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The total area of a cylinder consists of the lateral area plus two circular bases: $A_T = 2\pi r h + 2\pi r^2 = 2\pi r (h + r)$. Substituting $r = 3$ and $h = 7$: $A_T = 2\pi (3)(7 + 3) = 6\pi (10) = 60\pi\text{ cm}^2$.
The lateral area of a cone is given by $A_L = \pi r g$, where $g$ is the slant height (generatriz). Using the Pythagorean theorem: $g = \sqrt{r^2 + h^2} = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10\text{ cm}$. Thus, $A_L = \pi (6)(10) = 60\pi\text{ cm}^2$.
The volume of a sphere is given by $V = \frac{4}{3}\pi R^3$. Substituting $R = 3$: $V = \frac{4}{3}\pi (3^3) = \frac{4}{3}\pi (27) = 4 \cdot 9\pi = 36\pi\text{ cm}^3$.
The volume of any pyramid is $V = \frac{1}{3} A_b \cdot h$. For a regular square base, the base area is $A_b = a^2 = 6^2 = 36\text{ cm}^2$. Thus, $V = \frac{1}{3} (36)(10) = 12 \cdot 10 = 120\text{ cm}^3$.
Using $\cos(\alpha + \beta) = \cos\alpha \cos\beta - \sin\alpha \sin\beta$ with $\alpha = 45^\circ$ and $\beta = 30^\circ$: $\cos(75^\circ) = \cos 45^\circ \cos 30^\circ - \sin 45^\circ \sin 30^\circ = \left(\frac{\sqrt{2}}{2}\right)\left(\frac{\sqrt{3}}{2}\right) - \left(\frac{\sqrt{2}}{2}\right)\left(\frac{1}{2}\right) = \frac{\sqrt{6} - \sqrt{2}}{4}$.
The double-angle formula is $\sin(2\theta) = 2\sin\theta \cos\theta$. In Quadrant II, sine is positive and cosine is negative: $\cos\theta = -\sqrt{1 - \sin^2\theta} = -\sqrt{1 - 9/25} = -\sqrt{16/25} = -\frac{4}{5}$. Thus, $\sin(2\theta) = 2 \left(\frac{3}{5}\right) \left(-\frac{4}{5}\right) = -\frac{24}{25}$.
By the Law of Cosines: $c^2 = a^2 + b^2 - 2ab \cos C$. Substituting the given values: $c^2 = 5^2 + 7^2 - 2(5)(7)\cos 60^\circ = 25 + 49 - 70(1/2) = 74 - 35 = 39$. Taking the positive square root gives $c = \sqrt{39} \approx 6.24$.
The space diagonal of a rectangular parallelepiped is found by applying the Pythagorean theorem twice: $d = \sqrt{a^2 + b^2 + c^2}$. Substituting the edges: $d = \sqrt{3^2 + 4^2 + 12^2} = \sqrt{9 + 16 + 144} = \sqrt{169} = 13\text{ cm}$.
A regular tetrahedron consists of 4 congruent equilateral triangles. The area of an equilateral triangle with side $a$ is $A_1 = \frac{\sqrt{3}}{4} a^2$. Thus, the total area is $A_T = 4 A_1 = 4 \left(\frac{\sqrt{3}}{4} a^2\right) = a^2 \sqrt{3}$. For $a = 4\text{ cm}$: $A_T = 4^2 \sqrt{3} = 16\sqrt{3}\text{ cm}^2$.
A regular hexagon of side $a$ consists of 6 equilateral triangles, so its base area is $A_b = 6 \left(\frac{\sqrt{3}}{4} a^2\right) = \frac{3\sqrt{3}}{2} a^2$. For $a = 2$: $A_b = \frac{3\sqrt{3}}{2} (4) = 6\sqrt{3}\text{ cm}^2$. The volume of a prism is $V = A_b \cdot h = (6\sqrt{3})(5) = 30\sqrt{3}\text{ cm}^3$.
By fundamental spatial geometry definitions, two lines that do not intersect and are not parallel cannot lie in the same plane; they are skew lines (rectas que se cruzan o alabeadas).
The Theorem of the Three Perpendiculars states: A straight line lying in a plane is perpendicular to an oblique line intersecting that plane if and only if it is perpendicular to the orthogonal projection of the oblique line onto that plane.
