text
stringlengths
9
16.2k
-18 + 40 = 22
\log_{3}\left(7 + 2 y\right) = 1 \implies y = -2
f(c) = 5 + 4 c + c^2, \quad g(z) = 1 + 3 z \implies f(g(z)) = 10 + 18 z + 9 z^2
(-5, -1, -1) \times (1, 3, -1) = (4, -6, -14)
\frac{-23}{23} = -1
f(x) = 3 + 3 x + x^2, \quad g(t) = 3 + 3 t \implies f(g(t)) = 21 + 27 t + 9 t^2
-3 + u = 0 \implies u = 3
(-1, -2, 2) \times (1, -2, -1) = (6, 1, 4)
6 - 5 z + z^2 = 0 \implies z \in \{ 2, 3 \}
18 - 9 x - 2 x^2 + x^3 = 0 \implies x \in \{ -3, 2, 3 \}
-3 + 26 = 23
\lim_{u \to 1} \left(2 u\right) = 2
49 \cdot 25 = 1225
(-1, -1, 0) \times (3, 3, -4) = (4, -4, 0)
\log_{4}\left(8 + 2 y\right) = 3 \implies y = 28
f(x) = 3 + 5 x + x^2, \quad g(v) = 2 + 2 v \implies f(g(v)) = 17 + 18 v + 4 v^2
(-3, 4, -3) \times (-5, -1, -2) = (-11, 9, 23)
\lim_{c \to 4} \left(c\right) = 4
21 + 42 = 63
35 + 41 = 76
7 + y = 0 \implies y = -7
\frac{37}{47} = \frac{37}{47}
-48 + 2 x + x^2 = 0 \implies x \in \{ -8, 6 \}
-6 + 11 = 5
\log_{4}\left(5 + u\right) = 1 \implies u = -1
-46 \cdot 32 = -1472
6 + 7 w + w^2 = 0 \implies w \in \{ -6, -1 \}
-40 - 37 z + 4 z^2 + z^3 = 0 \implies z \in \{ -8, -1, 5 \}
\lim_{z \to -1} \left(2 z\right) = -2
18 + 9 b + b^2 = 0 \implies b \in \{ -6, -3 \}
f(c) = 4 + 2 c + c^2, \quad g(x) = 4 + 2 x \implies f(g(x)) = 28 + 20 x + 4 x^2
f(t) = 3 + t + t^2, \quad g(x) = 3 + 2 x \implies f(g(x)) = 15 + 14 x + 4 x^2
16 + 6 v - 9 v^2 + v^3 = 0 \implies v \in \{ -1, 2, 8 \}
8 + 9 c + c^2 = 0 \implies c \in \{ -8, -1 \}
f(x) = 2 + 3 x + x^2, \quad g(c) = 3 + c \implies f(g(c)) = 20 + 9 c + c^2
-28 - 3 t + t^2 = 0 \implies t \in \{ -4, 7 \}
-5 + 36 = 31
\log_{2}\left(4 + 3 t\right) = 1 \implies t = \frac{-2}{3}
f(y) = 1 + 3 y + y^2, \quad g(v) = 2 + v \implies f(g(v)) = 11 + 7 v + v^2
-18 - 3 a + a^2 = 0 \implies a \in \{ -3, 6 \}
5 + u = 0 \implies u = -5
6 - 19 = -13
\log_{5}\left(5 + x\right) = 2 \implies x = 20
\log_{4}\left(2 + 2 c\right) = 2 \implies c = 7
\int \left(2\right) dz = 2 z + C
f(v) = 3 + v + v^2, \quad g(u) = 1 + 2 u \implies f(g(u)) = 5 + 6 u + 4 u^2
f(x) = 4 + x + x^2, \quad g(t) = 1 + t \implies f(g(t)) = 6 + 3 t + t^2
(-1, 1, -4) \times (-3, 0, 1) = (1, 13, 3)
-41 \cdot 29 = -1189
40 + 13 x + x^2 = 0 \implies x \in \{ -8, -5 \}
