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4 years ago

D= V(x2 – x1)2 + (y2 – yy)2

Mathematics

1 answer:

I am Lyosha [343]4 years ago

3 0

Answer:

y = 2

Step-by-step explanation: divided both from each side

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Can someone explain this and help me write it and write the list of steps

Naddik [55]

113 = 72 + m < B ( by the external angle of a triangle theorem)

Subtract 72 from both sides:-

113 - 72 = 72 - 72 + m < b

m<b = 41 degrees

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4 years ago

10.48 divided by 1.31

fenix001 [56]

The answer is 8 thank me later ;)

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3 years ago

Read 2 more answers

Let x represent any year. Write an inequality in terms of x and 1610

SCORPION-xisa [38]

The year of the empire fell = 1610.

We are given x as a variable that represents the year after the given year 1610.

We need to setup an inequality for this situation.

Please note: The year after the year number 1610 will have a number that greater than 1610.

Let us make a statement for inequality to be written.

"The year after the 1610 is greater than 1610".

The year after the 1610 is x and gerater than symbol is >.

So, we can setup an inequality as:

x > 1610 : Read as x is greater than 1610.

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3 years ago

How many permutations in the word calculus

JulijaS [17]

The word CALCULUS contains

2 C's + 1 A + 2 L's + 2 U's + 1 S = 8 letters, in all.

The number of distinct permutations of all these 8 letters

taken together is

... ( 8! ) / [ (2!) (1!) (2!) (2!) (1!) ]

= [ 8 · (7!) ] / 8

= 7 · (6!)

= 7 ( 720 )

= 5040 ........................................... .

6 0

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