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3 years ago

Find the distance between 2 - 4 and 6 + i

Mathematics

1 answer:

saveliy_v [14]3 years ago

3 0

Answer:

Step-by-step explanation:

hello :

let : Z1 = 2-4i Z2 = 6+i

the distance between Z1 and Z2 is :

/Z2 - Z1 / = /(6+i)-(2-4i)/ =/4+5i/ = √(4²+5²) =√41

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Jk,kl, and lj are all tangent to circle o, ja=13,al=7, and ck=10 what is the perimeter of angle JKL

Alex73 [517]

see the picture attached to better understand the problem

we know that

two tangent segments drawn from the same exterior point are congruent

JA=JB ,

LA=LC,

KC=KB

JA=13 units

LA=7 units

kC=10 units

hence

perimeter = JA+JB+LA+LC+KC+KB------> 13+13+7+7+10+10------> =60 units

therefore

the answer is

the perimeter of triangle JKL is 60 units

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Step-by-step explanation:

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Dominik [7]

Answer:

$\frac{(x+4)^2}{25}+\frac{(y+3)^2}{9}=1$

Step-by-step explanation:

Given:

Center of the ellipse is, $(h,k)=(-4,-3)$

Minor axis length is, $2b=6$

A vertex of the ellipse is at (1, -3)

Now, distance between the center and the vertex is half of the length of the major axis.

Using distance formula for (-4, -3) and (1, -3), we get:

$a=\sqrt{(1+4)^2+(-3+3)^2}=5$

Therefore, the value of half of major axis is, $a=5$ . Also,

$2b=6\\b=\frac{6}{2}=3$

Now, equation of an ellipse with center $(h,k)$ is given as:

$\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1$

Plug in $h=-4,k=-3,a=5,b=3$ and determine the equation.

$\frac{(x-(-4))^2}{5^2}+\frac{(y-(-3))^2}{3^2}=1\\\\\frac{(x+4)^2}{25}+\frac{(y+3)^2}{9}=1$

Therefore, the equation of the ellipse is:

$\frac{(x+4)^2}{25}+\frac{(y+3)^2}{9}=1$

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2 years ago