Given: △ABC, AB = CB AK ⊥ BC , MC ⊥ AB Prove: AK = MC
1 answer:
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Answer:
<h2>For c = 5 → two solutions</h2><h2>For c = -10 → no solutions</h2>Step-by-step explanation:
We know
for any real value of <em>a</em>.
|a| = b > 0 - <em>two solutions: </em>a = b or a = -b
|a| = 0 - <em>one solution: a = 0</em>
|a| = b < 0 - <em>no solution</em>
<em />
|x + 6| - 4 = c
for c = 5:
|x + 6| - 4 = 5 <em>add 4 to both sides</em>
|x + 6| = 9 > 0 <em>TWO SOLUTIONS</em>
for c = -10
|x + 6| - 4 = -10 <em>add 4 to both sides</em>
|x + 6| = -6 < 0 <em>NO SOLUTIONS</em>
<em></em>
Calculate the solutions for c = 5:
|x + 6| = 9 ⇔ x + 6 = 9 or x + 6 = -9 <em>subtract 6 from both sides</em>
x = 3 or x = -15