High School Math 1 — every item

The $x$ term is $-4x$, so the coefficient of $x$ is $\mathbf{-4}$. ($7$ is the coefficient of $x^2$; $9$ is the constant term.)

$2(3x - 4) - 5x = 6x - 8 - 5x = \mathbf{x - 8}$.

$4 \cdot \frac{1}{4} - 6 \cdot \frac{1}{2} + 1 = 1 - 3 + 1 = \mathbf{-1}$.

Sum: $x + 3$. Twice the sum: $2(x + 3)$. Decreased by $5$: $\mathbf{2(x + 3) - 5}$. ($2x + 3 - 5$ doubles only $x$; $5 - 2(x+3)$ subtracts in the wrong direction; $2(x - 5) + 3$ groups the wrong pair.)

Length $= 3w + 4$. Perimeter $= 2(3w + 4) + 2w = 6w + 8 + 2w = \mathbf{8w + 8}$. ($4w + 4$ adds one length and one width; $6w + 8$ forgets the two widths; $3w^2 + 4w$ is the area.)

Three terms make a trinomial; the highest power is $x^3$, so the degree is $3$. Answer: trinomial, degree 3. ($7$ is the constant, not the degree; degree $2$ would need an $x^2$ term.)

$(2x - 3)(x + 5) = 2x^2 + 10x - 3x - 15 = \mathbf{2x^{2} + 7x - 15}$.

$(x + 6)(x + 2) = x^2 + 2x + 6x + 12 = x^2 + 8x + 12$, so $b = \mathbf{8}$ (and $c = 12$).

$(x + 5)^2 = x^2 + 5x + 5x + 25 = \mathbf{x^{2} + 10x + 25}$. ($x^2 + 25$ squares termwise and drops the middle; $x^2 + 5x + 25$ counts the cross product once; $x^2 + 10x + 10$ forgets to square the $5$.)

$x^2 + y^2 = (x + y)^2 - 2xy = 49 - 20 = \mathbf{29}$. Check: $x = 5, y = 2$ gives $25 + 4 = 29$. ($49$ forgets to subtract $2xy$; $39$ subtracts $xy$ only once; $69$ adds $2xy$ instead of subtracting.)

$5x - 7 = 3x + 9 \Rightarrow 2x = 16 \Rightarrow x = \mathbf{8}$. Check: $40 - 7 = 33 = 24 + 9$. ✓

$A = \frac{1}{2}bh \Rightarrow 2A = bh \Rightarrow \mathbf{h = \dfrac{2A}{b}}$. Check with $b = 4, h = 3$: $A = 6$ and $\frac{2 \cdot 6}{4} = 3$. ✓

Left side simplifies to $6x + 1$, which is identical to the right side. Subtracting $6x$ leaves $1 = 1$, true for every $x$: an identity with infinitely many solutions. (“No solution” would be a false statement like $1 = 4$; a single value appears only when the $x$ terms do not cancel.)

$-3 < 2x + 1 \le 7 \Rightarrow -4 < 2x \le 6 \Rightarrow \mathbf{-2 < x \le 3}$, a segment open at $-2$ and closed at $3$. (“$x < -2$ or $x \ge 3$” is two rays — an OR, the wrong connective; $-1 < x \le 4$ adds $1$ instead of subtracting; $-2 \le x < 3$ swaps which end is closed.)

$ax - 2a = 3x + b$ is an identity when $a = 3$ and $b = -2a = -6$. So $a + b = 3 + (-6) = \mathbf{-3}$. ($9$ takes $b = +6$; $-6$ reports $b$ alone; $3$ reports $a$ alone.)

$4y = -3x + 12 \Rightarrow y = -\frac{3}{4}x + 3$. Slope $= \mathbf{-\dfrac{3}{4}}$. (Shortcut for $Ax + By = C$: slope $= -\frac{A}{B}$.)

