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3 years ago

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The population of a city has increased by 35% since it was last measured. If the current population is 86,400, what was the prev

ious population?

1 answer:

Vsevolod [243]3 years ago

6 0

Answer:

Step-by-step explanation:

86,400*.35=30240

86400-30240=56160

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Nina and three friends at lunch at the cafeteria cafeteria. They decided to split the bill evenly the total bill was $17.84 how

Sedaia [141]

Answer: 4.46 for each

Step-by-step explanation: All that you have to do is do 17.84/4

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alexandr1967 [171]

The answer is -67.24

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3 years ago

Step 3: Using multi-step equations to solve problems a) You are using the equation 4(p - 7) = 44 to determine how many pictures

mrs_skeptik [129]

Answer:

p = 18

Explanation:

$\hookrightarrow \sf 4(p - 7) = 44$

$\sf change \ sides$

$\hookrightarrow \sf p-7\right}=\frac{44}{4}$

$\sf simplify$

$\hookrightarrow \sf p-7\right}= 11$

$\sf change \ sides$

$\hookrightarrow \sf p\right}= 11+7$

$\sf final \ answer$

$\hookrightarrow \sf p\right}= 18$

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3 years ago

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Acute, obtuse, or right?

dimaraw [331]

It would be obtuse.

Hope this helps :)

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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