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creativ13 [48]
3 years ago
14

Solve the equation x^2-4x+85=0

Mathematics
1 answer:
lara [203]3 years ago
3 0
Can't factor
use quadratic
for
ax^2+bx+c=0
x=\frac{-b+/- \sqrt{b^2-4ac} }{2a}

given
1x^2-4x+85=0
a=1
b=-4
c=85

x=\frac{-(-4)+/- \sqrt{(-4)^2-4(1)(85)} }{2(1)}
x=\frac{4+/- \sqrt{16-340} }{2}
x=\frac{4+/- \sqrt{-324} }{2}
x=\frac{4+/- (\sqrt{-1})(\sqrt{324}) }{2}
remember that √-1=i
x=\frac{4+/- i(\sqrt{324}) }{2}
x=\frac{4+/- i(18) }{2}
x=2+/-9i

x=2+9i or x=2-9i
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torisob [31]

Answer:

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Step-by-step explanation:

tan 3x=\frac{\sqrt{3} }{3} =\frac{1}{\sqrt{3} } =tan\frac{\pi }{6} =tan (n\pi +\frac{\pi }{6} )\\3x=n\pi +\frac{\pi }{6} =\frac{(6n+1)\pi }{6} \\x=\frac{(6n+1)\pi }{18} \\

where~x~is~an~integer.

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3 years ago
The green triangle is a dilation of the red triangle with a scale factor of s=1/3 and the center of dilation is at the point (4,
klasskru [66]

Given:

The scale factor is s=\dfrac{1}{3} and the center of dilation is at the point (4,2).

Red is original figure and green is dilated figure.

To find:

The coordinates of point C' and point A.

Solution:

Rule of dilation: If a figure is dilated with a scale factor k and the center of dilation is at the point (a,b), then

(x,y)\to (k(x-a)+a,k(y-b)+b)

According to the given information, the scale factor is \dfrac{1}{3} and the center of dilation is at (4,2).

(x,y)\to (\dfrac{1}{3}(x-4)+4,\dfrac{1}{3}(y-2)+2)            ...(i)

Let us assume the vertices of red triangle are A(m,n), B(10,14) and C(-2,11).

Using (i), we get

C(-2,11)\to C'(\dfrac{1}{3}(-2-4)+4,\dfrac{1}{3}(11-2)+2)

C(-2,11)\to C'(\dfrac{1}{3}(-6)+4,\dfrac{1}{3}(9)+2)

C(-2,11)\to C'(-2+4,3+2)

C(-2,11)\to C'(2,5)

Therefore, the coordinates of Point C' are C'(2,5).

We assumed that point A is A(m,n).

Using (i), we get

A(m,n)\to A'(\dfrac{1}{3}(m-4)+4,\dfrac{1}{3}(n-2)+2)

From the given figure it is clear that the image of point A is (8,4).

A'(\dfrac{1}{3}(m-4)+4,\dfrac{1}{3}(n-2)+2)=A'(8,4)

On comparing both sides, we get

\dfrac{1}{3}(m-4)+4=8

\dfrac{1}{3}(m-4)=8-4

(m-4)=3(4)

m=12+4

m=16

And,

\dfrac{1}{3}(n-2)+2=4

\dfrac{1}{3}(n-2)=4-2

(n-2)=3(2)

n=6+2

n=8

Therefore, the coordinates of point A are (16,8).

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Answer:

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