Question

1. given the table assuming its linear, estimate the value when x= 10?

x	y
20	35
30	55
50	75

1. given the table, find y when x=7

x	y
2	6
3	18
4	54

1. which quadrants is the solution in y> -2 + 3x
2. find the value k and p in the following expression

\\(\frac{(x^3 y^k)^p}{x^4 y^6} = x^3 y^3\\)

Explanation:

Problem 19:

Step1: Calculate slope m

$m=\frac{y_2-y_1}{x_2-x_1}=\frac{35-5}{20-5}=\frac{30}{15}=2$

Step2: Find linear equation

Use point (5,5): $y-5=2(x-5)$ → $y=2x-5$

Step3: Substitute x=10

$y=2(10)-5=20-5=15$

Problem 20:

Step1: Identify pattern

$y$ follows $y=2\times3^{x-1}$ (each term ×3)

Step2: Substitute x=7

$y=2\times3^{7-1}=2\times3^6=2\times729=1458$

Problem 21:

Step1: Analyze line $y=3x-2$

Y-intercept: (0,-2); X-intercept: $(\frac{2}{3},0)$

Step2: Locate solution region

The inequality $y>3x-2$ is above the line, covering Quadrants I, II, IV (and part of III, but main quadrants are I, II, IV)

Problem 22:

Step1: Simplify left side

$\frac{(x^3y^k)^p}{x^4y^6}=\frac{x^{3p}y^{kp}}{x^4y^6}=x^{3p-4}y^{kp-6}$

Step2: Equate exponents for x

$3p-4=3$ → $3p=7$ → $p=\frac{7}{3}$

Step3: Equate exponents for y

$kp-6=3$ → $k\times\frac{7}{3}=9$ → $k=\frac{27}{7}$

Answer:

1. When $x=10$, $y=15$
2. When $x=7$, $y=1458$
3. The solution is in Quadrants I, II, and IV
4. $k=\frac{27}{7}$, $p=\frac{7}{3}$