Реши . в парке астаны посадили 3 ряда берёз по 8 деревьев в каждом и 2 ряда елей по 4 дерева в каждом. каких деревьев посадили больше и во сколько раз? ​

8×4=32 березы

4×2=8 елей

32: 8=4 раз больше

ответ: берез в 4 раза больше елей

z=tg(xy

2

)

\dfrac{\partial z}{\partial y}=\dfrac{1}{\cos^2(xy^2)} \cdot(xy^2)'_y=\dfrac{1}{\cos^2(xy^2)} \cdot2xy=\dfrac{2xy}{\cos^2(xy^2)}

∂y

∂z

=

cos

2

(xy

2

)

1

⋅(xy

2

)

y

′

=

cos

2

(xy

2

)

1

⋅2xy=

cos

2

(xy

2

)

2xy

\begin{gathered}\dfrac{\partial^2z}{\partial y^2}=\dfrac{(2xy)'_y\cdot\cos^2(xy^2)-2xy\cdot(\cos^2(xy^2))'_y}{(\cos^2(xy^2))^2} ==\dfrac{2x\cdot\cos^2(xy^2)-2xy\cdot 2\cos(xy^2)\cdot(\cos(xy^2))'_y}{\cos^4(xy^2)} ==\dfrac{2x\cos^2(xy^2)-4xy\cos(xy^2)\cdot(-\sin(xy^2))\cdot(xy^2)'_y}{\cos^4(xy^2)} ==\dfrac{2x\cos^2(xy^2)+4xy\cos(xy^2)\sin(xy^2)\cdot2xy}{\cos^4(xy^2)} =\dfrac{2x\cos(xy^2)+8x^2y^2\sin(xy^2)}{\cos^3(xy^2)}\end{gathered}

∂y

2

∂

2

z

=

(cos

2

(xy

2

))

2

(2xy)

y

′

⋅cos

2

(xy

2

)−2xy⋅(cos

2

(xy

2

))

y

′

=

=

cos

4

(xy

2

)

2x⋅cos

2

(xy

2

)−2xy⋅2cos(xy

2

)⋅(cos(xy

2

))

y

′

=

=

cos

4

(xy

2

)

2xcos

2

(xy

2

)−4xycos(xy

2

)⋅(−sin(xy

2

))⋅(xy

2

)

y

′

=

=

cos

4

(xy

2

)

2xcos

2

(xy

2

)+4xycos(xy

2

)sin(xy

2

)⋅2xy

=

cos

3

(xy

2

)

2xcos(xy

2

)+8x

2

y

2

sin(xy

2

)

\dfrac{\partial z}{\partial x}=\dfrac{1}{\cos^2(xy^2)} \cdot(xy^2)'_x=\dfrac{1}{\cos^2(xy^2)} \cdot y^2=\dfrac{y^2}{\cos^2(xy^2)}

∂x

∂z

=

cos

2

(xy

2

)

1

⋅(xy

2

)

x

′

=

cos

2

(xy

2

)

1

⋅y

2

=

cos

2

(xy

2

)

y

2

\begin{gathered}\dfrac{\partial^2z}{\partial x^2}=\dfrac{(y^2)'_x\cdot\cos^2(xy^2)-y^2\cdot(\cos^2(xy^2))'_x}{(\cos^2(xy^2))^2} ==\dfrac{0\cdot\cos^2(xy^2)-y^2\cdot2\cos(xy^2)\cdot(\cos(xy^2))'_x}{\cos^4(xy^2)} ==\dfrac{-2y^2\cos(xy^2)\cdot(-\sin(xy^2))\cdot(xy^2)'_x}{\cos^4(xy^2)} ==\dfrac{2y^2\cos(xy^2)\sin(xy^2)\cdot y^2}{\cos^4(xy^2)} =\dfrac{2y^4\sin(xy^2)}{\cos^3(xy^2)}\end{gathered}

∂x

2

∂

2

z

=

(cos

2

(xy

2

))

2

(y

2

)

x

′

⋅cos

2

(xy

2

)−y

2

⋅(cos

2

(xy

2

))

x

′

=

=

cos

4

(xy

2

)

0⋅cos

2

(xy

2

)−y

2

⋅2cos(xy

2

)⋅(cos(xy

2

))

x

′

=

=

cos

4

(xy

2

)

−2y

2

cos(xy

2

)⋅(−sin(xy

2

))⋅(xy

2

)

x

′

=

=

cos

4

(xy

2

)

2y

2

cos(xy

2

)sin(xy

2

)⋅y

2

=

cos

3

(xy

2

)

2y

4

sin(xy

2

)

ответ:

пошаговое объяснение:

первая страница

2.

3.

3.

[tex]r=\sqrt{(-5)^2+(-5)^2}=\sqrt{50}=5\sqrt{2} =arctg(\frac{-5}{-5})=arctg(1)=\frac{\pi}{4}+\pi=\frac{5\pi}{4} /tex]

[tex]-5-5i=5\sqrt{2} (cos(\frac{5\pi}{4}+i*sin(\frac{5\pi}{4} ) /tex]