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3 years ago

Which of the following is considered a model for enzyme action

Mathematics

1 answer:

sergij07 [2.7K]3 years ago

8 0

Answer choices? I can try and figure it out if there are choices

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A coin weights about 14^-2 pounds. Find the weight of 1000 of coins. Round your answer to the nearest tenth.

yuradex [85]

Answer:

5.1

Step-by-step explanation:

14^-2*1000

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2 years ago

Helppppppppppppp meeee plsssssss.

maria [59]

Answer:

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

What is the equation of a line that passes through (4, 7) and has a slope of −14?

deff fn [24]

Slope = -1/4
(4, 7)

y - 7 = -1/4 (x - 4)
y - 7 = -1/4x + 1
y = -1/4 + 8

answer is C

5 0

2 years ago

What is the CIRCUMFERENCE of a circle with a radius of 18m?

Lyrx [107]

Answer:

the first one is 113.04m

the second one is 615.44 cm

the third one is 530.66 ft

the fourth one is 163.28 in

the fifth one is 31.4 m

the sixth one is 28.26 m

the seventh one is 314

the eightth one is 50,24

finally finished:)))

7 0

2 years ago

Need help with Calculus 1 inverse trig functions

lidiya [134]

Answer:

$\displaystyle y'=3\frac{1+\frac{x}{\sqrt{1+x^2}}}{2+2x^2+2x\sqrt{1+x^2}}$

Step-by-step explanation:

<u>The Derivative of a Function</u>

The derivative of f, also known as the instantaneous rate of change, or the slope of the tangent line to the graph of f, can be computed by the definition formula

$\displaystyle f'(x)=\lim\limits_{\Delta x \rightarrow 0}\frac{f(x+\Delta x)-f(x)}{\Delta x}$

There are tables where the derivative of all known functions are provided for an easy calculation of specific functions.

The derivative of the inverse tangent is given as

$\displaystyle (tan^{-1}u)'=\frac{u'}{1+u^2}$

Where u is a function of x as provided:

$y=3tan^{-1}(x+\sqrt{1+x^2})$

If we set

$u=(x+\sqrt{1+x^2})$

Then

$\displaystyle u'=1+\frac{2x}{2\sqrt{1+x^2}}$

$\displaystyle u'=1+\frac{x}{\sqrt{1+x^2}}$

Taking the derivative of y

$y'=3[tan^{-1}(x+\sqrt{1+x^2})]'$

Using the change of variables

$\displaystyle y'=3[tan^{-1}u]'=3\frac{u'}{1+u^2}$

$\displaystyle y'=3\frac{u'}{1+u^2}=3\frac{1+\frac{x}{\sqrt{1+x^2}}}{1+(x+\sqrt{1+x^2})^2}$

Operating

$\displaystyle y'=3\frac{1+\frac{x}{\sqrt{1+x^2}}}{1+x^2+2x\sqrt{1+x^2}+1+x^2}$

$\boxed{\displaystyle y'=3\frac{1+\frac{x}{\sqrt{1+x^2}}}{2+2x^2+2x\sqrt{1+x^2}}}$

8 0

2 years ago