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2 years ago

Please helpare these functions or not?

Mathematics

1 answer:

crimeas [40]2 years ago

5 0

Answer:

1. Function

2. Not a function

3. Function

4. Not a function

Step-by-step explanation:

A function just means that for each input it outputs only 1 output. This output isn't necessarily unique. So for example if you're just given: f(2) = 3 and f(1) = 3, this is a function since each input only outputs 1 value, even though the output isn't unique, but if you're given: f(3) = 2, f(2) = 1, f(3) = 3. that's not a function since f(3) outputs both 2 and 3.

Anyways now that you hopefully understand this, let's look at each image.

Relation 1: (Function)

So for each input (the domain) it's only pointing to one output (range), even if multiple input (the domain) are pointing to the same output (the range), it's still a function.

Relation 2: (Not a function)

This is not a function, since if you look at the input 7 (the range), you'll see it outputs two things (range). It outputs -6 and -7. So this is not a function

Relation 3: (Function)

This is a function since each x-value (input) has only one y-value (output). So it's a function

Relation 4: (Not a functio)

This is not a function, since there are multiply coordinates with the same x-coordinate (input) and different y-coordinates (output). The x-coordinate 2 has the output b, y, and m, and since no value is given for these variables, it can be assumed they're different values, thus it's not a function.

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In right triangle RST, if cos R = 0.6, what is the length of RT?

svet-max [94.6K]

Answer:

1) In right triangle ABC with <A = 90, the ratio for sinB= 5/13. 2) Given ... a) sin (40) = 32/CA b) Cos(50) = 32/ CA ... In triangle RST, angle R is a right angle. If TR = 6 . and ... Determine the length of AC to the nearest 10th.

8 0

3 years ago

What is 5.316 - 1.942 (show ur work)

Fiesta28 [93]

Answer:

3.374

Step-by-step explanation:

$\mathrm{Write\:the\:numbers\:one\:under\:the\:other,\:line\:up\:the\:decimal\:points.}$

$\mathrm{Add\:trailing\:zeroes\:so\:the\:numbers\:have\:the\:same\:length.}$

$\begin{matrix}\:\:&5&.&3&1&6\\ -&1&.&9&4&2\end{matrix}$

$\mathrm{Subtract\:each\:column\:of\:digits,\:starting\:from\:the\:right\:and\:working\:left}$

$\mathrm{In\:the\:bolded\:column,\:subtract\:the\:second\:digit\:from\:the\:first}:\quad \:6-2=4$

$\frac{\begin{matrix}\:\:&5&.&3&1&\textbf{6}\\ -&1&.&9&4&\textbf{2}\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\:\:&\:\:&\textbf{4}\end{matrix}}$

$\mathrm{In\:the\:bolded\:column,\:subtract\:the\:second\:digit\:from\:the\:first}$

$\frac{\begin{matrix}\:\:&5&.&3&\textbf{1}&6\\ -&1&.&9&\textbf{4}&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\:\:&\textbf{\:\:}&4\end{matrix}}$

$\mathrm{The\:bottom\:number\:is\:larger\:than\:the\:upper\:number.\:\:Try\:to\:'borrow'\:a\:digit\:from\:the\:left.}$

$\mathrm{The\:top\:digit\:is\:not\:bigger\:than\:the\:bottom\:one.\:\:Try\:to\:'borrow'\:a\:digit\:from\:the\:left.}$

$\frac{\begin{matrix}\:\:&5&.&\textbf{3}&1&6\\ -&1&.&\textbf{9}&4&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\textbf{\:\:}&\:\:&4\end{matrix}}$

$\mathrm{Borrow\:}1\mathrm{\:from\:}5\mathrm{.\:\:The\:remainder\:is\:}4$

$\frac{\begin{matrix}\:\:&\textbf{4}&\:\:&10&\:\:&\:\:\\ \:\:&\textbf{\linethrough{5}}&.&3&1&6\\ -&\textbf{1}&.&9&4&2\end{matrix}}{\begin{matrix}\:\:&\textbf{\:\:}&\:\:&\:\:&\:\:&4\end{matrix}}$

$\mathrm{Add\:}1\mathrm{\:ten\:to\:}3:\quad \:10+3=13$

$\frac{\begin{matrix}\:\:&4&\:\:&\textbf{13}&\:\:&\:\:\\ \:\:&\linethrough{5}&.&\textbf{\linethrough{3}}&1&6\\ -&1&.&\textbf{9}&4&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\textbf{\:\:}&\:\:&4\end{matrix}}$

$\mathrm{Borrow\:}1\mathrm{\:from\:}13\mathrm{.\:\:The\:remainder\:is\:}12$

$\frac{\begin{matrix}\:\:&4&\:\:&\textbf{12}&10&\:\:\\ \:\:&\linethrough{5}&.&\textbf{\linethrough{13}}&1&6\\ -&1&.&\textbf{9}&4&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\textbf{\:\:}&\:\:&4\end{matrix}}$