Perpendicular slope: $-\frac{1}{2}$. Point-slope: $y + 1 = -\frac{1}{2}(x - 2) = -\frac{1}{2}x + 1$. Slope-intercept: $\mathbf{y = -\dfrac{1}{2}x}$ — the line passes through the origin. Check: $x = 2$ gives $y = -1$. ✓

$m = \frac{-9}{3} = -3$. Using $(1, 5)$: $5 = -3(1) + b \Rightarrow b = 8$. The line is $y = -3x + 8$, with $y$-intercept $\mathbf{8}$. Check: $x = 4$ gives $-12 + 8 = -4$. ✓

$(3, 1)$: $1 < 2(3) - 3 = 3$ is true, so $\mathbf{(3, 1)}$ is in the region. ($(0,0)$ gives $0 < -3$, false; $(1,2)$ gives $2 < -1$, false; $(2,1)$ lies on the dashed boundary, which a strict inequality excludes.)

Slope of the given line: $\frac{4}{3}$. New line: $1 = \frac{4}{3}(6) + b = 8 + b$, so $b = \mathbf{-7}$. The line is $y = \frac{4}{3}x - 7$. (The given line's own intercept, $-3$, is a different line.)

$3x + 2x - 1 = 14 \Rightarrow 5x = 15 \Rightarrow x = \mathbf{3}$ (and $y = 5$). Check: $9 + 5 = 14$. ✓

$2 \times (3x + 2y = 16)$ gives $6x + 4y = 32$. Adding $5x - 4y = 1$: $11x = 33$, $x = 3$. Then $9 + 2y = 16 \Rightarrow y = \mathbf{\dfrac{7}{2}}$. Check: $15 - 14 = 1$. ✓

$8a + 600 - 5a = 780 \Rightarrow 3a = 180 \Rightarrow a = \mathbf{60}$. Then $s = 60$; check $8(60) + 5(60) = 480 + 300 = 780$. ✓

The first equation is $x - 2y = 4$; the second is $x - 2y = 6$. Same left side, different constants: the lines are parallel and distinct, so there is no solution. (“Infinitely many” would need the constants to match too; two lines can never cross exactly twice.)

Adding: $5p + 5n = 25$, so $p + n = \mathbf{5}$ dollars. (Solving fully gives $p = 1$, $n = 4$: $1 and $4 are the separate prices, not the sum.)

$f(4) = 3(4) - 5 = 12 - 5 = \mathbf{7}$.

$\frac{x + 4}{2} = 11 \Rightarrow x + 4 = 22 \Rightarrow x = \mathbf{18}$. Check: $f(18) = \frac{22}{2} = 11$. ✓

In $\{(1, 2), (2, 2), (3, 2)\}$ the inputs $1, 2, 3$ are all different, so each has exactly one output — it is a function even though every output is $2$. Answer: $\mathbf{\{(1,2),\ (2,2),\ (3,2)\}}$. (The first set sends $1$ to both $2$ and $4$; the second sends $0$ to three outputs; the third sends $2$ to both $1$ and $3$.)

$x - 1 \ge 0 \Rightarrow x \ge 1$, so the domain is $\mathbf{[1, \infty)}$ with a square bracket because $x = 1$ gives $\sqrt{0} = 0$. (All reals ignores the root; $(1, \infty)$ wrongly excludes $1$; $[0, \infty)$ forgets the $-1$ inside.)

$25 + 0.15m = 61 \Rightarrow 0.15m = 36 \Rightarrow m = \frac{36}{0.15} = \mathbf{240}$ miles. Check: $25 + 0.15(240) = 25 + 36 = 61$. ✓

GCF $= 6x^2$, so $12x^3 - 18x^2 = \mathbf{6x^{2}(2x - 3)}$. (The other three are all true products, but each leaves a common factor inside the bracket — $x$, $2$, or $3$ — so they are not complete.)

$6x^2 - 5x - 6 = 6x^2 + 4x - 9x - 6 = 2x(3x + 2) - 3(3x + 2) = \mathbf{(2x - 3)(3x + 2)}$. Check the middle term: $2x \cdot 2 + (-3)(3x) = 4x - 9x = -5x$. ✓

$x^2 + 7x + 12 = (x + 3)(x + 4)$. The width is $x + 3$, so the length is $\mathbf{x + 4}$.

