}{3} = 4[/tex]
[tex](-3)^3 = -27[/tex]
[tex]\text{Coefficient} = 4 \times -27 = -108[/tex]
Answer: C. -108
- From the equation [tex]x^2 + y^2 + 6x - 8y + 9 = 0[/tex], find the radius.
Rearrange to form a circle equation:)
Complete the square:
[tex](x^2 + 6x) + (y^2 - 8y) = -9[/tex]
Complete the square for [tex]x[/tex]:
[tex]x^2 + 6x = (x+3)^2 - 9[/tex]
Complete the square for [tex]y[/tex]:
[tex]y^2 - 8y = (y-4)^2 - 16[/tex]
Substitute back:
[tex](x+3)^2 - 9 + (y-4)^2 - 16 = -9[/tex]
[tex](x+3)^2 + (y-4)^2 = 16[/tex]
The radius is [tex]\sqrt{16} = 4[/tex].
Answer: A. 4
- What is the distance between the points (3, 4) and (7, 1)?
Use the distance formula:
[tex]\text{Distance} = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}[/tex]
[tex]= \sqrt{(7-3)^2 + (1-4)^2}[/tex]
[tex]= \sqrt{4^2 + (-3)^2}[/tex]
[tex]= \sqrt{16 + 9}[/tex]
[tex]= \sqrt{25}[/tex]
[tex]= 5[/tex]
Answer: A. 5
- What is the range of the relation [tex]R = \{(2, 5), (3, 7), (4, 5), (6, 8)\}[/tex]?
The range of a relation is the set of all second elements (y-values).
Thus, the range is [tex]\{5, 7, 8\}[/tex].