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Lana71 [14]
2 years ago
13

I don’t know if the answer i put is correct,, can someone help me?

Mathematics
1 answer:
joja [24]2 years ago
5 0
R=60pi/(2pi) (second one)
The circumference of an arbitrary circle is 2*pi*r, where r is the radius, so we can set the equation up like so, based on the info given:
2*pi*r=60*pi
r=60pi/(2pi)
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Explain the relationship between the range of values and quartiles.
vitfil [10]
I think you meant qualities
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Find angle sum therom ​
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Sum of the interior angles of any triangle is 180° .

Thus ;

y + 48° + 90° = 180°

y + 138° = 180°

Subtract sides 138°

y + 138° - 138° = 180° - 138° \\

y = 42°

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4 0
3 years ago
The points A BC :(2, 2,1), :(1,1,3), :(2,0,5) − are the vertices of a right triangle. The radius of the sphere with center at th
ElenaW [278]

Answer:

r=1

Step-by-step explanation:

First we need to know the length of each side of the triangle, so we use the formula of the vector modulus:

|AB|= \sqrt{(b_{1}-a_{1})^{2}+(b_{2}-a_{2})^{2}+(b_{3}-a_{3})^{2}

By doing so, we find:

|AB|=\sqrt {6}\\|BC|=\sqrt {6}\\|AC|=2\sqrt {5}

With this we know that the triangle is not right, but, we assume the longest side as the hypotenuse of the problem.

As we have two equal sides, we know that the line between point |AB| and the center of the hypotenuse is perpendicular, therefore, we can calculate it using Pythagoras theorem:

|BC|^{2}=r^{2}+(\frac{|AC|}{2})^{2}\\\\r^{2}=|BC|^{2}-(\frac{|AC|}{2})^{2}\\\\r^{2}=(\sqrt{6})^{2}- (\frac{2\sqrt{5}}{2})^{2}\\\\r^{2}=6-5=1\\r=1

4 0
3 years ago
Explain how to draw a line segments that measures 2 7/16 inches
viktelen [127]

9514 1404 393

Answer:

  draw the line using a ruler on which you have identified points 2 7/16 inches apart

Step-by-step explanation:

On your ruler graduated in inches with marks at 1/16-inch intervals, locate the 0 mark and the mark 1/16 inch before the half-inch mark between 2 and 3 inches.

Draw your line along the edge of the ruler between the two marks you have identified: 0 and 2 7/16. The line will be 2 7/16 inches long.

6 0
3 years ago
Find an equation of the plane through the point (−5,−1,2) with normal vector ????=⟨−1,−5,2⟩.
Ganezh [65]

Answer:

x + 5y - 2z + 14 = 0

Step-by-step explanation:

A plane that passes through the point (x_{0}, y_{0}, z_{0}) with a normal vector of (a,b,c) has the following equation, initially:

a(x - x_{0}) + b(y - y_{0}) + c(z - z_{0}) = 0

After we solve this, we have

ax + by + cz + d = 0

In this problem, we have that:

(x_{0}, y_{0}, z_{0}) = (-5,-1,2)

(a,b,c) = (−1,−5,2)

So

a(x - x_{0}) + b(y - y_{0}) + c(z - z_{0}) = 0

-1(x - (-5)) - 5(y - (-1)) + 2(z - 2) = 0

-(x + 5) - 5(y + 1) + 2(z - 2) = 0

-x - 5 - 5y - 5 + 2z - 4 = 0

-x - 5y + 2z - 14 = 0

Multplying everything by -1

x + 5y - 2z + 14 = 0

3 0
4 years ago
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