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Zepler [3.9K]
3 years ago
7

Whats carible usually shown on y axis

Mathematics
1 answer:
love history [14]3 years ago
3 0
Independent variable 
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What is the quotient of ÷
Svetlanka [38]
There isn't a quotient of the division sign
6 0
3 years ago
PLS help me Order four and seven tenths repeating, negative four and eight ninths, 490%, and −4.9 from least to greatest.
Digiron [165]

The order from least to greatest negative four and seven tenths repeating, negative four and eight ninths, 490%, and −4.9 from least to greatest is; option A

  • −4.9,
  • negative four and eight ninths
  • four and seven tenths repeating
  • 490%,

<h3>Ascending order</h3>

four and seven tenths repeating

= 4.77777777

negative four and eight ninths

= -4 8/9

= -4.888888888888888

490%

= 490/100

= 4.9

−4.9

Rearranging from least to greatest (ascending order)

  • -4.9
  • -4.888888888888888
  • 4.77777777
  • 4.9

Learn more about ascending order;

brainly.com/question/12783355

#SPJ1

6 0
2 years ago
Find the area of the shape shown below
ra1l [238]

Answer:

18 MARK AS BRAINLIEST OR ELSE!!!!

Step-by-step explanation:

area = 1/2h(sum of bases)

area = 1/2x3(3+9)

area = 1/2 of 36

area = 18

3 0
2 years ago
PLease answer as soon as possible
yulyashka [42]

Answer:

No

Step-by-step explanation:

Every x value in the domain should only have 1 y value. In the graph, for x=5 (the input) there are 2 y values (6 and -6) so this is not a function.

3 0
3 years ago
Read 2 more answers
If f(x)=2x+sinx and the function g is the inverse of f then g'(2)=
Alexxx [7]
\bf f(x)=y=2x+sin(x)&#10;\\\\\\&#10;inverse\implies x=2y+sin(y)\leftarrow f^{-1}(x)\leftarrow g(x)&#10;\\\\\\&#10;\textit{now, the "y" in the inverse, is really just g(x)}&#10;\\\\\\&#10;\textit{so, we can write it as }x=2g(x)+sin[g(x)]\\\\&#10;-----------------------------\\\\

\bf \textit{let's use implicit differentiation}\\\\&#10;1=2\cfrac{dg(x)}{dx}+cos[g(x)]\cdot \cfrac{dg(x)}{dx}\impliedby \textit{common factor}&#10;\\\\\\&#10;1=\cfrac{dg(x)}{dx}[2+cos[g(x)]]\implies \cfrac{1}{[2+cos[g(x)]]}=\cfrac{dg(x)}{dx}=g'(x)\\\\&#10;-----------------------------\\\\&#10;g'(2)=\cfrac{1}{2+cos[g(2)]}

now, if we just knew what g(2)  is, we'd be golden, however, we dunno

BUT, recall, g(x) is the inverse of f(x), meaning, all domain for f(x) is really the range of g(x) and, the range for f(x), is the domain for g(x)

for inverse expressions, the domain and range is the same as the original, just switched over

so, g(2) = some range value
that  means if we use that value in f(x),   f( some range value) = 2

so... in short, instead of getting the range from g(2), let's get the domain of f(x) IF the range is 2

thus    2 = 2x+sin(x)

\bf 2=2x+sin(x)\implies 0=2x+sin(x)-2&#10;\\\\\\&#10;-----------------------------\\\\&#10;g'(2)=\cfrac{1}{2+cos[g(2)]}\implies g'(2)=\cfrac{1}{2+cos[2x+sin(x)-2]}

hmmm I was looking for some constant value... but hmm, not sure there is one, so I think that'd be it
5 0
3 years ago
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