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\log_{5}\left(3 + 3 c\right) = 2 \implies c = \frac{22}{3}
-8 t + t^2 = 0 \implies t \in \{ 0, 8 \}
\cos{\left(t\right)} \left(-9 + t\right) = t \cos{\left(t\right)} - 9 \cos{\left(t\right)}
\lim_{v \to 1} \left(0\right) = 0
27 \cdot 39 = 1053
35 - 7 = 28
33 + 14 = 47
2 + a = 0 \implies a = -2
\log_{4}\left(1 + 2 t\right) = 1 \implies t = \frac{3}{2}
2 y + y^2 = 0 \implies y \in \{ -2, 0 \}
\int \left(1\right) dt = t + C
30 + 11 x + x^2 = 0 \implies x \in \{ -6, -5 \}
\int \left(\cos{\left(z\right)}\right) dz = \sin{\left(z\right)} + C
10 + 7 x + x^2 = 0 \implies x \in \{ -5, -2 \}
\frac{d}{da}\left(2 \sin{\left(a\right)}\right) = 2 \cos{\left(a\right)}
(2, -4, -5) \times (-3, -2, -1) = (-6, 17, -16)
3 y + y^2 = 0 \implies y \in \{ -3, 0 \}
\log_{5}\left(2 + 3 t\right) = 1 \implies t = 1
\log_{4}\left(6 + 4 z\right) = 3 \implies z = \frac{29}{2}
\frac{d}{dz}\left(0\right) = 0
4 + u = 0 \implies u = -4
\log_{5}\left(4 + 3 y\right) = 2 \implies y = 7
\log_{4}\left(7 + 2 b\right) = 3 \implies b = \frac{57}{2}
f(u) = 5 + 3 u + u^2, \quad g(y) = 2 + y \implies f(g(y)) = 15 + 7 y + y^2
2 b \left(a + b\right) = 2 a b + 2 b^2
\frac{-24}{37} = \frac{-24}{37}
\frac{d}{dv}\left(2 v + v^3\right) = 2 + 3 v^2
-70 - 31 w + 4 w^2 + w^3 = 0 \implies w \in \{ -7, -2, 5 \}
41 + 21 = 62
\frac{d}{dc}\left(c\right) = 1
f(u) = 2 + u + u^2, \quad g(z) = 3 + 2 z \implies f(g(z)) = 14 + 14 z + 4 z^2
t^4 \left(-9 + t\right) = -9 t^4 + t^5
\log_{2}\left(5 + 4 v\right) = 3 \implies v = \frac{3}{4}
\int \left(3 x^2 \sin{\left(x\right)} + x^3 \cos{\left(x\right)}\right) dx = x^3 \sin{\left(x\right)} + C
\frac{d}{da}\left(0\right) = 0
\frac{d}{dz}\left(z^2\right) = 2 z
\log_{3}\left(8 + 4 z\right) = 3 \implies z = \frac{19}{4}
\lim_{y \to 1} \left(y^2\right) = 1
\frac{-36}{4} = -9
f(w) = 3 + 2 w + w^2, \quad g(z) = 4 + z \implies f(g(z)) = 27 + 10 z + z^2
\int \left(1\right) dc = c + C
f(c) = 5 + 3 c + c^2, \quad g(b) = 1 + 2 b \implies f(g(b)) = 9 + 10 b + 4 b^2
(1, 5, 2) \times (3, 0, 2) = (10, 4, -15)
-32 + 44 t - 13 t^2 + t^3 = 0 \implies t \in \{ 1, 4, 8 \}
f(t) = 5 + 4 t + t^2, \quad g(z) = 3 + z \implies f(g(z)) = 26 + 10 z + z^2
\frac{d}{dw}\left(7\right) = 0
\log_{5}\left(2 + x\right) = 2 \implies x = 23
f(x) = 2 + 2 x + x^2, \quad g(u) = 3 + u \implies f(g(u)) = 17 + 8 u + u^2
\frac{d}{du}\left(u^2\right) = 2 u
