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DerKrebs [107]
3 years ago
13

Help Plz you awesome 69 points btw

Mathematics
1 answer:
monitta3 years ago
4 0

\\ \sf\longmapsto \sqrt{5}

  • We can't find the square-root as a whole simple no hence its irrational .
  • root 9 is seemed to be irrational but its square root is 3
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-20n^2-39n-15

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6 0
4 years ago
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}&#10;\\r =  \sqrt{7^2+(-1)^2}&#10;\\r= \sqrt{50} &#10;\\  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)&#10;\\y-6= \frac{5-6}{3-(-4)}  (x-(-4))&#10;\\y-6=- \frac{1}{7} (x+4)&#10;\\7y-42=-x-4&#10;\\x+7y-38=0 

The distance from B(x, y) to C(3, 5) is 5√2.
\sqrt{(x-3)^2+(y-5)^2}=5 \sqrt{2} &#10; \\(x-3)^2+(y-5)^2=(5 \sqrt{2})^2 &#10;\\(x-3)^2+(y-5)^2=50

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

x_1=38-7y_1=38-7 \times 6 = 38-42=-4&#10;\\x_2=38-7y_2=38-7 \times 4 = 38-28=10&#10;

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

But (–4, 6) are the coordinates of point A.
Therefore, point B has coordinates (10,4).
8 0
3 years ago
What Am I - using what you know
eduard
Hi there!

The missing angle is 23 degrees.

Remember every triangle has 180 degrees no matter what type. So you add 75 and 82 and subtract it from 180 giving you 23.

The first name is acute and the last name is scalene.

A acute triangle is when all angles are less than 90 and an obtuse triangle is when there is 1 obtuse angle and 2 acute angles.

Hope this helps !
6 0
3 years ago
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