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german
3 years ago
14

PLEASE HELP! The question is in the picture, i need to answer URGENTLY!

Mathematics
1 answer:
Andrej [43]3 years ago
4 0
Kim is 43 Jordan is 50

200÷4=50
50-7=43

200÷4-7=x
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Answer: is 5.2

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3 erasers cost $4.41 which equations would help determine the cost of 4 erasers
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In a dilation the image is always similar to its pre image true or false
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18. Let f be a function with domain the set of all real numbers and having the following
Mashcka [7]

The derivative of the given function is f'(x) = k f(x) where k= \lim_{h \to 0} \frac{f(h)-1}{h}.

<h3>What is the derivative of a function?</h3>

Let f be a function defined on a neighborhood of a real number a. Then f is said to be differentiable or derivable at 'a' if \lim_{x \to a} \frac{f(x)-f(a)}{x-a} exists finitely. The limit is called the derivative or differential coefficient of f at 'a'. It is denoted by f'(a).

If f is differentiable at 'a', then

f'(a) = \lim_{h\to 0} \frac{f(a+h)-f(a)}{h}

<h3>Calculation:</h3>

The given properties are:

(i) f(x + y) = f(x)f(y) for all real numbers x and y.

(ii) \lim_{h \to 0} \frac{f(h)-1}{h} = k; where k is a nonzero real number.

Then, the derivative of the function f(x) is,

f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}

From property (i), f(x + h) = f(x)f(h)

On substituting,

f'(x) = \lim_{h \to 0} \frac{f(x)f(h)-f(x)}{h}

      = \lim_{h \to 0} \frac{f(x)[f(h) - 1]}{h}

From property (ii), \lim_{h \to 0} \frac{f(h)-1}{h} = k;

f'(x) = \lim_{h \to 0} \frac{f(x)[f(h) - 1]}{h}

      = f(x). \lim_{h \to 0} \frac{f(h)-1}{h}

      = f(x). k

      = kf(x)

Therefore, f'(x) = k f(x); where f'(x) exists for all real numbers of x.

Learn more about the derivative of a function here:

brainly.com/question/5313449

#SPJ9

6 0
2 years ago
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