The intersecting secant-tangent theorem says that
where T is the point at which line segment PO touches the circle. Similarly,
so we have
OP is of course as long as itself, and AO and BO are radii of the same circle so they have the same lengths.
This means triangles APO and BPO are congruent, which means angles APO and BPO are congruent, so angle APB is bisected by PO.
Answer:
No, but i would like to know who they are
Step-by-step explanation:
Answer:
53$
Step-by-step explanation:
6% of 50 is 3, 50+3=53
The answer is: " 91 " .
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→ " B = 91 " .
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Explanation:
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Given:
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" A + B = 180 " ;
"A = -2x + 115 " ; ↔ A = 115 − 2x ;
"B = - 6x + 169 " ; ↔ B = 169 − 6x ;
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METHOD 1)
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Solve for "x" ; and then plug the solved value for "x" into the expression given for "B" ; to solve for "B"
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(115 − 2x) + (169 − 6x) =
115 − 2x + 169 − 6x = ?
→ Combine the "like terms" ; as follows:
+ 115 + 169 = + 284 ;
− 2x − 6x = − 8x ;
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And rewrite as:
" − 8x + 284 " ;
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→ " - 8x + 284 = 180 " ;
Subtract: "284" from each side of the equation:
→ " - 8x + 284 − 284 = 180 − 284 " ;
to get:
→ " -8x = -104 ;
Divide EACH SIDE of the equation by "-8 " ;
to isolate "x" on one side of the equation; & to solve for "x" ;
→ -8x / -8 = -104/-8 ;
→ x = 13
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Now, to find the value of "B" :
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"B = - 6x + 169 " ; ↔ B = 169 − 6x ;
↔ B = 169 − 6x ;
= 169 − 6(13) ; ===========> Plug in our "solved value, "13", for "x" ;
= 169 − (78) ;
= 91 ;
B = " 91 " .
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The answer is: " 91 " .
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→ " B = 91 " .
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Now; let us check our answer:
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→ A + B = 180 ;
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Plug in our "solved answer" ; which is "91", for "B" ; as follows:
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→ A + 91 = ? 180? ;
↔ A = ? 180 − 91 ? ;
→ A = ? -89 ? Yes!
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→ " A = -2x + 115 " ; ↔ A = 115 − 2x ;
Plug in our solved value for "x"; which is: "13" ;
" A = 115 − 2x " ;
→ A = ? 115 − 2(13) ? ;
→ A = ? 115 − (26) ? ;
→ A = ? 29 ? Yes!
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METHOD 2)
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Given:
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" A + B = 180 " ;
"A = -2x + 115 " ; ↔ A = 115 − 2x ;
"B = - 6x + 169 " ; ↔ B = 169 − 6x ;
→ Solve for the value of "B" :
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A + B = 180 ;
→ B = 180 − A ;
→ B = 180 − (115 − 2x) ;
→ B = 180 − 1(115 − 2x) ; ==========> {Note the "implied value of "1" } ;
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Note the "distributive property" of multiplication:__________________________________________________ a(b + c) = ab + ac ; <u><em>AND</em></u>:
a(b − c) = ab − ac .________________________________________________________
Let us examine the following part of the problem:
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→ " − 1(115 − 2x) " ;
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→ " − 1(115 − 2x) " = (-1 * 115) − (-1 * 2x) ;
= -115 − (-2x) ;
= -115 + 2x ;
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So we can bring down the: " {"B = 180 " ...}" portion ;
→and rewrite:
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→ B = 180 − 115 + 2x ;
→ B = 65 + 2x ;
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Now; given: "B = - 6x + 169 " ; ↔ B = 169 − 6x ;
→ " B = 169 − 6x = 65 + 2x " ;
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→ " 169 − 6x = 65 + 2x "
Subtract "65" from each side of the equation; & Subtract "2x" from each side of the equation:
→ 169 − 6x − 65 − 2x = 65 + 2x − 65 − 2x ;
to get:
→ " - 8x + 104 = 0 " ;
Subtract "104" from each side of the equation:
→ " - 8x + 104 − 104 = 0 − 104 " ;
to get:
→ " - 8x = - 104 ;
Divide each side of the equation by "-8" ;
to isolate "x" on one side of the equation; & to solve for "x" ;
→ -8x / -8 = -104 / -8 ;
to get:
→ x = 13 ;
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Now, let us solve for: " B " ; → {for which this very question/problem asks!} ;
→ B = 65 + 2x ;
Plug in our solved value, " 13 ", for "x" ;
→ B = 65 + 2(13) ;
= 65 + (26) ;
→ B = " 91 " .
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Also, check our answer:
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Given: "B = - 6x + 169 " ; ↔ B = 169 − 6x = 91 ;
When "x = 13 " ; does: " B = 91 " ?
→ Plug in our "solved value" of " 13 " for "x" ;
→ to see if: "B = 91" ; (when "x = 13") ;
→ B = 169 − 6x ;
= 169 − 6(13) ;
= 169 − (78)______________________________________________________
→ B = " 91 " .
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