1) <span>28x^2-36х+11=0
D=b</span>²-4ac=1296-4×28×11=64
x₁=<u>-b+√D </u> = <u>11</u>
2a 4
x₂= <u>-b-√D</u> = 0,5
2a
2) -49х^2+21х-2=0;
D = b2 - 4ac = 212<span> - 4·(-49)·(-2) = 441 - 392 = 49
</span>x₁=<u>2</u>
7
x₂=<u>1
</u> 7
<span>3) -7х^2-4x+11=0
</span>D = b2 - 4ac = (-4)2<span> - 4·(-7)·11 = 16 + 308 = 324
</span>x₁= <u>-11</u>
7
x₂= 1
4) -23х^2-22х+1=0
D = b2 - 4ac = (-22)2<span> - 4·(-23)·1 = 484 + 92 = 576
</span>x₁= <u>1</u>
23
x₂=-1
<span>5) 3х^2-14х+16=0.
</span>D = b2 - 4ac = (-14)2 - 4·3·16 = 196 - 192 = 4
x₁=<span> 2
</span>x₂= <u>8
</u> 3
TgA=BC/AC=5/3⇒AC=0,6BC
AC²+BC²=AB²
0,36BC²+BC²=289
1,36BC²=289
BC²=28900/136
BC=170/2√34=85/√34
AC=3/5*85/34=51/√34=51√34/34=3√34/2
cos²A=1:(1+tg²A)=1:(1+25/9)=1:34/9=9/34
sin²A=1-cos²A=1-9/34=25/34
sinA=5/√34
sinA=CH/AC⇒CH=AC*sinA=3√34/2*5/√34=15/2=7,5
<span>16-9,5у=3у+21
16 -21 = 3y +9.5y
-5 = 12.5y
y =-5 /12.5 = -0.4</span>
400 г раствора составляют 100%
36 г соли составляют х %
х = 36 · 100 : 400 = 9%
Ответ: 9 %