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Lynna [10]
3 years ago
13

The diagram shows XPOR. Which term describes point S? A circumcenter B Incenter C centroid D Orthocender​

Mathematics
1 answer:
podryga [215]3 years ago
7 0

Answer: The answer is B incenter

Step-by-step explanation: hope it helps

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Name the property illustrated.<br><br> 4 + 3 + 9 = 3 + 4 + 9<br><br><br> Yall please answer ASAP
sattari [20]

Answer:

true

Step-by-step explanation:

4+3+9=3+4+9

use the commutative property to rearrange the terms.

4+3+9=4+3+9

equality is true because both members are identical.

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If one pound of turkey costs 1.89 how many will x pounds cost
Mekhanik [1.2K]
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It would be 1.89 times how ever many pounds there are:
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PilotLPTM [1.2K]

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Step-by-step explanation:

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3 years ago
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BabaBlast [244]

Answer:

\displaystyle \frac{(a - b)^{2} - {c}^{2} }{{a}^{2} - {(b+ c)}^{2} } + \frac{(b - c)^{2} - {a}^{2} }{{b}^{2} - {(c+ a)}^{2} } + \frac{(c - a)^{2} - {b}^{2} }{{c}^{2} - {(a+ b)}^{2} }=1

Step-by-step explanation:

<u>Algebra Simplifying</u>

We are given the expression:

\displaystyle T=\frac{(a - b)^{2} - {c}^{2} }{{a}^{2} - {(b+ c)}^{2} } + \frac{(b - c)^{2} - {a}^{2} }{{b}^{2} - {(c+ a)}^{2} } + \frac{(c - a)^{2} - {b}^{2} }{{c}^{2} - {(a+ b)}^{2} }

We need to repeatedly use the following identity:

(a^2-b^2)=(a-b)(a+b)

For example, the first numerator has a difference of squares thus we factor as:

(a - b)^{2} - {c}^{2} =(a-b-c)(a-b+c)

The first denominator can be factored also:

{a}^{2} - {(b+ c)}^{2} = (a-b-c)(a+b+c)

Applying the same procedure to all the expressions:

\displaystyle T=\frac{(a - b- c)(a-b+c) }{a-b-c)(a+b+c) } + \frac{(b - c - a)(b-c+a) }{(b-c-a)(b+c+a) } + \frac{(c - a-b)(c-a+b) }{(c-a-b)(c+a+b)}

Simplifying all the fractions:

\displaystyle T=\frac{a - b- c}{a+b+c} + \frac{b-c+a}{b+c+a } + \frac{c-a+b }{c+a+b}

Since all the denominators are equal:

\displaystyle T=\frac{a - b+ c+b-c+a+c-a+b}{a+b+c}

Simplifying:

\displaystyle T=\frac{a +c+b}{a+b+c}

Simplifying again:

T = 1

Thus:

\boxed{\displaystyle \frac{(a - b)^{2} - {c}^{2} }{{a}^{2} - {(b+ c)}^{2} } + \frac{(b - c)^{2} - {a}^{2} }{{b}^{2} - {(c+ a)}^{2} } + \frac{(c - a)^{2} - {b}^{2} }{{c}^{2} - {(a+ b)}^{2} }=1}

8 0
3 years ago
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