$\mathrm{Add\:}1\mathrm{\:ten\:to\:}1:\quad \:10+1=11$

$\frac{\begin{matrix}\:\:&4&\:\:&12&\textbf{11}&\:\:\\ \:\:&\linethrough{5}&.&\linethrough{13}&\textbf{\linethrough{1}}&6\\ -&1&.&9&\textbf{4}&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\:\:&\textbf{\:\:}&4\end{matrix}}$

$\mathrm{In\:the\:bolded\:column,\:subtract\:the\:second\:digit\:from\:the\:first}:\quad \:11-4=7$

$\mathrm{Place\:the\:decimal\:point\:in\:the\:answer\:directly\:below\:the\:decimal\:points\:in\:the\:terms}$

$\frac{\begin{matrix}\:\:&4&\textbf{\:\:}&12&11&\:\:\\ \:\:&\linethrough{5}&\textbf{.}&\linethrough{13}&\linethrough{1}&6\\ -&1&\textbf{.}&9&4&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\textbf{.}&3&7&4\end{matrix}}$

$\mathrm{In\:the\:bolded\:column,\:subtract\:the\:second\:digit\:from\:the\:first}:\quad \:4-1=3$

$\frac{\begin{matrix}\:\:&\textbf{4}&\:\:&12&11&\:\:\\ \:\:&\textbf{\linethrough{5}}&.&\linethrough{13}&\linethrough{1}&6\\ -&\textbf{1}&.&9&4&2\end{matrix}}{\begin{matrix}\:\:&\textbf{3}&.&3&7&4\end{matrix}}$

$=3.374$

Hence the correct answer is 3.374

7 0

2 years ago

creativ13 [48]

Answer:

$\displaystyle \lim_{h \to 0} \frac{\sqrt{3} \Delta (\frac{\pi}{3})}{h^2} = \boxed{ 144 \sqrt{3} }$

General Formulas and Concepts:
Pre-Calculus

2x2 Matrix Determinant:
$\displaystyle \left| \begin{array}{ccc} a & b \\ c & d \end{array} \right| = ad - bc$

3x3 Matrix Determinant:
$\displaystyle \left| \begin{array}{ccc} a & b & c \\ d & e & f \\ g & h & i \end{array} \right| = a \left| \begin{array}{ccc} e & f \\ h & i \end{array} \right| - b \left| \begin{array}{ccc} d & f \\ g & i \end{array} \right| + c \left| \begin{array}{ccc} d & e \\ g & h \end{array} \right|$

Calculus

Limits

Limit Rule [Variable Direct Substitution]:
$\displaystyle \lim_{x \to c} x = c$

Limit Property [Multiplied Constant]:
$\displaystyle \lim_{x \to c} bf(x) = b \lim_{x \to c} f(x)$

Special Limit Rule [L’Hopital’s Rule]:
$\displaystyle \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)}$

Derivatives

Derivatives
Derivative Notation

Derivative Property [Addition/Subtraction]:
$\displaystyle (u + v)' = u' + v'$

Derivative Rule [Chain Rule]:
$\displaystyle [u(v)]' = u'(v)v'$

Step-by-step explanation:

*Note:

I will not be able to fit in all the derivative work and will assume you can take derivatives with ease.

Step 1: Define

Identify given.

$\displaystyle \Delta (x) = \left| \begin{array}{ccc} \tan x & \tan (x + h) & \tan (x + 2h) \\ \tan (x + 2h) & \tan x & \tan (x + h) \\ \tan (x + h) & \tan (x + 2h) & \tan x \end{array} \right|$

$\displaystyle \lim_{h \to 0} \frac{\sqrt{3} \Delta (\frac{\pi}{3})}{h^2}$

Step 2: Find Limit Pt. 1

[Function] Simplify [3x3 and 2x2 Matrix Determinant]:
$\displaystyle \Delta (x) = \tan^3 (2h + x) + \tan^3 (h + x) + \tan^3 x - 3 \tan x \tan (h + x) \tan (2h + x)$
[Function] Substitute in x:
$\displaystyle \Delta \bigg( \frac{\pi}{3} \bigg) = \tan^3 \bigg( 2h+ \frac{\pi}{3} \bigg) + \tan^3 \bigg( h + \frac{\pi}{3} \bigg) + 3\sqrt{3} - 3\sqrt{3} \tan \bigg( h + \frac{\pi}{3} \bigg) \tan \bigg( 2h+ \frac{\pi}{3} \bigg)$

