Question

unit 3: rational numbers

1. which of these numbers are between \\(\frac{4}{6}\\) and \\(\frac{7}{5}\\)?

\\(\frac{5}{6}, \frac{1}{5}, \frac{7}{8}, \frac{4}{5}\\)
a. \\(\frac{5}{6}\\) and \\(\frac{7}{8}\\) b. \\(\frac{5}{6}, \frac{7}{8}\\), and \\(\frac{4}{5}\\) c. \\(\frac{1}{5}\\) and \\(\frac{7}{8}\\) d. \\(\frac{5}{6}\\) and \\(\frac{4}{5}\\)
unit 4: linear relations

1. which graph represents the equation \\(y = 2x + 3\\)?

graph with x-axis from 0 to 9, y-axis from 0 to 11, lines p, q, r, s
a. line s b. line q c. line p d. line r
unit 5: polynomials

1. subtract: \\((3x - 7x^2 + 2) - (4x^2 - 5 + 6x)\\)

a. \\(-11x^2 + 3x - 7\\) b. \\(-11x^2 - 9x - 3\\) c. \\(-11x^2 - 3x + 7\\) d. \\(11x^2 + 3x - 7\\)

1. subtract: \\((3y^2 - 5x^2 + 4) - (2x - 8 + 4y^2)\\)

a. \\(-1y^2 - 5x^2 - 2x - 4\\) b. \\(3y^2 - 7x^2 + 12\\) c. \\(-4x + 12\\) d. \\(-1y^2 - 5x^2 - 2x + 12\\)

Explanation:

P, Q, R, S. Let's re - examine. The equation \(y=2x + 3\) has a slope of 2 and y - intercept 3. So when \(x = 0\), \(y = 3\), and when \(x = 1\), \(y=2(1)+3 = 5\). Let's check the lines:

• Line P: If we look at the graph, Line P passes through \((0,4)\) and \((1,6)\), slope is 2, but y - intercept is 4.
• Line Q: Let's take two points. Let's say \((0,3)\) and \((1,5)\), slope \(m=\frac{5 - 3}{1-0}=2\), which matches the slope of \(y = 2x + 3\) and y - intercept 3. Wait, maybe I misread the graph. Wait, the correct line should have a y - intercept of 3 and slope 2. So Line Q: Let's check the points. If \(x = 0\), \(y = 3\), and \(x = 1\), \(y = 5\) (since \(y=2(1)+3 = 5\)). So Line Q has a slope of 2 and y - intercept of 3.

Wait, maybe I made a mistake earlier. Let's re - calculate:
The equation \(y = 2x+3\)

• When \(x = 0\), \(y = 3\)
• When \(x = 1\), \(y=2(1)+3 = 5\)
• When \(x = 2\), \(y=2(2)+3 = 7\)

Looking at the graph, Line Q: Let's see, if we take \(x = 0\), \(y = 3\) (matches the y - intercept) and \(x = 1\), \(y = 5\) (matches \(y=2(1)+3\)). So Line Q has a slope of 2 and y - intercept of 3, which matches the equation \(y = 2x+3\).

Step1: Distribute the negative sign

We have \((3x-7x^{2}+2)-(4x^{2}-5 + 6x)\). Distribute the negative sign to each term inside the second parentheses: \(3x-7x^{2}+2-4x^{2}+5 - 6x\)

Step2: Combine like terms

• For the \(x^{2}\) terms: \(-7x^{2}-4x^{2}=-11x^{2}\)
• For the \(x\) terms: \(3x-6x=-3x\)
• For the constant terms: \(2 + 5=7\)

So the result is \(-11x^{2}-3x + 7\)

Step1: Distribute the negative sign

We have \((3y^{2}-5x^{2}+4)-(2x - 8 + 4y^{2})\). Distribute the negative sign: \(3y^{2}-5x^{2}+4-2x + 8-4y^{2}\)

Step2: Combine like terms

• For the \(y^{2}\) terms: \(3y^{2}-4y^{2}=-y^{2}\)
• For the \(x^{2}\) terms: \(-5x^{2}\) (no other \(x^{2}\) terms)
• For the \(x\) terms: \(-2x\) (no other \(x\) terms)
• For the constant terms: \(4 + 8 = 12\)

So the result is \(-y^{2}-5x^{2}-2x + 12\) or \(-1y^{2}-5x^{2}-2x + 12\)

Answer:

b. Line Q

Question 12