The fundamental double-angle identity is $\cos(2x) = \cos^2 x - \sin^2 x$. Substituting $\cos^2 x = 1 - \sin^2 x$ gives $\cos(2x) = (1 - \sin^2 x) - \sin^2 x = 1 - 2\sin^2 x$. (Similarly, using $\sin^2 x = 1 - \cos^2 x$ yields $2\cos^2 x - 1$).
By the Law of Sines: $\frac{a}{\sin A} = \frac{b}{\sin B}$. Substituting the known values: $\frac{6}{\sin 30^\circ} = \frac{b}{\sin 45^\circ} \implies \frac{6}{1/2} = \frac{b}{\sqrt{2}/2} \implies 12 = \frac{2b}{\sqrt{2}} = b\sqrt{2}$. Solving for $b$: $b = \frac{12}{\sqrt{2}} = \frac{12\sqrt{2}}{2} = 6\sqrt{2}\text{ cm}$.
The perpendicularity criterion between a line and a plane requires perpendicularity to TWO lines of the plane that intersect each other. Two intersecting lines determine the plane, so a line perpendicular to both is perpendicular to every line of the plane. Once this holds, $r$ is perpendicular to $\pi$ and its foot is the orthogonal projection of any of its points onto the plane.
The distance from a point $P(x_0, y_0, z_0)$ to the plane $Ax + By + Cz + D = 0$ is $d = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}}$. Substituting into the numerator: $|2(1) - 2(2) + (-1) + 5| = |2 - 4 - 1 + 5| = |2| = 2$. The normal vector is $(2, -2, 1)$, so the denominator is $\sqrt{4 + 4 + 1} = \sqrt{9} = 3$. Hence $d = \frac{2}{3}$.
A regular triangular prism has 3 rectangular lateral faces, each with base $a = 5$ and height $h = 8$. The lateral area is the perimeter of the base multiplied by the height: $A_L = P_b \cdot h = (3 \times 5) \cdot 8 = 15 \times 8 = 120\text{ cm}^2$.
A sphere is the locus of points whose distance to the centre equals the radius. Applying the distance formula in space and squaring gives the canonical equation $(x - x_c)^2 + (y - y_c)^2 + (z - z_c)^2 = R^2$. With $C(2, -1, 3)$ the coordinates are SUBTRACTED, so $y - (-1)$ becomes $y + 1$, and the right-hand side is $R^2 = 4^2 = 16$.
For two parallel planes $Ax + By + Cz + D_1 = 0$ and $Ax + By + Cz + D_2 = 0$, the distance between them is $d = \frac{|D_1 - D_2|}{\sqrt{A^2 + B^2 + C^2}}$. Here $A = 2, B = -1, C = 2$, $D_1 = -4$, and $D_2 = 5$. Thus, $d = \frac{|-4 - 5|}{\sqrt{2^2 + (-1)^2 + 2^2}} = \frac{|-9|}{\sqrt{4 + 1 + 4}} = \frac{9}{\sqrt{9}} = \frac{9}{3} = 3$.
The lateral area of a regular pyramid is $A_L = \frac{P \cdot a_p}{2}$, where $P$ is the base perimeter and $a_p$ is the SLANT height (apotema de la pirámide), not the vertical height. The slant height is found from the right triangle formed by the vertical height, the base apothem ($\frac{10}{2} = 5\text{ cm}$) and the slant height itself: $a_p = \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13\text{ cm}$. With $P = 4(10) = 40\text{ cm}$: $A_L = \frac{40 \cdot 13}{2} = 260\text{ cm}^2$.
The base diameter is 12, so the radius is $r = 6\text{ cm}$. Using the right triangle formed by radius, altitude, and slant height: $h = \sqrt{g^2 - r^2} = \sqrt{10^2 - 6^2} = \sqrt{100 - 36} = \sqrt{64} = 8\text{ cm}$. The volume of a cone is $V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi (6^2)(8) = \frac{1}{3}\pi (36)(8) = 12 \cdot 8\pi = 96\pi\text{ cm}^3$.
The sine function equals zero at all integer multiples of $\pi$: $\sin(\theta) = 0 \iff \theta = k\pi$ for $k \in \mathbb{Z}$. Setting the argument $\theta = 3x$: $3x = k\pi \implies x = \frac{k\pi}{3}$ for all $k \in \mathbb{Z}$.
