Question

14.
unit 3: rational numbers

____ 10. 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

____ 11. which graph represents the equation \\(y = 2x + 3\\)?

graph with x-axis from 0 to 9, y-axis from 0 to 11, with lines p, q, r, s. line s starts at (0,0), line q starts at (0,2), line p starts at (0,3), line r starts at (3,0)

a. line s
b. line q
c. line p
d. line r

unit 5: polynomials

____ 12. 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\\)

____ 13. 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\\)

questions i should practice: /14

Explanation:

Question 10

First, convert the fractions to decimals for easier comparison. $\frac{4}{6}\approx0.6667$, $\frac{7}{5} = 1.4$. Now check each option:

• Option a: $\frac{5}{6}\approx0.833$, $\frac{7}{8}=0.875$. Both are between $0.6667$ and $1.4$.
• Option b: $\frac{4}{5}=0.8$, but $\frac{7}{5}=1.4$ is not between (it's equal to the upper bound, but the question is "between", so this is incorrect).
• Option c: $\frac{1}{5}=0.2$ is below $0.6667$, so incorrect.
• Option d: $\frac{4}{5}=0.8$ is between, but $\frac{5}{6}\approx0.833$ is also between, but this option is less accurate than a as a has two numbers both between. Wait, no, re - check: The original numbers to check are $\frac{5}{6},\frac{1}{5},\frac{7}{8},\frac{4}{5}$. We need to see which of these are between $\frac{4}{6}$ and $\frac{7}{5}$. $\frac{5}{6}\approx0.833$ (between), $\frac{1}{5}=0.2$ (not), $\frac{7}{8}=0.875$ (between), $\frac{4}{5}=0.8$ (between). Wait, maybe I made a mistake earlier. Wait the options are:

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 $\frac{4}{5}$

Wait, let's recalculate: $\frac{4}{6}\approx0.6667$, $\frac{7}{5}=1.4$. $\frac{5}{6}\approx0.833$ (between), $\frac{1}{5}=0.2$ (not), $\frac{7}{8}=0.875$ (between), $\frac{4}{5}=0.8$ (between). So the numbers between are $\frac{5}{6},\frac{7}{8},\frac{4}{5}$. So option b. Wait, maybe my initial thought was wrong. Let's do fraction comparison instead of decimals. $\frac{4}{6}=\frac{2}{3}\approx0.666$, $\frac{7}{5}=1\frac{2}{5}=1.4$. $\frac{5}{6}$: compare to $\frac{2}{3}$, $\frac{5}{6}-\frac{2}{3}=\frac{5 - 4}{6}=\frac{1}{6}>0$, so $\frac{5}{6}>\frac{2}{3}$; compare to $\frac{7}{5}$, $\frac{5}{6}=\frac{25}{30}$, $\frac{7}{5}=\frac{42}{30}$, so $\frac{5}{6}<\frac{7}{5}$. $\frac{7}{8}$: $\frac{7}{8}=0.875$, $\frac{2}{3}\approx0.666$, so $\frac{7}{8}>\frac{2}{3}$; $\frac{7}{8}=\frac{35}{40}$, $\frac{7}{5}=\frac{56}{40}$, so $\frac{7}{8}<\frac{7}{5}$. $\frac{4}{5}=0.8$, $\frac{4}{5}-\frac{2}{3}=\frac{12 - 10}{15}=\frac{2}{15}>0$, so $\frac{4}{5}>\frac{2}{3}$; $\frac{4}{5}=\frac{4}{5}$, $\frac{7}{5}=\frac{7}{5}$, so $\frac{4}{5}<\frac{7}{5}$. $\frac{1}{5}=0.2<\frac{2}{3}$. So $\frac{5}{6},\frac{7}{8},\frac{4}{5}$ are between. So option b.

The equation is $y = 2x+3$. The slope - intercept form is $y=mx + b$, where $m$ is the slope and $b$ is the y - intercept. Here, $m = 2$ (slope) and $b = 3$ (y - intercept, the value of $y$ when $x = 0$).

• Line S: When $x = 0$, $y = 0$, so y - intercept is 0, not 3.
• Line Q: Let's check the slope. For a line, slope $m=\frac{y_2 - y_1}{x_2 - x_1}$. If we take two points, say when $x = 0$, $y = 3$ (y - intercept), and when $x = 1$, $y=2(1)+3 = 5$. So the line should pass through $(0,3)$ and $(1,5)$. Looking at the graph, Line P: Let's check its slope. If $x = 0$, $y = 2$ (no, wrong). Wait, Line P: when $x = 0$, $y = 3$? Wait, the graph: Line P has a steeper slope. Wait, the slope of $y = 2x+3$ is 2. Let's calculate the slope of each line:
• Line S: Passes through $(0,0)$ and $(1,2)$, slope $=\frac{2 - 0}{1 - 0}=2$, but y - intercept is 0, not 3.
• Line Q: Passes through $(0,3)$ and $(1,5)$, slope $=\frac{5 - 3}{1 - 0}=2$, y - intercept is 3. This matches $y = 2x+3$.
• Line P: Passes through $(0,2)$ and $(1,4)$, slope $=2$, but y - intercept is 2, not 3.
• Line R: Passes through $(3,0)$ and $(4,2)$, slope $=\frac{2 - 0}{4 - 3}=2$, but y - intercept is - 6 (when $x = 0$, $y=-6$), not 3.

So Line Q has slope 2 and y - intercept 3, so it represents $y = 2x+3$.

Step 1: Distribute the negative sign

We have $(3x-7x^{2}+2)-(4x^{2}-5 + 6x)=3x-7x^{2}+2-4x^{2}+5 - 6x$

Step 2: Combine like terms

For $x^{2}$ terms: $-7x^{2}-4x^{2}=-11x^{2}$
For $x$ terms: $3x-6x=-3x$
For constant terms: $2 + 5=7$
So the result is $-11x^{2}-3x + 7$

Answer:

b. $\frac{5}{6},\frac{7}{8}$, and $\frac{4}{5}$

Question 11