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f(a) = 3 + 4 a + a^2, \quad g(t) = 4 + 2 t \implies f(g(t)) = 35 + 24 t + 4 t^2
f(c) = 5 + 3 c + c^2, \quad g(u) = 4 + 2 u \implies f(g(u)) = 33 + 22 u + 4 u^2
f(a) = 5 + 3 a + a^2, \quad g(z) = 3 + z \implies f(g(z)) = 23 + 9 z + z^2
41 - 8 = 33
(-2, 5, -3) \times (-2, -1, 1) = (2, 8, 12)
f(c) = 3 + 3 c + c^2, \quad g(v) = 4 + v \implies f(g(v)) = 31 + 11 v + v^2
\int \left(-\sin{\left(y\right)}\right) dy = -1 + \cos{\left(y\right)} + C
z^4 \left(x - z\right) = x z^4 - z^5
30 - 31 c + c^3 = 0 \implies c \in \{ -6, 1, 5 \}
15 \cdot 49 = 735
-30 - 29 x + 2 x^2 + x^3 = 0 \implies x \in \{ -6, -1, 5 \}
\int \left(-12 y\right) dy = -6 y^2 + C
1 + b = 0 \implies b = -1
-2 \cdot 27 = -54
2 + z = 0 \implies z = -2
f(t) = 4 + 2 t + t^2, \quad g(z) = 2 + 2 z \implies f(g(z)) = 12 + 12 z + 4 z^2
\frac{d}{da}\left(7 + a - \left(-4 + a\right)\right) = 0
-47 + 4 = -43
f(x) = 5 + x + x^2, \quad g(c) = 4 + c \implies f(g(c)) = 25 + 9 c + c^2
\log_{5}\left(3 + 2 b\right) = 1 \implies b = 1
\frac{-8}{15} = \frac{-8}{15}
\lim_{w \to -5} \left(3 + w\right) = -2
\lim_{x \to 5} \left(\sin{\left(x\right)}\right) = \sin{\left(5\right)}
f(x) = 4 + 2 x + x^2, \quad g(a) = 2 + 2 a \implies f(g(a)) = 12 + 12 a + 4 a^2
\log_{4}\left(5 + 2 a\right) = 1 \implies a = \frac{-1}{2}
\lim_{v \to -2} \left(-2\right) = -2
-9 \cdot 43 = -387
(4, 5, -5) \times (-3, 0, -3) = (-15, 27, 15)
-4 + c = 0 \implies c = 4
\log_{2}\left(7 + c\right) = 1 \implies c = -5
f(u) = 3 + 3 u + u^2, \quad g(x) = 1 + 3 x \implies f(g(x)) = 7 + 15 x + 9 x^2
\frac{d}{dw}\left(w^2\right) = 2 w
\frac{d}{dx}\left(x\right) = 1
\int \left(-\sin{\left(t\right)}\right) dt = -1 + \cos{\left(t\right)} + C
-21 - 4 a + a^2 = 0 \implies a \in \{ -3, 7 \}
\int \left(-z \sin{\left(z\right)} + \cos{\left(z\right)}\right) dz = z \cos{\left(z\right)} + C
f(y) = 2 + 3 y + y^2, \quad g(w) = 3 + 2 w \implies f(g(w)) = 20 + 18 w + 4 w^2
\lim_{x \to 5} \left(x^2\right) = 25
\int \left(-6 b\right) db = -3 b^2 + C
8 - 6 y + y^2 = 0 \implies y \in \{ 2, 4 \}
\log_{4}\left(4 + 3 w\right) = 3 \implies w = 20
f(a) = 5 + 2 a + a^2, \quad g(w) = 3 + 3 w \implies f(g(w)) = 20 + 24 w + 9 w^2
\log_{4}\left(5 + y\right) = 2 \implies y = 11
2 + z = 0 \implies z = -2
\frac{d}{db}\left(b - b^4\right) = 1 - 4 b^3
a \left(-10 + w\right) = -10 a + a w
(-4, 5, -3) \times (4, -4, 3) = (3, 0, -4)