$x^2 - 9 = (x - 3)(x + 3)$; the middle terms $3x$ and $-3x$ cancel. Answer: $\mathbf{(x - 3)(x + 3)}$. ($(x-3)^2 = x^2 - 6x + 9$ and $(x+3)^2 = x^2 + 6x + 9$ both have a middle term; $(x - 9)(x + 1) = x^2 - 8x - 9$.)

The sums are $49, 26, 19, 16, 14$ — all different — so there are $\mathbf{5}$ possible values of $b$. ($10$ counts each unordered pair twice; $4$ misses a pair.)

$(x - 2)(x - 3) = 0 \Rightarrow x = 2$ or $x = 3$. The larger solution is $\mathbf{3}$.

$x = \dfrac{3 \pm \sqrt{25}}{4} = \dfrac{3 \pm 5}{4}$, so $x = 2$ or $x = \dfrac{-2}{4} = \mathbf{-\dfrac{1}{2}}$. Factoring check: $(2x + 1)(x - 2) = 2x^2 - 3x - 2$. ✓

Vertex at $t = -\frac{64}{-32} = 2$. Then $h(2) = -16(4) + 64(2) + 80 = -64 + 128 + 80 = \mathbf{144}$ feet. ($2$ is the time, not the height; $80$ is the starting height.)

$x = -\dfrac{b}{2a} = -\dfrac{-12}{6} = \mathbf{2}$. ($6$ drops the $a$ and computes $-\frac{b}{2}$; $-2$ loses the sign; $12$ is just $-b$.)

$r + s = 7$ and $r - s = 3$ give $r = 5$, $s = 2$. So $k = rs = \mathbf{10}$. Check: $x^2 - 7x + 10 = (x - 5)(x - 2)$, roots $5$ and $2$, which differ by $3$. ✓ ($12$ comes from guessing roots $3$ and $4$, which differ by $1$.)

$f(3) = 800 \cdot \frac{1}{8} = \mathbf{100}$. Each step halves: $800 \to 400 \to 200 \to 100$.

$V(2) = 24000 \cdot 0.85^2 = 24000 \cdot 0.7225 = \mathbf{17340}$ dollars. (Subtracting $15\%$ of the original twice, $24000 - 2 \cdot 3600 = 16800$, is linear decay, not exponential.)

Start value $3$, ratio $2$: $\mathbf{y = 3 \cdot 2^{x}}$. Check $x = 3$: $3 \cdot 8 = 24$. ✓ ($y = 2 \cdot 3^x$ swaps $a$ and $b$ — it gives $2$ at $x = 0$; $y = 3x + 3$ adds $3$ each step instead of doubling; $6^x$ gives $1$ at $x = 0$.)

A bacteria count doubles every hour multiplies by a constant factor $2$, so it is exponential: $N(t) = N_0 \cdot 2^t$. (The plant, taxi, and jar each add a constant $2$ per step — linear, even though the number $2$ appears.)

$A = 2000(1.03)^2 = 2000 \cdot 1.0609 = \mathbf{2121.80}$ dollars. (Simple $6\%$ for one year would give $2,120; compounding twice earns the extra $1.80.)

$-2 < 0$, so $f(-2) = -2 + 3 = \mathbf{1}$. (Using $2x$ would give $-4$, but that rule is only for $x \ge 0$.)

$\dfrac{f(4) - f(1)}{4 - 1} = \dfrac{8 - (-1)}{3} = \dfrac{9}{3} = \mathbf{3}$.

The graph rises between the low point and the high point, so $f$ is increasing on $\mathbf{(-1, 2)}$. ($(-3, 5)$ lists the $y$-values of those two points, not an $x$-interval; on $(-4, -1)$ and $(2, 5)$ the graph is falling.)

$\dfrac{h(5) - h(2)}{5 - 2} = \dfrac{44 - 20}{3} = \mathbf{8}$ m/s. ($24$ is $h(5) - h(2)$ without dividing by the time; $\frac{32}{3}$ uses the height change from $t = 0$; $6.4$ is $\frac{32}{5}$, the rate over the whole table.)

Vertex at $x = 3$; $f(3) = -9 + 18 - 5 = \mathbf{4}$. The maximum value is $4$, reached at $x = 3$. (Reporting $3$ answers “where,” not “what.”)