The volume of a pyramid frustum is $V = \frac{h}{3} (A_1 + A_2 + \sqrt{A_1 A_2})$. The lower base area is $A_1 = 6^2 = 36\text{ cm}^2$ and upper base area is $A_2 = 2^2 = 4\text{ cm}^2$. The geometric mean is $\sqrt{A_1 A_2} = \sqrt{36 \times 4} = \sqrt{144} = 12\text{ cm}^2$. Thus: $V = \frac{9}{3} (36 + 4 + 12) = 3 (52) = 156\text{ cm}^3$.
Using $\tan(\alpha - \beta) = \frac{\tan\alpha - \tan\beta}{1 + \tan\alpha \tan\beta}$ with $\alpha = 45^\circ$ and $\beta = 30^\circ$: $\tan(15^\circ) = \frac{1 - 1/\sqrt{3}}{1 + 1/\sqrt{3}} = \frac{\sqrt{3} - 1}{\sqrt{3} + 1}$. Rationalizing the denominator: $\frac{(\sqrt{3} - 1)^2}{(\sqrt{3} + 1)(\sqrt{3} - 1)} = \frac{3 - 2\sqrt{3} + 1}{3 - 1} = \frac{4 - 2\sqrt{3}}{2} = 2 - \sqrt{3}$.
Two planes are parallel when their normal vectors are proportional. The normal of $\pi$ is $\vec{n} = (3, -1, 2)$. For $6x - 2y + 4z + 1 = 0$ the normal is $(6, -2, 4) = 2(3, -1, 2)$, so the two planes share a direction and are parallel. They are not coincident because the constant terms are not in the same ratio: dividing the second equation by 2 gives $3x - y + 2z + 0.5 = 0$, whose independent term differs from $-7$.
In a regular triangular pyramid, the apex projects onto the circumcenter of the equilateral base. The circumradius of an equilateral triangle with side $a$ is $R = \frac{a}{\sqrt{3}} = \frac{6}{\sqrt{3}} = 2\sqrt{3}\text{ cm}$. In the right triangle formed by lateral edge $b$, altitude $h$, and circumradius $R$: $h = \sqrt{b^2 - R^2} = \sqrt{5^2 - (2\sqrt{3})^2} = \sqrt{25 - 12} = \sqrt{13}\text{ cm}$.
A regular octahedron has 8 equilateral triangular faces. Let two adjacent faces share an edge. The angle $\alpha$ between their outward unit normal vectors $\mathbf{n}_1$ and $\mathbf{n}_2$ satisfies $\mathbf{n}_1 \cdot \mathbf{n}_2 = -\frac{1}{3}$. Thus, the dihedral angle between adjacent faces is obtuse, with $\cos\alpha = -\frac{1}{3}$ (yielding $\alpha \approx 109^\circ 28'$).
The angle between two planes equals the angle between their normal vectors, taken as the acute value. Here $\vec{n_1} = (1, 1, 0)$ and $\vec{n_2} = (0, 1, 1)$. Their scalar product is $1(0) + 1(1) + 0(1) = 1$, and each has norm $\sqrt{1 + 1 + 0} = \sqrt{2}$. Therefore $\cos\theta = \frac{|1|}{\sqrt{2} \cdot \sqrt{2}} = \frac{1}{2}$, giving $\theta = 60^\circ$.
Use the intercept form (ecuación segmentaria) of a plane: $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$. Substituting the intercepts $a = 2, b = 3, c = 6$: $\frac{x}{2} + \frac{y}{3} + \frac{z}{6} = 1$. Multiply through by the least common multiple (6): $3x + 2y + z = 6 \implies 3x + 2y + z - 6 = 0$.
Heron's formula is $A = \sqrt{s(s - a)(s - b)(s - c)}$, where $s = \frac{a + b + c}{2}$ is the semiperimeter. Here $s = \frac{7 + 8 + 9}{2} = \frac{24}{2} = 12\text{ cm}$. The differences are: $s - a = 12 - 7 = 5$; $s - b = 12 - 8 = 4$; $s - c = 12 - 9 = 3$. Thus: $A = \sqrt{12 \times 5 \times 4 \times 3} = \sqrt{12 \times 5 \times 12} = \sqrt{144 \times 5} = 12\sqrt{5}\text{ cm}^2$.
The powers of $i$ repeat in a cycle of length four: $i^1 = i$, $i^2 = -1$, $i^3 = -i$, $i^4 = 1$. To evaluate $i^n$, divide the exponent by 4 and use the remainder. Here $2026 = 4 \cdot 506 + 2$, so the remainder is 2 and $i^{2026} = i^2 = -1$.