\log_{4}\left(4 + v\right) = 1 \implies v = 0
\int \left(2 y\right) dy = y^2 + C
(0, -2, 2) \times (5, 0, 2) = (-4, 10, 10)
-28 + 39 t - 12 t^2 + t^3 = 0 \implies t \in \{ 1, 4, 7 \}
\log_{4}\left(6 + 2 w\right) = 3 \implies w = 29
-38 + 7 = -31
49 - 49 v - v^2 + v^3 = 0 \implies v \in \{ -7, 1, 7 \}
b = 0 \implies b = 0
44 - 33 = 11
4 - 5 z + z^2 = 0 \implies z \in \{ 1, 4 \}
-18 - 15 = -33
-42 - 29 x - 2 x^2 + x^3 = 0 \implies x \in \{ -3, -2, 7 \}
\log_{2}\left(8 + 4 c\right) = 3 \implies c = 0
f(z) = 5 + 3 z + z^2, \quad g(y) = 4 + y \implies f(g(y)) = 33 + 11 y + y^2
\frac{48}{15} = \frac{16}{5}
-3 + u = 0 \implies u = 3
\log_{3}\left(7 + 4 v\right) = 1 \implies v = -1
(-2, -5, 0) \times (-4, 5, 0) = (0, 0, -30)
\log_{2}\left(2 + 4 u\right) = 1 \implies u = 0
f(v) = 5 + 4 v + v^2, \quad g(x) = 1 + 2 x \implies f(g(x)) = 10 + 12 x + 4 x^2
3 - 28 = -25
\log_{5}\left(6 + 3 w\right) = 1 \implies w = \frac{-1}{3}
\lim_{b \to -2} \left(\cos{\left(b\right)}\right) = \cos{\left(2\right)}
\int \left(2 y\right) dy = y^2 + C
\log_{4}\left(6 + 4 z\right) = 3 \implies z = \frac{29}{2}
f(b) = 5 + 2 b + b^2, \quad g(u) = 1 + 2 u \implies f(g(u)) = 8 + 8 u + 4 u^2
\log_{4}\left(1 + x\right) = 3 \implies x = 63
(-4, 2, 4) \times (-2, -4, -3) = (10, -20, 20)
3 \cdot 31 = 93
f(y) = 4 + 2 y + y^2, \quad g(u) = 1 + 3 u \implies f(g(u)) = 7 + 12 u + 9 u^2
f(z) = 3 + z + z^2, \quad g(x) = 1 + x \implies f(g(x)) = 5 + 3 x + x^2
\frac{-8}{1} = -8
7 - 8 t + t^2 = 0 \implies t \in \{ 1, 7 \}
\log_{2}\left(5 + z\right) = 3 \implies z = 3
\int \left(4 u^3\right) du = u^4 + C
\log_{5}\left(8 + z\right) = 3 \implies z = 117
\frac{d}{dc}\left(-c^2\right) = -2 c
(2, -5, 0) \times (-5, 3, 1) = (-5, -2, -19)
16 + 28 = 44
7 + c = 0 \implies c = -7
(-2, 4, 1) \times (0, 4, 4) = (12, 8, -8)
\left(6 - y\right) \left(-2 + y\right) = -12 + 8 y - y^2
\log_{3}\left(6 + c\right) = 2 \implies c = 3
\frac{d}{da}\left(a\right) = 1
-7 + x = 0 \implies x = 7
\frac{d}{dy}\left(-\left(1 + y\right) + \sin{\left(y\right)}\right) = -1 + \cos{\left(y\right)}
f(u) = 3 + 3 u + u^2, \quad g(z) = 3 + z \implies f(g(z)) = 21 + 9 z + z^2
f(y) = 1 + 2 y + y^2, \quad g(t) = 4 + t \implies f(g(t)) = 25 + 10 t + t^2
-49 + u^2 = 0 \implies u \in \{ -7, 7 \}
-13 + 3 = -10
f(v) = 2 + v + v^2, \quad g(z) = 2 + z \implies f(g(z)) = 8 + 5 z + z^2
\log_{4}\left(7 + 3 t\right) = 2 \implies t = 3
\frac{42}{1} = 42