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natima [27]
4 years ago
9

What is the square root of 16 over 81

Mathematics
2 answers:
jek_recluse [69]4 years ago
5 0

Answer:0.4

Step-by-step explanation:

Dafna1 [17]4 years ago
3 0

Answer:

0.04938271604

Step-by-step explanation:

Hope this helps :)

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Is it possible to prove the triangles congruent? if so, what postulate would you use?
amid [387]

Answer:

Is in the attached image.

Hope this helped

-Adan


8 0
3 years ago
You are told that you will have to wait for 5 hours in a line with a group of other people. Determine if:
Gnoma [55]

Answer:

Step-by-step explanation:

You that 60minutes makes 1 hour

So you will multiply the 60minutes by 5

And u will get 300

So your 5 hours will be 300 minutes

7 0
3 years ago
The incorrect work of a student to solve an equation 2(y + 8) = 4y is shown below:
PolarNik [594]

Answer:

Step 2 involved distributive property and the value of y is equal to 8.

Step-by-step explanation:

In step 2, there has been an error in applying the Distributive Property correctly.

Distributive Property-  a(b+c)

                                       = a x b + a x c

Step 2: 2y+16=4y

Step 3:2y=16

Step 4: y=8

y=8

8 0
3 years ago
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)
\\\\\\
inverse\implies x=2y+sin(y)\leftarrow f^{-1}(x)\leftarrow g(x)
\\\\\\
\textit{now, the "y" in the inverse, is really just g(x)}
\\\\\\
\textit{so, we can write it as }x=2g(x)+sin[g(x)]\\\\
-----------------------------\\\\

\bf \textit{let's use implicit differentiation}\\\\
1=2\cfrac{dg(x)}{dx}+cos[g(x)]\cdot \cfrac{dg(x)}{dx}\impliedby \textit{common factor}
\\\\\\
1=\cfrac{dg(x)}{dx}[2+cos[g(x)]]\implies \cfrac{1}{[2+cos[g(x)]]}=\cfrac{dg(x)}{dx}=g'(x)\\\\
-----------------------------\\\\
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
\\\\\\
-----------------------------\\\\
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
What is the value of x if 3x - 6 = 21 ?
Anestetic [448]

Answer:

<h2>x = 9</h2>

Step-by-step explanation:

3x-6=21\qquad\text{add 6 to both sides}\\\\3x-6+6=21+6\\\\3x=27\qquad\text{divide both sides by 3}\\\\\dfrac{3x}{3}=\dfrac{27}{3}\\\\x=9

3 0
3 years ago
Read 2 more answers
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