In the binomic form $z = a + bi$ the coefficients $a$ and $b$ are both real numbers. For $z = -4 + 9i$ the real part is $a = -4$ and the imaginary part is $b = 9$. The imaginary part is the real coefficient that multiplies $i$, not the product $9i$ itself.
Multiply as ordinary binomials and then replace $i^2$ by $-1$: $(3 + 2i)(4 - 5i) = 12 - 15i + 8i - 10i^2$. Combining the imaginary terms gives $12 - 7i - 10i^2$, and since $i^2 = -1$ the last term becomes $+10$. Therefore the product is $12 + 10 - 7i = 22 - 7i$.
Multiply numerator and denominator by the conjugate of the denominator, $1 + i$. The denominator becomes $(1 - i)(1 + i) = 1^2 + 1^2 = 2$, a real number. The numerator becomes $(2 + 3i)(1 + i) = 2 + 2i + 3i + 3i^2 = 2 + 5i - 3 = -1 + 5i$. Hence the quotient is $\frac{-1 + 5i}{2} = -\frac{1}{2} + \frac{5}{2}i$.
The conjugate of $z = a + bi$ is $\bar{z} = a - bi$: only the sign of the imaginary part changes, while the real part is left untouched. For $z = -6 - 11i$ the real part $-6$ stays as it is and $-11i$ becomes $+11i$, so $\bar{z} = -6 + 11i$. Geometrically, conjugation reflects the point across the real axis of the Argand plane.
The modulus of $z = a + bi$ is $|z| = \sqrt{a^2 + b^2}$, the distance from the origin to the point $(a, b)$ in the Argand plane. Here $|z| = \sqrt{(-8)^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10$. Note that squaring removes the negative sign, so the modulus is always non-negative.
Equality of complex numbers requires the real parts to match and the imaginary parts to match, giving two independent real equations. From the real parts: $2x - 6 = 4$, hence $2x = 10$ and $x = 5$. From the imaginary parts: $3y + 2 = 5$, hence $3y = 3$ and $y = 1$.
For any complex number, $z \cdot \bar{z} = (a + bi)(a - bi) = a^2 - (bi)^2 = a^2 + b^2 = |z|^2$, always a non-negative real number. With $a = 3$ and $b = -4$: $z \cdot \bar{z} = 3^2 + (-4)^2 = 9 + 16 = 25$. This identity is exactly what makes the conjugate the right multiplier for clearing an imaginary denominator.
A complex number $z = a + bi$ is plotted as the point $(a, b)$, with the real part on the horizontal axis and the imaginary part on the vertical axis. For $z = -5 + 2i$ the coordinates are $(-5, 2)$: the abscissa is negative and the ordinate is positive, which places the point in the second quadrant.
First compute the modulus: $r = \sqrt{(-1)^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = 2$. Next locate the point $(-1, \sqrt{3})$: the real part is negative and the imaginary part positive, so $z$ lies in the second quadrant. The reference angle satisfies $\tan\alpha = \left|\frac{\sqrt{3}}{-1}\right| = \sqrt{3}$, so $\alpha = 60^\circ$, and the argument in the second quadrant is $\theta = 180^\circ - 60^\circ = 120^\circ$. Hence $z = 2(\cos 120^\circ + i\,\text{sen}\,120^\circ)$.
The point $(7, -7)$ has a positive real part and a negative imaginary part, so $z$ lies in the fourth quadrant. The reference angle satisfies $\tan\alpha = \left|\frac{-7}{7}\right| = 1$, giving $\alpha = 45^\circ$. In the fourth quadrant the argument measured counter-clockwise from the positive real axis is $\theta = 360^\circ - 45^\circ = 315^\circ$ (equivalently $\frac{7\pi}{4}$ radians).
To multiply two complex numbers in trigonometric form, multiply the moduli and add the arguments: $z_1 z_2 = r_1 r_2 [\cos(\theta_1 + \theta_2) + i\,\text{sen}(\theta_1 + \theta_2)]$. Here the moduli give $3 \cdot 5 = 15$ and the arguments give $40^\circ + 80^\circ = 120^\circ$, so $z_1 z_2 = 15(\cos 120^\circ + i\,\text{sen}\,120^\circ)$.
To divide in trigonometric form, divide the moduli and subtract the arguments: $\frac{z_1}{z_2} = \frac{r_1}{r_2}[\cos(\theta_1 - \theta_2) + i\,\text{sen}(\theta_1 - \theta_2)]$. Here $\frac{12}{4} = 3$ and $150^\circ - 30^\circ = 120^\circ$, so the quotient is $3(\cos 120^\circ + i\,\text{sen}\,120^\circ)$.
