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MrRissso [65]
3 years ago
11

What is the solution to the equation below

Mathematics
1 answer:
Tasya [4]3 years ago
8 0

Answer:

Theres no question-

Step-by-step explanation:

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For exercises 4–9, find the value of x. round to the nearest tenth.[
WITCHER [35]

Answer:

Step-by-step explanation:

where is exercise 4-9?

or at least give us the equation

3 0
3 years ago
Elena needs to cut a square piece of wood with an area of 7 square inches. How long should the sides of the square be, rounded t
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3.5 because you take 7 and divide by 2 since you have to multiply you base times width to get area I think I might be wrong
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3 years ago
If Mike has m pencils and Sarah has three times as many pencils as Mike, how many more pencils does Sarah has?
svetlana [45]

Answer:

3m

Step-by-step explanation:

the answer is 3m becaus three represents the times. And m represents mike.

4 0
3 years ago
Read 2 more answers
152 kids went to six flags, 65% had costumes on how many didn't have costume
Romashka-Z-Leto [24]

152(1 - .65) = 53.2
so about 53 kids didn't have costumes.
5 0
3 years ago
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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
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