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yaroslaw [1]
3 years ago
9

Factor the trinomial below.

Mathematics
2 answers:
vovikov84 [41]3 years ago
8 0

Answer:

its D

Step-by-step explanation:

Dont even trip I just did the test ;)

liberstina [14]3 years ago
5 0

D

after you use the FOIL method (just multiplying everything) you'll get x^2 - 8x - 5x + 40. now just add the like terms

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sladkih [1.3K]

\huge{\color{magenta}{\fbox{\textsf{\textbf{Answer}}}}}

\frak {\huge{ \frac{1}{1 +  {x}^{2} } }}

Step-by-step explanation:

\sf let \: f(x) =  { \tan }^{ - 1} x \\  \\  \sf f(x + h) =  { \tan}^{ - 1} (x + h)

\sf f'(x) =  \frac{f(x+h)  - f(x) }{h}

\sf \implies \lim_{  h \to 0  } \frac{ { \tan }^{ - 1}(x + h) -  { \tan}^{ - 1}x  }{h}  \\  \\  \\  \sf  \implies  \lim_ {h \to 0}    \frac{  { \tan}^{ - 1} \frac{x + h - x}{1 + (x + h)x} }{h}

By using

\sf { \tan}^{ - 1} x -  { \tan}^{ - 1} y   = \\   \sf { \tan}^{ - 1}  \frac{x - y}{1 + xy} formula

\sf  \implies  \large \lim_{h \to0 }   \frac{  { \tan}^{ - 1}  \frac{h}{1 + hx +  {x}^{2} } }{h}  \\  \\  \\  \sf  \implies   \large{\lim_{h \to0}   } \frac{ { \tan}^{ - 1}  \frac{h}{1 + hx +  {x}^{2} } }{ \frac{h}{1 + hx  +  {x}^{2} }  \times (1 + hx +  {x}^{2} )}  \\  \\  \\  \sf  \implies \large  \lim_{h \to0} \frac{ { \tan}^{ - 1} \frac{h}{1 + hx +  {x}^{2} }  }{ \frac{h}{1 + hx +  {x}^{2} } }  +  \lim_{h \to0} \frac{1}{1 + hx +  {x}^{2} }

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<u>\sf  \large  \implies 0 +  \frac{1}{1 + 0 +  {x}^{2} }  \\  \\  \\  \purple{ \boxed  { \implies  \frac{1}{1 +  {x}^{2} } }}</u>

6 0
2 years ago
Order the side lengths from least to greatest
Liula [17]

Answer:

BC < CE < BE < ED < BD

Step-by-step explanation:

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m∠BEC = 45°

Order of the angles from least to greatest,

m∠BEC < m∠CBE > mBCE

Sides opposite to these sides will be in the same ratio,

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Now in ΔBED,

m∠BEC + m∠BED = 180°

m∠BED = 180 - 45

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3 years ago
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Contact [7]
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3 years ago
Please help this is timed
elena-14-01-66 [18.8K]

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option d is correct

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Anna [14]

Answer:

See attached

Step-by-step explanation:

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So zeros are (0, 8), (3.608, 0) and (-1.108, 0)

3 0
3 years ago
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