De Moivre's formula states $[r(\cos\theta + i\,\text{sen}\,\theta)]^n = r^n(\cos n\theta + i\,\text{sen}\,n\theta)$. Here $r^4 = 2^4 = 16$ and $n\theta = 4 \cdot 15^\circ = 60^\circ$, so the result is $16(\cos 60^\circ + i\,\text{sen}\,60^\circ)$. Since $\cos 60^\circ = \frac{1}{2}$ and $\text{sen}\,60^\circ = \frac{\sqrt{3}}{2}$, this becomes $16\left(\frac{1}{2} + i\frac{\sqrt{3}}{2}\right) = 8 + 8i\sqrt{3}$.
Write $32 = 32(\cos 0^\circ + i\,\text{sen}\,0^\circ)$. The $n$-th roots of a non-zero complex number are given by $z_k = \sqrt[n]{r}\left[\cos\frac{\theta + 360^\circ k}{n} + i\,\text{sen}\frac{\theta + 360^\circ k}{n}\right]$ for $k = 0, 1, \dots, n - 1$. Every root shares the same modulus $\sqrt[5]{32} = 2$, and consecutive arguments differ by $\frac{360^\circ}{5} = 72^\circ$. The five roots therefore sit at the vertices of a regular pentagon inscribed in the circle of radius 2 centred at the origin.
Factor out the common factor: $x^3 + 4x = x(x^2 + 4) = 0$. This gives $x = 0$ or $x^2 + 4 = 0$. The second equation has no real solution, but in $\mathbb{C}$ it yields $x^2 = -4$, so $x = \pm\sqrt{-4} = \pm 2i$. Checking: $(2i)^2 = 4i^2 = -4$, so $(2i)^3 + 4(2i) = -8i + 8i = 0$. The complete solution set is $S = \{0, 2i, -2i\}$, three roots for a cubic, as the Fundamental Theorem of Algebra requires.
Write $-8$ in trigonometric form: $r = 8$ and $\theta = 180^\circ$, so $-8 = 8(\cos 180^\circ + i\,\text{sen}\,180^\circ)$. The cube roots have modulus $\sqrt[3]{8} = 2$ and arguments $\frac{180^\circ + 360^\circ k}{3}$ for $k = 0, 1, 2$, that is $60^\circ$, $180^\circ$ and $300^\circ$. Converting each: $2(\cos 60^\circ + i\,\text{sen}\,60^\circ) = 1 + i\sqrt{3}$; $2(\cos 180^\circ + i\,\text{sen}\,180^\circ) = -2$; $2(\cos 300^\circ + i\,\text{sen}\,300^\circ) = 1 - i\sqrt{3}$.
Apply the quadratic formula with $a = 1$, $b = -4$, $c = 13$: the discriminant is $b^2 - 4ac = 16 - 52 = -36$. In $\mathbb{C}$ this is not an obstacle, since $\sqrt{-36} = 6i$. Therefore $x = \frac{4 \pm 6i}{2} = 2 \pm 3i$. Verifying: $(2 + 3i)^2 = 4 + 12i + 9i^2 = -5 + 12i$, and $-5 + 12i - 4(2 + 3i) + 13 = -5 + 12i - 8 - 12i + 13 = 0$.
If $P(x)$ has real coefficients and $P(z) = 0$, then taking conjugates of the whole equation gives $P(\bar{z}) = \overline{P(z)} = 0$, because conjugation leaves every real coefficient unchanged and distributes over sums and products. Hence $\bar{z} = 3 + 2i$ is also a root. The two conjugate roots combine into the real quadratic factor $x^2 - 6x + 13$, whose coefficients are the sum $2\,\text{Re}(z) = 6$ and the product $|z|^2 = 13$.
The modulus is multiplicative: $|z_1 z_2| = |z_1| \cdot |z_2|$, so the product does not have to be expanded. Here $|z_1| = \sqrt{3^2 + 4^2} = \sqrt{25} = 5$ and $|z_2| = \sqrt{1^2 + (-1)^2} = \sqrt{2}$. Therefore $|z_1 z_2| = 5\sqrt{2}$. Checking directly, $z_1 z_2 = (3 + 4i)(1 - i) = 3 - 3i + 4i + 4 = 7 + i$, whose modulus is $\sqrt{49 + 1} = \sqrt{50} = 5\sqrt{2}$.