12 + 5 a - 6 a^2 + a^3 = 0 \implies a \in \{ -1, 3, 4 \}
\lim_{x \to 4} \left(x^3\right) = 64
\frac{-20}{12} = \frac{-5}{3}
4 + 5 a + a^2 = 0 \implies a \in \{ -4, -1 \}
\frac{d}{da}\left(a^4\right) = 4 a^3
-4 v + v^2 = 0 \implies v \in \{ 0, 4 \}
-30 - y + y^2 = 0 \implies y \in \{ -5, 6 \}
7 + v = 0 \implies v = -7
-13 - 28 = -41
-32 + 4 u + u^2 = 0 \implies u \in \{ -8, 4 \}
\lim_{x \to -5} \left(x^2\right) = 25
12 - 8 t + t^2 = 0 \implies t \in \{ 2, 6 \}
\frac{d}{db}\left(-8 + b^2\right) = 2 b
f(c) = 4 + 3 c + c^2, \quad g(y) = 4 + y \implies f(g(y)) = 32 + 11 y + y^2
2 \left(-a + y\right) a = 2 a y - 2 a^2
-16 - 22 v - 5 v^2 + v^3 = 0 \implies v \in \{ -2, -1, 8 \}
6 + c = 0 \implies c = -6
\lim_{u \to -5} \left(2 u\right) = -10
\int \left(\cos{\left(u\right)}\right) du = \sin{\left(u\right)} + C
f(y) = 2 + 4 y + y^2, \quad g(b) = 4 + b \implies f(g(b)) = 34 + 12 b + b^2
\frac{d}{da}\left(a^3\right) = 3 a^2
f(a) = 2 + 5 a + a^2, \quad g(c) = 1 + c \implies f(g(c)) = 8 + 7 c + c^2
\frac{d}{dt}\left(-28 + t\right) = 1
\log_{5}\left(3 + b\right) = 2 \implies b = 22
\frac{16}{6} = \frac{8}{3}
\log_{4}\left(6 + 3 w\right) = 3 \implies w = \frac{58}{3}
\frac{-33}{40} = \frac{-33}{40}
-21 t + 4 t^2 + t^3 = 0 \implies t \in \{ -7, 0, 3 \}
\lim_{a \to 3} \left(a\right) = 3
f(a) = 1 + 5 a + a^2, \quad g(v) = 4 + 3 v \implies f(g(v)) = 37 + 39 v + 9 v^2
(5, 0, 2) \times (-5, 1, 1) = (-2, -15, 5)
-27 \cdot 13 = -351
\lim_{a \to -4} \left(0\right) = 0
-50 - 25 w + 2 w^2 + w^3 = 0 \implies w \in \{ -5, -2, 5 \}
(3, -4, 5) \times (0, 2, 0) = (-10, 0, 6)
(2, 3, 3) \times (-3, -2, -2) = (0, -5, 5)
-6 t + t^2 = 0 \implies t \in \{ 0, 6 \}
-19 - 23 = -42
\lim_{w \to 5} \left(w^4\right) = 625
(4, 1, 2) \times (2, -5, 0) = (10, 4, -22)
\left(u + v\right) \left(9 + u\right) = 9 u + 9 v + u v + u^2
-56 a - a^2 + a^3 = 0 \implies a \in \{ -7, 0, 8 \}
\log_{3}\left(8 + c\right) = 1 \implies c = -5
\log_{2}\left(6 + 2 v\right) = 1 \implies v = -2
\int \left(4 v^3\right) dv = v^4 + C
\frac{d}{dz}\left(2 z - \left(1 + z\right)\right) = 1
f(a) = 2 + a + a^2, \quad g(b) = 1 + b \implies f(g(b)) = 4 + 3 b + b^2
(2, -5, 3) \times (-4, -1, -5) = (28, -2, -22)
-5 z \left(-5 + z\right) = 25 z - 5 z^2
\int \left(4 x^3\right) dx = x^4 + C