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anzhelika [568]

3 years ago

6

In a right triangle the length of a hypotenuse is c and the length of one leg is a, and the length of the other leg is b, what i

s the value of b, if
c=13, a=12?

1 answer:

mart [117]3 years ago

5 0

Answer:

5

Step-by-step explanation:

Square of 13 minus square of 12 and then under root.. You will get 5

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lana [24]

Answer:

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I might be wrong but:

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3 years ago

Need help adding these<br><br> √54 + √24

Gennadij [26K]

Answer:

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Step-by-step explanation:

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8 0

4 years ago

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Identify the center and radius from the equation of the circle given below. x^2+y^2+121-20y=-10x

Tcecarenko [31]

Answer:

Center: (-5,10)

Radius: 2

Step-by-step explanation:

The equation of the circle in center-radius form is:

$(x-h)^2+(y-k)^2=r^2$

Where the point (h,k) is the center of the circle and "r" is the radius.

Subtract 121 from both sides of the equation:

$x^2+y^2+121-20y-121=-10x-121\\x^2+y^2-20y=-10x-121$

Add 10x to both sides:

$x^2+y^2-20y+10x=-10x-121+10x\\x^2+y^2-20y+10x=-121$

Make two groups for variable "x" and variable "y":

$(x^2+10x)+(y^2-20y)=-121$

Complete the square:

Add $(\frac{10}{2})^2=5^2$ inside the parentheses of "x".

Add $(\frac{20}{2})^2=10^2$ inside the parentheses of "y".

Add $5^2$ and $10^2$ to the right side of the equation.

Then:

$(x^2+10x+5^2)+(y^2-20y+10^2)=-121+5^2+10^2\\(x^2+10x+5^2)+(y^2-20y+10^2)=4$

Rewriting, you get that the equation of the circle in center-radius form is:

$(x+5)^2+(y-10)^2=2^2$

You can observe that the radius of the circle is:

$r=2$

And the center is:

$(h,k)=(-5,10)$

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