text
stringlengths
9
16.2k
\lim_{y \to 5} \left(-6 y\right) = -30
\lim_{b \to 5} \left(4 - b\right) = -1
50 \cdot 48 = 2400
f(x) = 4 + 2 x + x^2, \quad g(u) = 3 + 3 u \implies f(g(u)) = 19 + 24 u + 9 u^2
\frac{d}{dt}\left(t\right) = 1
\log_{5}\left(4 + t\right) = 3 \implies t = 121
\log_{2}\left(2 + a\right) = 2 \implies a = 2
f(c) = 5 + c + c^2, \quad g(u) = 2 + 2 u \implies f(g(u)) = 11 + 10 u + 4 u^2
a^4 \left(-4 + a\right) = -4 a^4 + a^5
7 - 25 = -18
18 \cdot 27 = 486
-6 + b = 0 \implies b = 6
\int \left(-\sin{\left(x\right)}\right) dx = -1 + \cos{\left(x\right)} + C
\int \left(1\right) dv = v + C
f(z) = 3 + 2 z + z^2, \quad g(x) = 2 + x \implies f(g(x)) = 11 + 6 x + x^2
29 \cdot 47 = 1363
49 - 43 = 6
1 + w = 0 \implies w = -1
\frac{d}{du}\left(-10 + u + \sin{\left(u\right)}\right) = 1 + \cos{\left(u\right)}
\frac{-43}{22} = \frac{-43}{22}
-37 - 48 = -85
\lim_{y \to -3} \left(2 y\right) = -6
\log_{3}\left(1 + 2 v\right) = 3 \implies v = 13
-32 + 20 = -12
\lim_{b \to 1} \left(4 + b\right) = 5
6 x + 5 x^2 + x^3 = 0 \implies x \in \{ -3, -2, 0 \}
-21 \cdot 40 = -840
\frac{-2}{12} = \frac{-1}{6}
\int \left(2 y + \cos{\left(y\right)}\right) dy = y^2 + \sin{\left(y\right)} + C
25 + 7 = 32
6 + y = 0 \implies y = -6
\log_{5}\left(4 + w\right) = 3 \implies w = 121
\frac{d}{dz}\left(10 z\right) = 10
15 + 14 = 29
\frac{-31}{1} = -31
\lim_{b \to 3} \left(b^4\right) = 81
24 + 10 y + y^2 = 0 \implies y \in \{ -6, -4 \}
-18 b - 3 b^2 + b^3 = 0 \implies b \in \{ -3, 0, 6 \}
\log_{4}\left(7 + 3 x\right) = 2 \implies x = 3
\frac{d}{dv}\left(\cos{\left(v\right)}\right) = -\sin{\left(v\right)}
\int \left(1 + 3 t^2\right) dt = t + t^3 + C
10 - 13 b + 2 b^2 + b^3 = 0 \implies b \in \{ -5, 1, 2 \}
\lim_{z \to -2} \left(z\right) = -2
\log_{5}\left(6 + 3 b\right) = 3 \implies b = \frac{119}{3}
-7 + b = 0 \implies b = 7
\cos{\left(t\right)} \left(6 + t\right) = t \cos{\left(t\right)} + 6 \cos{\left(t\right)}
-7 + t = 0 \implies t = 7
f(b) = 3 + 5 b + b^2, \quad g(w) = 2 + w \implies f(g(w)) = 17 + 9 w + w^2
120 - 49 u + u^3 = 0 \implies u \in \{ -8, 3, 5 \}
\log_{5}\left(7 + 3 v\right) = 2 \implies v = 6
-168 - 46 a + 3 a^2 + a^3 = 0 \implies a \in \{ -6, -4, 7 \}
\left(c + x\right) \left(3 + x\right) = 3 c + 3 x + c x + x^2
-33 + 37 = 4
f(z) = 2 + z + z^2, \quad g(u) = 4 + 3 u \implies f(g(u)) = 22 + 27 u + 9 u^2
\left(-u + w\right) \cos{\left(w\right)} = -u \cos{\left(w\right)} + w \cos{\left(w\right)}
3 - 4 y + y^2 = 0 \implies y \in \{ 1, 3 \}
\log_{4}\left(6 + 4 x\right) = 1 \implies x = \frac{-1}{2}
16 - 10 a + a^2 = 0 \implies a \in \{ 2, 8 \}
-46 + 17 = -29
4 + z = 0 \implies z = -4
\frac{d}{du}\left(u\right) = 1
f(u) = 1 + 3 u + u^2, \quad g(w) = 1 + 2 w \implies f(g(w)) = 5 + 10 w + 4 w^2
\int \left(-\sin{\left(x\right)}\right) dx = -1 + \cos{\left(x\right)} + C
\int \left(-1 - \sin{\left(t\right)}\right) dt = -1 - t + \cos{\left(t\right)} + C
\frac{d}{dt}\left(t^4 - \sin{\left(t\right)}\right) = 4 t^3 - \cos{\left(t\right)}
\log_{5}\left(4 + 2 w\right) = 3 \implies w = \frac{121}{2}
f(u) = 3 + u + u^2, \quad g(y) = 4 + 2 y \implies f(g(y)) = 23 + 18 y + 4 y^2
-14 + 5 c + c^2 = 0 \implies c \in \{ -7, 2 \}
\log_{2}\left(5 + 4 x\right) = 2 \implies x = \frac{-1}{4}
\frac{d}{db}\left(b - \left(3 + b\right)\right) = 0
\log_{5}\left(4 + 3 x\right) = 1 \implies x = \frac{1}{3}
\log_{3}\left(7 + 2 u\right) = 3 \implies u = 10
\int \left(-\sin{\left(b\right)}\right) db = -1 + \cos{\left(b\right)} + C
\log_{5}\left(1 + 2 a\right) = 3 \implies a = 62
14 - 19 a + 4 a^2 + a^3 = 0 \implies a \in \{ -7, 1, 2 \}
\left(u + v\right) \left(9 + v\right) = 9 u + 9 v + u v + v^2
-11 + 26 = 15
\frac{30}{30} = 1
\sin{\left(a\right)} \left(-3 + a\right) = a \sin{\left(a\right)} - 3 \sin{\left(a\right)}
-7 + z = 0 \implies z = 7
36 - 7 v^2 + v^3 = 0 \implies v \in \{ -2, 3, 6 \}
\lim_{u \to -5} \left(1 + u\right) = -4
-41 - 21 = -62
45 - 8 = 37
-15 - 2 w + w^2 = 0 \implies w \in \{ -3, 5 \}
0 \cdot 30 = 0
5 + t = 0 \implies t = -5
2 + z = 0 \implies z = -2
\int \left(-10 - 2 u\right) du = -10 u - u^2 + C
16 - 10 x + x^2 = 0 \implies x \in \{ 2, 8 \}
\log_{3}\left(8 + 2 b\right) = 1 \implies b = \frac{-5}{2}
\log_{5}\left(2 + u\right) = 2 \implies u = 23
\left(4 + c\right) \left(-10 + t\right) = -40 - 10 c + 4 t + c t
-44 - 33 = -77
-48 - 2 t + t^2 = 0 \implies t \in \{ -6, 8 \}
c^2 \left(-10 + u\right) = c^2 u - 10 c^2
(-1, 4, 0) \times (-2, 0, -2) = (-8, -2, 8)
-5 + a = 0 \implies a = 5
\frac{-3}{3} = -1
f(u) = 4 + 4 u + u^2, \quad g(w) = 3 + 2 w \implies f(g(w)) = 25 + 20 w + 4 w^2