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2 years ago

6

What’s the difference in these two questions and why did we divide in one and multiple in one ( also why did we divide by8 in qu

estion 2)

1 answer:

kari74 [83]2 years ago

5 0

Answer:

times and divide

they are both diffrent equations

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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

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

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

<u>Now</u><u> </u><u>putting</u><u> </u><u>the</u><u> </u><u>value</u><u> </u><u>of</u><u> </u><u>h</u><u> </u><u>=</u><u> </u><u>0</u>

<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

Question 4 options:

andrew11 [14]

I. The value of x is equal to 8.

II. The dimensions of the rectangle are 13 and 5 inches respectively.

<u>Given the following data:</u>

Area of a rectangle = 65 square inches.

Length of rectangle = x + 5 inches

Width of rectangle = x - 3 inches

To find the value of x, and the dimensions of the rectangle:

Mathematically, the area of a rectangle is given by the formula:

$A = LW$

Substituting the values, we have:

$65 = (x+5)(x-3)\\\\65=x^2-3x+5x-15\\\\65=x^2+2x-15-65\\\\x^2 +2x-80=0\\\\x^2 +10x-8x-80=0\\\\x(x+10)-8(x+10)=0\\\\(x-8)(x+10)=0$

x = 8 or x = -10

<u>For the</u><u> length:</u>

when x = 8

$Length = 8 +5$

Length = 13 inches.

<u>For the</u><u> width:</u>

when x = 8

$Length = 8 -3$

Length = 5 inches.

Find more information: brainly.com/question/11037225

4 0

2 years ago

On Jerry's text message plan she pays 0.80 for every 20 message. what is cost per message?

dmitriy555 [2]

Jerry’s cost is $.04 per message

3 0

3 years ago

The perimeter of a building is 900 feet. If the length of the building is 400 feet, find the width. How many lights will be need

Helen [10]

Answer:

The width of the building is 50 lamps and 60 lamps will be needed.

Step-by-step explanation:

The perimeter is computed by the sum of the sides of the building, since the length and width are equal on the oposite sides of the building, the perimeter is given by:

perimeter = 2*length + 2*width

900 = 2*(400) + 2*width

2*width + 800 = 900

2*width = 900 - 800

2*width = 100

width = 100/2 = 50 feets

To know how many lights bulbs will be needed we need to take the perimeter of the building and divide it by the space between the lamps. We have:

number o lamps = 900/15 = 60 lamps

8 0

3 years ago

Read 2 more answers

I still don't get it.

AlexFokin [52]

Ok? If there a question?

7 0

3 years ago

Read 2 more answers