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\frac{-45}{41} = \frac{-45}{41}
-21 - 16 = -37
f(x) = 1 + 5 x + x^2, \quad g(z) = 1 + 3 z \implies f(g(z)) = 7 + 21 z + 9 z^2
f(c) = 3 + 5 c + c^2, \quad g(y) = 4 + 2 y \implies f(g(y)) = 39 + 26 y + 4 y^2
\log_{2}\left(7 + 3 x\right) = 3 \implies x = \frac{1}{3}
\cos{\left(c\right)} \left(6 - c\right) = -c \cos{\left(c\right)} + 6 \cos{\left(c\right)}
f(x) = 5 + 4 x + x^2, \quad g(t) = 4 + t \implies f(g(t)) = 37 + 12 t + t^2
12 - 22 = -10
\frac{2}{47} = \frac{2}{47}
-280 - 61 c + 4 c^2 + c^3 = 0 \implies c \in \{ -7, -5, 8 \}
7 \left(-2 + c\right) = -14 + 7 c
y = 0 \implies y = 0
\int \left(2 b \cos{\left(b\right)} + 2 \sin{\left(b\right)}\right) db = 2 b \sin{\left(b\right)} + C
\log_{4}\left(4 + 3 c\right) = 3 \implies c = 20
-32 + 44 v - 13 v^2 + v^3 = 0 \implies v \in \{ 1, 4, 8 \}
\lim_{x \to 4} \left(-2 - x\right) = -6
48 + 46 b + 13 b^2 + b^3 = 0 \implies b \in \{ -8, -3, -2 \}
-46 + 47 = 1
\frac{d}{dv}\left(v^2\right) = 2 v
f(c) = 3 + 2 c + c^2, \quad g(v) = 4 + 3 v \implies f(g(v)) = 27 + 30 v + 9 v^2
-4 + w = 0 \implies w = 4
\int \left(2 - \sin{\left(u\right)}\right) du = -1 + 2 u + \cos{\left(u\right)} + C
-31 - 8 = -39
\log_{2}\left(7 + 4 u\right) = 3 \implies u = \frac{1}{4}
-5 + t = 0 \implies t = 5
-5 + u = 0 \implies u = 5
\int \left(-\sin{\left(x\right)}\right) dx = -1 + \cos{\left(x\right)} + C
-20 - 21 = -41
-8 + v = 0 \implies v = 8
(-1, -1, -5) \times (-3, -2, -3) = (-7, 12, -1)
\log_{5}\left(5 + 3 u\right) = 1 \implies u = 0
f(a) = 5 + 4 a + a^2, \quad g(x) = 3 + 2 x \implies f(g(x)) = 26 + 20 x + 4 x^2
\log_{5}\left(3 + b\right) = 3 \implies b = 122
5 + c = 0 \implies c = -5
-7 - 26 = -33
\int \left(\cos{\left(u\right)}\right) du = \sin{\left(u\right)} + C
\frac{d}{da}\left(-10 - a^2\right) = -2 a
27 + 8 = 35
-\left(1 - b\right) = -1 + b
f(u) = 4 + 2 u + u^2, \quad g(c) = 3 + 3 c \implies f(g(c)) = 19 + 24 c + 9 c^2
-13 \cdot 29 = -377
\log_{3}\left(3 + 2 x\right) = 3 \implies x = 12
\frac{d}{db}\left(\cos{\left(b\right)}\right) = -\sin{\left(b\right)}
15 - 49 = -34
\frac{9}{19} = \frac{9}{19}
\frac{d}{dx}\left(x^2\right) = 2 x
8 t + t^2 = 0 \implies t \in \{ -8, 0 \}
\lim_{z \to 5} \left(z^2\right) = 25
\frac{40}{34} = \frac{20}{17}
\log_{5}\left(1 + 3 t\right) = 1 \implies t = \frac{4}{3}
6 + w = 0 \implies w = -6
(-3, -1, -4) \times (1, 0, 5) = (-5, 11, 1)
54 \left(b + u\right) = 54 b + 54 u
-41 \cdot 2 = -82
-7 + y = 0 \implies y = 7
(1, -1, -5) \times (-4, 1, 5) = (0, 15, -3)
-32 - 4 = -36
15 - 8 u + u^2 = 0 \implies u \in \{ 3, 5 \}
144 - 54 b - b^2 + b^3 = 0 \implies b \in \{ -8, 3, 6 \}
\lim_{z \to 0} \left(z^3\right) = 0
-24 - 32 = -56
20 + 9 u + u^2 = 0 \implies u \in \{ -5, -4 \}
\frac{-16}{35} = \frac{-16}{35}
\log_{2}\left(6 + 2 v\right) = 2 \implies v = -1
f(a) = 4 + 2 a + a^2, \quad g(b) = 4 + 3 b \implies f(g(b)) = 28 + 30 b + 9 b^2
-8 + x = 0 \implies x = 8
20 - 9 a + a^2 = 0 \implies a \in \{ 4, 5 \}
f(c) = 2 + 4 c + c^2, \quad g(a) = 2 + a \implies f(g(a)) = 14 + 8 a + a^2
\lim_{u \to -1} \left(u\right) = -1
\frac{d}{da}\left(0\right) = 0
42 + 55 u + 14 u^2 + u^3 = 0 \implies u \in \{ -7, -6, -1 \}
\int \left(-\sin{\left(c\right)}\right) dc = -1 + \cos{\left(c\right)} + C
f(x) = 4 + x + x^2, \quad g(z) = 3 + 3 z \implies f(g(z)) = 16 + 21 z + 9 z^2
21 + 11 = 32
\log_{4}\left(8 + 4 c\right) = 2 \implies c = 2
(0, 4, -2) \times (-3, -3, 0) = (-6, 6, 12)
\int \left(1\right) da = a + C
\log_{4}\left(2 + y\right) = 3 \implies y = 62
\frac{d}{dy}\left(8 y + y^2\right) = 8 + 2 y
3 + b = 0 \implies b = -3
\frac{d}{dt}\left(t^3 \cos{\left(t\right)}\right) = 3 t^2 \cos{\left(t\right)} - t^3 \sin{\left(t\right)}
7 + u = 0 \implies u = -7
(2, -5, 2) \times (4, 3, -2) = (4, 12, 26)
-6 v + v^2 = 0 \implies v \in \{ 0, 6 \}
-21 - 25 y - 3 y^2 + y^3 = 0 \implies y \in \{ -3, -1, 7 \}
\log_{5}\left(1 + 4 z\right) = 1 \implies z = 1
\lim_{y \to -4} \left(y^3\right) = -64
16 + 10 v + v^2 = 0 \implies v \in \{ -8, -2 \}
4 \cdot 21 = 84
\lim_{x \to 2} \left(x^2\right) = 4
\int \left(1\right) dv = v + C
8 + u = 0 \implies u = -8
\int \left(4 a^3\right) da = a^4 + C
(5, -1, 4) \times (-2, -2, 2) = (6, -18, -12)
\log_{5}\left(5 + 4 w\right) = 1 \implies w = 0
40 + 38 t + 11 t^2 + t^3 = 0 \implies t \in \{ -5, -4, -2 \}
\frac{d}{du}\left(\sin{\left(u\right)}\right) = \cos{\left(u\right)}
\lim_{t \to -1} \left(0\right) = 0
\lim_{v \to -2} \left(-9 + v\right) = -11
-3 - 2 c + c^2 = 0 \implies c \in \{ -1, 3 \}