The table shows the weight of bunches of bananas and the price of each bunch

Mathematics

1 answer:

skad [1K]3 years ago

6 0

Wow good to know. But I’m not seeing a picture

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Can someone please help me with this

pogonyaev

The geometric mean you would use is C

6 0

3 years ago

The equation above shows how temperature F, measured in degrees Fahrenheit, relates to a temperature C, measured in degrees Cels

Murrr4er [49]

Answer:

I (even though you said not to worry about dis question) will try and answerrrrrrrrrrrrr-

I would say- hmmm- B.) II only aka A temperature increase of 1 degree Celsius is equivalent to a temperature increase of 1.8 degrees Fahrenheit.

6 0

2 years ago

Let f(x) = (x ^ 2 - 5x + 4)/(x - 1) and g(x) = 2 ^ (x - 3). For which value of x does f(g(x)) have a discontinuity ?

polet [3.4K]

Answer:

There is a discontinuity at x=1

Step-by-step explanation:

6 0

3 years ago

Complete the generic rectangle (PLEASE HELP)

navik [9.2K]

I believe the area of the blank rectangle, the biggest one, is 2800. The whole rectangle all together is 3225.

Explanation:

Since the length of one of the rectangles is 70, the rectangle with an area of 210 will have a side length of 3. This means the rectangle with an area of 15 has a side length of 5. <em>That</em> means the rectangle with an area of 200 has a side length of 40. Therefore, the blank rectangle has an area of 2800, as it's 2 side lengths are 70 and 40.

4 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