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\frac{-2}{23} = \frac{-2}{23}
5 + x = 0 \implies x = -5
\int \left(2 t\right) dt = t^2 + C
\log_{3}\left(2 + 2 w\right) = 2 \implies w = \frac{7}{2}
-35 \cdot 45 = -1575
-3 + 37 = 34
\log_{2}\left(2 + x\right) = 3 \implies x = 6
\log_{5}\left(3 + 3 c\right) = 2 \implies c = \frac{22}{3}
-3 + t = 0 \implies t = 3
\lim_{w \to -4} \left(-w\right) = 4
(-4, -3, 4) \times (-1, 1, 4) = (-16, 12, -7)
\frac{-16}{41} = \frac{-16}{41}
32 - 4 w - 8 w^2 + w^3 = 0 \implies w \in \{ -2, 2, 8 \}
-27 \cdot 24 = -648
\lim_{y \to 4} \left(2 y\right) = 8
72 - 6 t - 7 t^2 + t^3 = 0 \implies t \in \{ -3, 4, 6 \}
35 \cdot 4 = 140
32 - 12 z + z^2 = 0 \implies z \in \{ 4, 8 \}
22 \cdot 14 = 308
12 - 13 u + u^3 = 0 \implies u \in \{ -4, 1, 3 \}
\log_{4}\left(3 + 2 v\right) = 3 \implies v = \frac{61}{2}
f(w) = 5 + 3 w + w^2, \quad g(t) = 4 + 3 t \implies f(g(t)) = 33 + 33 t + 9 t^2
\frac{d}{db}\left(b^3\right) = 3 b^2
\int \left(1\right) dz = z + C
\frac{27}{24} = \frac{9}{8}
-3 u \left(-u + x\right) = -3 u x + 3 u^2
\lim_{z \to -4} \left(0\right) = 0
\frac{-31}{5} = \frac{-31}{5}
(3, 4, -1) \times (4, -2, 1) = (2, -7, -22)
\log_{5}\left(2 + 3 y\right) = 2 \implies y = \frac{23}{3}
\lim_{z \to 4} \left(0\right) = 0
16 - 4 v - 4 v^2 + v^3 = 0 \implies v \in \{ -2, 2, 4 \}
\int \left(1\right) dv = v + C
\lim_{t \to -1} \left(10\right) = 10
\log_{4}\left(5 + 3 w\right) = 3 \implies w = \frac{59}{3}
\int \left(2 c\right) dc = c^2 + C
-4 - 3 u + u^2 = 0 \implies u \in \{ -1, 4 \}
\lim_{c \to 2} \left(5 - c\right) = 3
\frac{d}{dt}\left(\cos{\left(t\right)}\right) = -\sin{\left(t\right)}
\int \left(1\right) du = u + C
f(u) = 3 + u + u^2, \quad g(w) = 1 + 2 w \implies f(g(w)) = 5 + 6 w + 4 w^2
-24 - 5 w + w^2 = 0 \implies w \in \{ -3, 8 \}
\frac{d}{dc}\left(\sin{\left(c\right)}\right) = \cos{\left(c\right)}
f(b) = 3 + 3 b + b^2, \quad g(c) = 3 + c \implies f(g(c)) = 21 + 9 c + c^2
\frac{d}{dy}\left(0\right) = 0
\lim_{y \to 1} \left(2 y\right) = 2
-30 a + a^2 + a^3 = 0 \implies a \in \{ -6, 0, 5 \}
-40 + 25 = -15
32 \cdot 38 = 1216
\int \left(3 t^2\right) dt = t^3 + C
\frac{d}{dw}\left(2 w \sin{\left(w\right)}\right) = 2 w \cos{\left(w\right)} + 2 \sin{\left(w\right)}
