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210 - 47 b - 4 b^2 + b^3 = 0 \implies b \in \{ -7, 5, 6 \}
1 + u = 0 \implies u = -1
\lim_{w \to -3} \left(1 + w\right) = -2
6 + 5 z + z^2 = 0 \implies z \in \{ -3, -2 \}
(0, -2, 1) \times (3, 1, 2) = (-5, 3, 6)
\log_{2}\left(7 + a\right) = 1 \implies a = -5
\log_{2}\left(4 + 2 u\right) = 1 \implies u = -1
\log_{2}\left(5 + 3 a\right) = 2 \implies a = \frac{-1}{3}
\frac{8}{27} = \frac{8}{27}
-2 - 36 = -38
f(a) = 4 + 4 a + a^2, \quad g(c) = 2 + 3 c \implies f(g(c)) = 16 + 24 c + 9 c^2
28 - 11 z + z^2 = 0 \implies z \in \{ 4, 7 \}
\int \left(\cos{\left(x\right)}\right) dx = \sin{\left(x\right)} + C
f(y) = 5 + 3 y + y^2, \quad g(u) = 2 + 2 u \implies f(g(u)) = 15 + 14 u + 4 u^2
\int \left(4 w^3\right) dw = w^4 + C
5 + z = 0 \implies z = -5
-2 w + w^2 = 0 \implies w \in \{ 0, 2 \}
u \left(3 + c\right) = 3 u + c u
22 + 16 = 38
(-3, -2, 4) \times (-3, 4, 5) = (-26, 3, -18)
-16 + u^2 = 0 \implies u \in \{ -4, 4 \}
\log_{5}\left(8 + 4 w\right) = 3 \implies w = \frac{117}{4}
14 + 9 w + w^2 = 0 \implies w \in \{ -7, -2 \}
\lim_{x \to 4} \left(-10 + x\right) = -6
-32 - 45 = -77
-14 + 5 t + 8 t^2 + t^3 = 0 \implies t \in \{ -7, -2, 1 \}
4 + 10 = 14
-5 + v = 0 \implies v = 5
48 - 28 u + u^3 = 0 \implies u \in \{ -6, 2, 4 \}
\log_{5}\left(7 + 2 v\right) = 2 \implies v = 9
\frac{d}{du}\left(3 \cos{\left(u\right)}\right) = -3 \sin{\left(u\right)}
f(b) = 3 + 4 b + b^2, \quad g(t) = 2 + 3 t \implies f(g(t)) = 15 + 24 t + 9 t^2
-2 + b = 0 \implies b = 2
\frac{d}{dy}\left(y^2\right) = 2 y
2 + 3 t + t^2 = 0 \implies t \in \{ -2, -1 \}
40 + 13 t + t^2 = 0 \implies t \in \{ -8, -5 \}
\frac{-28}{33} = \frac{-28}{33}
f(z) = 2 + 4 z + z^2, \quad g(x) = 3 + 3 x \implies f(g(x)) = 23 + 30 x + 9 x^2
7 - 8 y + y^2 = 0 \implies y \in \{ 1, 7 \}
\lim_{x \to -2} \left(-7 x\right) = 14
\int \left(\cos{\left(v\right)}\right) dv = \sin{\left(v\right)} + C
\log_{5}\left(7 + 3 b\right) = 1 \implies b = \frac{-2}{3}
\int \left(2 v\right) dv = v^2 + C
\log_{3}\left(2 + 4 z\right) = 2 \implies z = \frac{7}{4}
-5 + t = 0 \implies t = 5
f(t) = 5 + 2 t + t^2, \quad g(z) = 3 + 3 z \implies f(g(z)) = 20 + 24 z + 9 z^2
\int \left(1 - 4 b^3\right) db = b - b^4 + C
f(a) = 4 + a + a^2, \quad g(t) = 2 + 2 t \implies f(g(t)) = 10 + 10 t + 4 t^2
f(w) = 2 + 5 w + w^2, \quad g(t) = 3 + t \implies f(g(t)) = 26 + 11 t + t^2
18 - 9 a - 2 a^2 + a^3 = 0 \implies a \in \{ -3, 2, 3 \}
24 + 11 w + w^2 = 0 \implies w \in \{ -8, -3 \}
-15 + 2 b + b^2 = 0 \implies b \in \{ -5, 3 \}
4 + v = 0 \implies v = -4
-20 - 6 = -26
-35 + 10 = -25
-210 - 23 w + 8 w^2 + w^3 = 0 \implies w \in \{ -7, -6, 5 \}
\lim_{z \to -3} \left(z^2\right) = 9
392 - 49 x - 8 x^2 + x^3 = 0 \implies x \in \{ -7, 7, 8 \}
f(y) = 4 + y + y^2, \quad g(z) = 2 + z \implies f(g(z)) = 10 + 5 z + z^2
(-2, 0, 0) \times (-2, -3, 2) = (0, 4, 6)
\lim_{t \to 0} \left(0\right) = 0
45 - 47 = -2
5 - 6 = -1
\int \left(-1 + 2 w\right) dw = -3 - \left(-3 + w\right) + w^2 + C
\frac{d}{dz}\left(2 z \sin{\left(z\right)}\right) = 2 z \cos{\left(z\right)} + 2 \sin{\left(z\right)}
-6 - 4 = -10
10 - 7 y + y^2 = 0 \implies y \in \{ 2, 5 \}
\log_{4}\left(8 + 2 c\right) = 1 \implies c = -2
-40 - 3 z + z^2 = 0 \implies z \in \{ -5, 8 \}
\log_{4}\left(8 + a\right) = 3 \implies a = 56
\log_{2}\left(3 + 2 w\right) = 1 \implies w = \frac{-1}{2}
-224 - 4 u + 11 u^2 + u^3 = 0 \implies u \in \{ -8, -7, 4 \}
\lim_{y \to 4} \left(y^2\right) = 16
-50 - 33 = -83
f(u) = 4 + 5 u + u^2, \quad g(w) = 4 + 2 w \implies f(g(w)) = 40 + 26 w + 4 w^2
42 - 23 = 19
\log_{4}\left(6 + 3 y\right) = 2 \implies y = \frac{10}{3}
-33 + 30 = -3
\log_{3}\left(7 + c\right) = 2 \implies c = 2
\frac{d}{dc}\left(10 c^4\right) = 40 c^3
-16 \cdot 1 = -16
25 - 6 = 19
\log_{3}\left(6 + 4 u\right) = 2 \implies u = \frac{3}{4}
-2 + a = 0 \implies a = 2
20 - 9 w + w^2 = 0 \implies w \in \{ 4, 5 \}
4 \cdot 30 = 120
-98 - 49 a + 2 a^2 + a^3 = 0 \implies a \in \{ -7, -2, 7 \}
-13 \cdot 21 = -273
\log_{3}\left(1 + 2 x\right) = 3 \implies x = 13
c \left(-8 - z\right) = -8 c - c z
50 \cdot 18 = 900
\log_{4}\left(4 + 4 t\right) = 2 \implies t = 3
\frac{d}{dc}\left(\cos{\left(c\right)}\right) = -\sin{\left(c\right)}
\lim_{b \to 2} \left(0\right) = 0
20 - 37 = -17
\log_{2}\left(1 + 4 u\right) = 3 \implies u = \frac{7}{4}
\lim_{x \to -2} \left(2 x\right) = -4
\frac{d}{dv}\left(2 v \left(5 - v\right)\right) = -2 v + 2 \left(5 - v\right)
2 + z = 0 \implies z = -2
-3 + 2 u + u^2 = 0 \implies u \in \{ -3, 1 \}