Step 3: Find Limit Pt. 2

[Limit] Rewrite [Limit Property - Multiplied Constant]:
$\displaystyle \lim_{h \to 0} \frac{\sqrt{3} \Delta (\frac{\pi}{3})}{h^2} = \sqrt{3} \lim_{h \to 0} \frac{\Delta (\frac{\pi}{3})}{h^2}$
[Limit] Apply Limit Rule [Variable Direct Substitution]:
$\displaystyle \lim_{h \to 0} \frac{\sqrt{3} \Delta (\frac{\pi}{3})}{h^2} = \sqrt{3} \bigg( \frac{0}{0} \bigg)$

Since we have an indeterminant form, we will have to use L'Hopital's Rule. We can differentiate using basic differentiation techniques listed above under "Calculus":

$\displaystyle \frac{d \Delta (\frac{\pi}{3})}{dh} = -3\sqrt{3} \bigg[ \tan^2 \bigg( h + \frac{\pi}{3} \bigg) + 1 \bigg] \tan \bigg( 2h + \frac{\pi}{3} \bigg) + tan^2 \bigg( h + \frac{\pi}{3} \bigg) \bigg[ 3 \tan^2 \bigg( h + \frac{\pi}{3} + 3 \bigg] - 3\sqrt{3} \tan \bigg( h + \frac{\pi}{3} \bigg) \bigg[ 2 \tan^2 \bigg( 2h + \frac{\pi}{3} \bigg) + 2 \bigg] + \tan^2 \bigg( 2h + \frac{\pi}{3} \bigg) \bigg[ 6 \tan^2 \bigg( 2h + \frac{\pi}{3} \bigg) + 6 \bigg]$

$\displaystyle \frac{d}{dh} h^2 = 2h$

Using L'Hopital's Rule, we can substitute the derivatives and evaluate again. When we do so, we should get another indeterminant form. We will need to use L'Hopital's Rule again:

$\displaystyle \frac{d^2 \Delta (\frac{\pi}{3})}{dh^2} = \tan \bigg( h + \frac{\pi}{3} \bigg) \bigg[ 2 \tan^2 \bigg( h + \frac{\pi}{3} \bigg) + 2 \bigg] \bigg[ \tan^2 \bigg( h + \frac{\pi}{3} \bigg) + 1 \bigg] - 2\sqrt{3} \bigg[ \tan^2 \bigg( h + \frac{\pi}{3} \bigg) + 1 \bigg] \bigg[ \tan^2 \bigg( 2h + \frac{\pi}{3} \bigg) + 1 \bigg] - \sqrt{3} \bigg[ \tan^2 \bigg( h + \frac{\pi}{3} \bigg) + 1 \bigg] \bigg[ 2 \tan^2 \bigg( 2h + \frac{\pi}{3} \bigg) + 2 \bigg]$

$\displaystyle + \tan^3 \bigg( h + \frac{\pi}{3} \bigg) \bigg[ 2 \tan^2 \bigg( h + \frac{\pi}{3} \bigg) + 2 \bigg] - \sqrt{3} \tan \bigg( h + \frac{\pi}{3} \bigg) \tan \bigg( 2h + \frac{\pi}{3} \bigg) \bigg[ 2 \tan^2 \bigg( h + \frac{\pi}{3} \bigg) + 2 \bigg] + \tan \bigg( 2h + \frac{\pi}{3} \bigg) \bigg[ 2 \tan^2 \bigg( 2h + \frac{\pi}{3} \bigg) + 2 \bigg] \bigg[ 4 \tan^2 \bigg( 2h + \frac{\pi}{3} \bigg) + 4 \bigg]$

$\displaystyle - 2\sqrt{3} \tan \bigg( h + \frac{\pi}{3} \bigg) \tan \bigg( 2h + \frac{\pi}{3} \bigg) \bigg[ 4 \tan^2 \bigg( 2h + \frac{\pi}{3} \bigg) + 4 \bigg] + 2 \tan^3 \bigg( 2h + \frac{\pi}{3} \bigg) \bigg[ 4 \tan^2 \bigg( 2h + \frac{\pi}{3} \bigg) + 4 \bigg]$

$\displaystyle \frac{d^2}{dh^2} h^2 = 2$

Substituting in the 2nd derivative found via L'Hopital's Rule should now give us a numerical value when evaluating the limit using limit rules and the unit circle:

$\displaystyle \lim_{h \to 0} \frac{\sqrt{3} \Delta (\frac{\pi}{3})}{h^2} = \boxed{ 144 \sqrt{3} }$

∴ we have evaluated the given limit.

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Learn more about limits: brainly.com/question/27438198

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3 0

2 years ago

Ppppppppeeeelelelelele help

kirill115 [55]

11/3*6/11=66/33=2
6 24/20-2 15/20=2 9/20

5 0

3 years ago

What equation can relate to this sentence 11 is the quotient of a number y and 6 ?

Natalija [7]

Answer: 11 = y ÷ 6

Step-by-step explanation: since the problem says 11 is that means the equation equals 11 and when it says quotient that means that the equation must include division.

4 0

4 years ago

Please helpare these functions or not?​

Please helpare these functions or not?