f(v) = 2 + 4 v + v^2, \quad g(z) = 4 + 2 z \implies f(g(z)) = 34 + 24 z + 4 z^2
(-1, -2, 4) \times (3, -2, 5) = (-2, 17, 8)
(1, 2, 2) \times (-5, -5, 2) = (14, -12, 5)
f(a) = 5 + 5 a + a^2, \quad g(t) = 4 + t \implies f(g(t)) = 41 + 13 t + t^2
f(b) = 1 + 3 b + b^2, \quad g(t) = 4 + 3 t \implies f(g(t)) = 29 + 33 t + 9 t^2
8 - 6 u + u^2 = 0 \implies u \in \{ 2, 4 \}
-10 \left(-8 + b\right) = 80 - 10 b
\lim_{a \to -1} \left(-12\right) = -12
\log_{3}\left(4 + 4 y\right) = 1 \implies y = \frac{-1}{4}
21 - 10 u + u^2 = 0 \implies u \in \{ 3, 7 \}
-6 + z = 0 \implies z = 6
18 + 9 c + c^2 = 0 \implies c \in \{ -6, -3 \}
6 - v - 6 v^2 + v^3 = 0 \implies v \in \{ -1, 1, 6 \}
\lim_{t \to -3} \left(t^2\right) = 9
\log_{5}\left(3 + 4 x\right) = 1 \implies x = \frac{1}{2}
17 - 9 = 8
-17 - 30 = -47
6 + z = 0 \implies z = -6
20 + 17 = 37
\int \left(-1\right) dw = -w + C
14 v + 9 v^2 + v^3 = 0 \implies v \in \{ -7, -2, 0 \}
f(x) = 1 + 3 x + x^2, \quad g(t) = 2 + 2 t \implies f(g(t)) = 11 + 14 t + 4 t^2
-42 - 29 w - 2 w^2 + w^3 = 0 \implies w \in \{ -3, -2, 7 \}
30 + 11 t + t^2 = 0 \implies t \in \{ -6, -5 \}
-18 + 3 c + c^2 = 0 \implies c \in \{ -6, 3 \}
-10 - 36 = -46
8 + a = 0 \implies a = -8
6 + t = 0 \implies t = -6
f(u) = 2 + 4 u + u^2, \quad g(x) = 2 + 2 x \implies f(g(x)) = 14 + 16 x + 4 x^2
f(z) = 1 + z + z^2, \quad g(v) = 4 + 2 v \implies f(g(v)) = 21 + 18 v + 4 v^2
\frac{d}{dv}\left(v\right) = 1
f(u) = 1 + 4 u + u^2, \quad g(a) = 4 + 2 a \implies f(g(a)) = 33 + 24 a + 4 a^2
f(v) = 2 + 2 v + v^2, \quad g(u) = 1 + 2 u \implies f(g(u)) = 5 + 8 u + 4 u^2
\frac{d}{db}\left(5 \cos{\left(b\right)}\right) = -5 \sin{\left(b\right)}
\log_{4}\left(7 + y\right) = 1 \implies y = -3
\frac{d}{dv}\left(v^4\right) = 4 v^3
\log_{3}\left(1 + 2 b\right) = 3 \implies b = 13
\frac{d}{db}\left(b - \left(-4 + b\right)\right) = 0
\log_{3}\left(1 + 3 y\right) = 3 \implies y = \frac{26}{3}
\int \left(2 x\right) dx = x^2 + C
\frac{d}{da}\left(a\right) = 1
(-1, 4, 4) \times (0, 2, 5) = (12, 5, -2)
\log_{2}\left(8 + y\right) = 1 \implies y = -6
\frac{42}{3} = 14
-22 + 4 = -18
15 - 46 = -31
\int \left(3 t^2\right) dt = t^3 + C
f(w) = 4 + 5 w + w^2, \quad g(a) = 2 + 2 a \implies f(g(a)) = 18 + 18 a + 4 a^2
\lim_{y \to 2} \left(\sin{\left(y\right)}\right) = \sin{\left(2\right)}