Ответы
Ответ дал:
0
В9
![f(x)=x^2+2x+c
\
f(a)=a^2+2a+c
\
f`(x)=2x+2
\
f`(a)=2a+2
\
y_k=a^2+2a+c+(2a+2)(x-a)=
\
=a^2+2a+c+2ax-2a^2+2x-2a=(2a+2)x-a^2+c
\
left { {{2a+2=-6} atop {-a^2+c=-6}} right.
\
2a=-8
\
a=-4
\
-(-4)^2+c=-6
\
c=-6+16
\
c=10](https://tex.z-dn.net/?f=f%28x%29%3Dx%5E2%2B2x%2Bc%0A%5C%0Af%28a%29%3Da%5E2%2B2a%2Bc%0A%5C%0Af%60%28x%29%3D2x%2B2%0A%5C%0Af%60%28a%29%3D2a%2B2%0A%5C%0Ay_k%3Da%5E2%2B2a%2Bc%2B%282a%2B2%29%28x-a%29%3D%0A%5C%0A%3Da%5E2%2B2a%2Bc%2B2ax-2a%5E2%2B2x-2a%3D%282a%2B2%29x-a%5E2%2Bc%0A%5C%0A+left+%7B+%7B%7B2a%2B2%3D-6%7D+atop+%7B-a%5E2%2Bc%3D-6%7D%7D+right.+%0A%5C%0A+2a%3D-8%0A%5C%0Aa%3D-4%0A%5C%0A-%28-4%29%5E2%2Bc%3D-6%0A%5C%0Ac%3D-6%2B16%0A%5C%0Ac%3D10 "f(x)=x^2+2x+c
\
f(a)=a^2+2a+c
\
f`(x)=2x+2
\
f`(a)=2a+2
\
y_k=a^2+2a+c+(2a+2)(x-a)=
\
=a^2+2a+c+2ax-2a^2+2x-2a=(2a+2)x-a^2+c
\
left { {{2a+2=-6} atop {-a^2+c=-6}} right.
\
2a=-8
\
a=-4
\
-(-4)^2+c=-6
\
c=-6+16
\
c=10")
Ответ: 10
В5
![l=P-2r
\
l=9.5-2cdot4=1.5
\
l= frac{ pi r alpha }{180}
\
alpha = frac{180l}{ pi r}
\
S= frac{ pi r^2 alpha }{360}
\
S=frac{ pi r^2 cdot 180l}{360 pi r} =frac{ rl}{2 } =frac{ 4cdot1.5}{2 } =3](https://tex.z-dn.net/?f=l%3DP-2r%0A%5C%0Al%3D9.5-2cdot4%3D1.5%0A%5C%0Al%3D+frac%7B+pi+r+alpha+%7D%7B180%7D+%0A%5C%0A+alpha+%3D+frac%7B180l%7D%7B+pi+r%7D+%0A%5C%0AS%3D+frac%7B+pi+r%5E2+alpha+%7D%7B360%7D+%0A%5C%0AS%3Dfrac%7B+pi+r%5E2+cdot+180l%7D%7B360+pi+r%7D+%3Dfrac%7B+rl%7D%7B2+%7D+%3Dfrac%7B+4cdot1.5%7D%7B2+%7D+%3D3 "l=P-2r
\
l=9.5-2cdot4=1.5
\
l= frac{ pi r alpha }{180}
\
alpha = frac{180l}{ pi r}
\
S= frac{ pi r^2 alpha }{360}
\
S=frac{ pi r^2 cdot 180l}{360 pi r} =frac{ rl}{2 } =frac{ 4cdot1.5}{2 } =3")
Ответ: 3
В6
![P(A)= frac{m}{n}
\
P(A)= frac{2+4}{3+2+5+1+4} = frac{6}{15} =0.4](https://tex.z-dn.net/?f=P%28A%29%3D+frac%7Bm%7D%7Bn%7D+%0A%5C%0AP%28A%29%3D+frac%7B2%2B4%7D%7B3%2B2%2B5%2B1%2B4%7D+%3D+frac%7B6%7D%7B15%7D+%3D0.4 "P(A)= frac{m}{n}
\
P(A)= frac{2+4}{3+2+5+1+4} = frac{6}{15} =0.4")
Ответ: 0,4
Ответ: 10
В5
Ответ: 3
В6
Ответ: 0,4
Вас заинтересует
2 года назад
8 лет назад
10 лет назад
10 лет назад
10 лет назад