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3 years ago

State whether the following statements are true or false, with reasoning:

Mathematics

1 answer:

zmey [24]3 years ago

3 0

Answer:

Attached is the complete working. Hope it helps.

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6. State the absolute maximum using the correct notation. Start answer with abs max: Please do not put any spaces in answer.

Montano1993 [528]

Answer:3.5:80

Step-by-step explanation:

common sense

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4 years ago

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How many terms are there in the expression 3x2 + 2x - 4? A:1 B:2 C:3 D:4

Nadya [2.5K]

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Step-by-step explanation:

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Standard form of the equation that goes through (3, -1) with a slope of -1, please!

11111nata11111 [884]

Answer:

x+y=2

Step-by-step explanation:

Standard Form: ax+by=c

For most convenience start with slope-intercept form: y=mx+b

1. -1 = -1(3)+b - solve for b

b=2

y= -1x+2 (slope-intercept form)

2. Now convert to standard form

Add 1x to both sides:

1x+y=2

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3 years ago

Which of the following numbers can be evenly divided by 4?

densk [106]

D is the answer it is divided by 4 evenly

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3 years ago

Find the derivative of <img src="https://tex.z-dn.net/?f=tan%5E%7B-1%7D%20x" id="TexFormula1" title="tan^{-1} x" alt="tan^{-1} x

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

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

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

Now putting the value of h = 0

$\sf \large \implies 0 + \frac{1}{1 + 0 + {x}^{2} } \\ \\ \\ \purple{ \boxed { \implies \frac{1}{1 + {x}^{2} } }}$

6 0

2 years ago