Answer:
x^2+8x+<u>1</u><u>6</u><u>=</u><u>(</u><u>x-4</u><u>)</u><u>^</u><u>2</u>
<em><u>EXPLANATION</u></em><em><u>:</u></em>
<u>(</u><u>a</u><u>+</u><u>b</u><u>)</u><u>^</u><u>2</u><u>=</u><u>a2</u><u>+</u><u>2</u><u>.</u><u>a</u><u>.</u><u>b</u><u>+</u><u>b2</u>
<u>we</u><u> </u><u>have</u><u> </u><u>to</u><u> </u><u>break</u><u> </u><u>the</u><u> </u><u>middle</u><u> </u><u>term</u><u> </u><u>i</u><u>n</u><u> </u><u>2</u><u>a</u><u>b</u><u> </u><u>here</u><u> </u><u>a</u><u> </u><u>is</u><u> </u><u>x</u><u> </u><u>then</u><u> </u><u>2</u><u>x</u><u>b</u><u>=</u><u>8</u><u>x</u><u>,</u><u> </u><u>=</u><u>></u><u> </u><u>b</u><u>=</u><u>4</u><u>,</u><u> </u><u>but</u><u> </u><u>value</u><u> </u><u>of</u><u> </u><u>a</u><u> </u><u>and</u><u> </u><u>b</u><u> </u><u>to</u><u> </u><u>get</u><u> </u><u>the</u><u> </u><u>req</u><u>uired</u><u> </u><u>equation</u><u>!</u>
- cos ( 1/2 x + 1/5 π ) = 0 ( and because if cos α = 0, α= π/2 + k π, k ∈ Z )
1/2 x + π/5 = π/2 + k π, k ∈ Z
1/2 x = π/2 - π/5 + k π / * 2
x = π - 2π/5 + 2 k π
x = 3/5 π + 2 k π = 0.6 π + 2 k π
Answer:
If k = 0: x 1 = 0.6 π = 3π/5
k = 1 : x 2 = 2.6 π = 13π/5
2x^(3)y+7xy-3x^(2)-12y
is in standard form
Answer: 20
Step-by-step explanation: Just took the quiz
X=5 because the denominator can never equal zero or it will be undefined. and 5-5=0