Answer:
<h2>
<1</h2>
option A is the right option.
Here,
<3=105°
<1=<3(being vertically opposite angles)
<1=105°
Explanation:
Please see the attached picture.
There,
<AOC and <BOD are formed by intersected line segments and they lie to the opposite side of common vertex.They are called vertically opposite angles.<AOD and <BOC are another pairs of vertically opposite angles. Vertically opposite angles are always equal.
Hope this helps..
Good luck on your assignment...
Answer:
185/9
Step-by-step explanation:
20 5/9
Multiply 20 by the 9 (the denominator), and you''ll get 180.
Add the 5 (the numerator) to 180, and you'll get 185.
Then, place 185 over the denominator which is 9 and you get 185/9!
check the pictured below.
now, notice that APB is just a flat-line and therefore those two angles there add up to 180°.
also notice that since ∡ACB is 90°, then those two angles add up to that much.
![\bf \stackrel{\measuredangle ACP}{(3x+2y)}+\stackrel{\measuredangle BCP}{(3x+4y)}=90\implies 6x+6y=90\implies \stackrel{\textit{common factor}}{6(x+y)}=90 \\\\\\ x+y=\cfrac{90}{6}\implies x+y=15\implies \boxed{x=15-y} \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Cbf%20%5Cstackrel%7B%5Cmeasuredangle%20ACP%7D%7B%283x%2B2y%29%7D%2B%5Cstackrel%7B%5Cmeasuredangle%20BCP%7D%7B%283x%2B4y%29%7D%3D90%5Cimplies%206x%2B6y%3D90%5Cimplies%20%5Cstackrel%7B%5Ctextit%7Bcommon%20factor%7D%7D%7B6%28x%2By%29%7D%3D90%0A%5C%5C%5C%5C%5C%5C%0Ax%2By%3D%5Ccfrac%7B90%7D%7B6%7D%5Cimplies%20x%2By%3D15%5Cimplies%20%5Cboxed%7Bx%3D15-y%7D%0A%5C%5C%5C%5C%5B-0.35em%5D%0A~%5Cdotfill)
![\bf \stackrel{\measuredangle APC}{(7x+3)}+\stackrel{\measuredangle BPC}{(16y)}=180\implies 7x+16y=177\implies 7\left( \boxed{15-y} \right)+16y=177 \\\\\\ 105-7y+16y=177\implies 9y=72\implies y=\cfrac{72}{9}\implies \blacktriangleright y=8 \blacktriangleleft \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \measuredangle BPC=16(8)\implies \measuredangle BPC=128](https://tex.z-dn.net/?f=%5Cbf%20%5Cstackrel%7B%5Cmeasuredangle%20APC%7D%7B%287x%2B3%29%7D%2B%5Cstackrel%7B%5Cmeasuredangle%20BPC%7D%7B%2816y%29%7D%3D180%5Cimplies%207x%2B16y%3D177%5Cimplies%207%5Cleft%28%20%5Cboxed%7B15-y%7D%20%5Cright%29%2B16y%3D177%0A%5C%5C%5C%5C%5C%5C%0A105-7y%2B16y%3D177%5Cimplies%209y%3D72%5Cimplies%20y%3D%5Ccfrac%7B72%7D%7B9%7D%5Cimplies%20%5Cblacktriangleright%20y%3D8%20%5Cblacktriangleleft%0A%5C%5C%5C%5C%5B-0.35em%5D%0A%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%0A%5Cmeasuredangle%20BPC%3D16%288%29%5Cimplies%20%5Cmeasuredangle%20BPC%3D128)
F(x) = x^(3/2) +3
f'(x) = (3/2)x^(1/2)
f''(x) = (1/2)*(3/2)*x^(-1/2)
f''(x) = 3/(4√x)
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If you mean f(x) = x√(x+3), parentheses are needed (or you need to typeset the expression).
