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Mathematics

2 answers:

Katen [24]2 years ago

8 0

Answer:

$36\sqrt{2}$

Explanation:

This radical can be simplified. You know that the number inside of the radical; 8, is the product of 4 and 2. You know that 4 is a perfect square (2 times 2). Therefore, you can leave the 2 on the inside and the 4 on the outside. You now must multiply the 18 on the outside of the radical by 2 because you must take the square root of the original 4 you had. Therefore, you answer will be $36\sqrt{2}$ .

Elena L [17]2 years ago

7 0

Sorry can’t help don’t know what you looking for

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Read 2 more answers

a circle has a center (3,5) and a diameter AB. The coordinates of A are (-4,6). what are the coordinates of B?

Kruka [31]

Find the length of the radius.

$r= \sqrt{(3-(-4))^2+(5-6)^2}
\\r = \sqrt{7^2+(-1)^2}
\\r= \sqrt{50} 
\\ r=5 \sqrt{2}$

Find the length of the diameter.

d = 2r = 2 × 5√2 = 10√2

Point B must lie on a line AC, where C is a center of a circle.
Find equation of line AC.
A(–4, 6), C(3, 5)

$y-y_1= \frac{y_2-y_1}{x_2-x_1}(x-x_1)
\\y-6= \frac{5-6}{3-(-4)} (x-(-4))
\\y-6=- \frac{1}{7} (x+4)
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The distance from B(x, y) to C(3, 5) is 5√2.
$\sqrt{(x-3)^2+(y-5)^2}=5 \sqrt{2} 
 \\(x-3)^2+(y-5)^2=(5 \sqrt{2})^2 
\\(x-3)^2+(y-5)^2=50$

Solve system of equations.
$x+7y-38=0
\\(x-3)^2+(y-5)^2=50
\\
\\x=38-7y
\\(38-7y-3)^2+(y-5)^2=50
\\(35-7y)^2+(y-5)^2=50
\\1225-490y+49y^2+y^2-10y+25-50=0
\\50y^2-500y+1200=0
\\y^2-10y+24=0
\\y^2-6y-4y+24=0
\\y(y-6)-4(y-6)=0 \\(y-6)(y-4)=0 \\y_1=6,y_2=4
$

$x_1=38-7y_1=38-7 \times 6 = 38-42=-4
\\x_2=38-7y_2=38-7 \times 4 = 38-28=10
$

Point B could have coordinates
$\\
\\(x_1,y_1)=(-4,6),(x_2,y_2)=(10,4)
$

But (–4, 6) are the coordinates of point A.
Therefore, point B has coordinates (10,4).

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3 years ago