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antoniya [11.8K]
3 years ago
10

The lengths of the sides of a triagle are 12, 13, and n. Which of the following must be true.

Mathematics
1 answer:
vampirchik [111]3 years ago
4 0
By the triangle inequality theorem we know that
b - a < c < a+b
where a and b are the given sides; c is the unknown side

In this case,
a = 12
b = 13
c = n

So,
b-a < n < b+a
13-12 < n < 13+12
1 < n < 25

So n is both greater than 1 and also less than 25
Broken up, this means n > 1 and n < 25

So it looks like the only answer is choice A. Choices C and D seem to have typos in them? Something seems off about them. 
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Find the standard form of the equation of the circle with endpoints of a diameter at the points (3,8) and (-7,2)
poizon [28]

(x+2)^{2}+(y-5)^{2}= 34

The equation of a circle in standard form is

(x-a)^{2}+(y-b)^{2}=r^{2}

where (a , b) are the coordinates of the centre and r is the radius

The centre is at the midpoint of the endpoints and the radius is the distance from the centre to either of the 2 endpoints

Using the midpoint formula

midpoint = [\frac{1}{2}(xx_{1}+x_{2}, \frac{1}{2}(y_{1}+y_{2}

where (x_{1},y_{1}=(3,8) and (x_{2},y_{2}=(-7,2)

centre = (\frac{1}{2}(3-7),\frac{1}{2}(8+2)) = (-2,5)

Calculate r using the distance formula

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equation of circle is : (x+2)²+(y-5)² = 34


8 0
3 years ago
What’s the correct answer for this?
VMariaS [17]

Answer:

Second option is the correct answer

(x+11)^2+(y+6)^2 = 324

Step-by-step explanation:

x^{2} +12y+22x+y^2-167=0\\x^{2}+22x +y^2 +12y-167=0\\(x^{2}+22x +121)-121 +(y^2 +12y+36)-36-167=0\\(x+11)^2+(y+6)^2-324 = 0\\\huge\red{\boxed{(x+11)^2+(y+6)^2 = 324}}\\

5 0
3 years ago
Read 2 more answers
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