\lim_{w \to 0} \left(w^2\right) = 0
\frac{d}{dw}\left(\cos{\left(w\right)}\right) = -\sin{\left(w\right)}
-4 + t = 0 \implies t = 4
\frac{d}{da}\left(7 a^2\right) = 14 a
\frac{d}{da}\left(\sin{\left(a\right)}\right) = \cos{\left(a\right)}
21 + 11 b - 9 b^2 + b^3 = 0 \implies b \in \{ -1, 3, 7 \}
-35 \cdot 49 = -1715
-17 \cdot 37 = -629
(-2, -2, -3) \times (1, -1, 4) = (-11, 5, 4)
-8 + z = 0 \implies z = 8
\log_{4}\left(3 + 3 b\right) = 2 \implies b = \frac{13}{3}
4 b + b^2 = 0 \implies b \in \{ -4, 0 \}
\frac{24}{4} = 6
120 + c - 10 c^2 + c^3 = 0 \implies c \in \{ -3, 5, 8 \}
(2, 4, -3) \times (-2, 4, -2) = (4, 10, 16)
\frac{15}{4} = \frac{15}{4}
-36 - 16 = -52
\lim_{b \to -1} \left(2 b\right) = -2
-11 \cdot 43 = -473
\lim_{w \to 3} \left(w^2\right) = 9
\lim_{c \to 1} \left(c^2\right) = 1
\frac{d}{du}\left(u^4\right) = 4 u^3
\lim_{y \to -3} \left(2 y\right) = -6
f(t) = 3 + 5 t + t^2, \quad g(c) = 2 + c \implies f(g(c)) = 17 + 9 c + c^2
5 \left(1 + a\right) = 5 + 5 a
(2, 2, 5) \times (5, 3, 2) = (-11, 21, -4)
42 + 29 v - 12 v^2 + v^3 = 0 \implies v \in \{ -1, 6, 7 \}
v = 0 \implies v = 0
\log_{4}\left(2 + 4 a\right) = 3 \implies a = \frac{31}{2}
f(b) = 1 + 3 b + b^2, \quad g(x) = 2 + 2 x \implies f(g(x)) = 11 + 14 x + 4 x^2
\log_{5}\left(6 + 2 v\right) = 1 \implies v = \frac{-1}{2}
f(w) = 5 + 2 w + w^2, \quad g(x) = 4 + 3 x \implies f(g(x)) = 29 + 30 x + 9 x^2
8 + 6 a + a^2 = 0 \implies a \in \{ -4, -2 \}
-40 + 47 = 7
f(y) = 4 + y + y^2, \quad g(a) = 2 + 2 a \implies f(g(a)) = 10 + 10 a + 4 a^2
a^4 \left(10 - a\right) = 10 a^4 - a^5
\frac{d}{du}\left(\sin{\left(u\right)}\right) = \cos{\left(u\right)}
15 \cdot 19 = 285
-126 + 3 z + 10 z^2 + z^3 = 0 \implies z \in \{ -7, -6, 3 \}
\log_{5}\left(4 + 4 c\right) = 3 \implies c = \frac{121}{4}
\lim_{c \to -2} \left(c\right) = -2
\log_{5}\left(1 + 2 x\right) = 1 \implies x = 2
-56 - 55 u + 2 u^2 + u^3 = 0 \implies u \in \{ -8, -1, 7 \}
\frac{d}{dv}\left(\cos{\left(v\right)}\right) = -\sin{\left(v\right)}
\lim_{v \to -5} \left(-7 - v\right) = -2
\frac{d}{db}\left(2 b + \cos{\left(b\right)}\right) = 2 - \sin{\left(b\right)}
f(u) = 5 + 3 u + u^2, \quad g(y) = 2 + 3 y \implies f(g(y)) = 15 + 21 y + 9 y^2
\log_{3}\left(4 + 2 w\right) = 3 \implies w = \frac{23}{2}
f(y) = 2 + 3 y + y^2, \quad g(z) = 4 + 3 z \implies f(g(z)) = 30 + 33 z + 9 z^2
-30 - 29 = -59
f(u) = 2 + 4 u + u^2, \quad g(t) = 2 + 2 t \implies f(g(t)) = 14 + 16 t + 4 t^2