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3 years ago

14.36 rounded to the nearest tenth

Mathematics

2 answers:

Alina [70]3 years ago

4 0

Answer:

14.4

Step-by-step explanation:

So we have to find the tenth value which is 3

So we have to use the value before it to round it up or down

Round up = 5 and above

Round down = 4 and below

The number before it is 6 which is greater than 5 so we Round up.

14.4

Finger [1]3 years ago

3 0

Answer:

14.4

Step-by-step explanation:

6 is closer to 40 than to 30

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3 years ago

Help with num 5 please.

Makovka662 [10]

Let's apply the derivative to both sides with respect to x. We'll use the chain rule.

$y = \ln\left(x + \sqrt{x^2+a^2}\right)\\\\\\\frac{dy}{dx} = \frac{d}{dx}\left[\ln\left(x + \sqrt{x^2+a^2}\right)\right]\\\\\\\frac{dy}{dx} = \frac{1}{x+\sqrt{x^2+a^2}}*\frac{d}{dx}\left[x + \sqrt{x^2+a^2}\right]\\\\\\\frac{dy}{dx} = \frac{1}{x+(x^2+a^2)^{1/2}}*\left(1 + 2x*\frac{1}{2}(x^2+a^2)^{-1/2}\right)\\\\\\\frac{dy}{dx} = \frac{1}{x+(x^2+a^2)^{1/2}}*\left(1 + x*\frac{1}{(x^2+a^2)^{1/2}}\right)\\\\\\\frac{dy}{dx} = \frac{1}{x+W}*\left(1 + x*\frac{1}{W}\right)\\\\\\$

where W = (x^2+a^2)^(1/2) = sqrt(x^2+a^2)

Let's simplify that a bit.

$\frac{dy}{dx} = \frac{1}{x+W}*\left(1 + x*\frac{1}{W}\right)\\\\\\\frac{dy}{dx} = \frac{1}{x+W}*\left(\frac{W}{W} + \frac{x}{W}\right)\\\\\\\frac{dy}{dx} = \frac{1}{x+W}*\frac{W+x}{W}\\\\\\\frac{dy}{dx} = \frac{1}{W}\\\\\\\frac{dy}{dx} = \frac{1}{\sqrt{x^2+a^2}}\\\\\\$

This concludes the first part of what your teacher wants.

-----------------------

Now onto the second part.

In this case, a = 3 so a^2 = 3^2 = 9.

Recall that if $g(x) = \frac{d}{dx}\left[f(x)\right]$

then $\displaystyle \int g(x)dx = f(x)+C$

We can say that f(x) is the antiderivative of g(x). The C is some constant.

So,

$\displaystyle \displaystyle \int \frac{1}{\sqrt{x^2+a^2}}dx = \ln\left(x + \sqrt{x^2+a^2}\right)+C\\\\\\\displaystyle \displaystyle \int \frac{1}{\sqrt{x^2+3^2}}dx = \ln\left(x + \sqrt{x^2+3^2}\right)+C\\\\\\\displaystyle \displaystyle \int \frac{1}{\sqrt{x^2+9}}dx = \ln\left(x + \sqrt{x^2+9}\right)+C\\\\\\$

Now let g(x) = ln(x + sqrt(x^2+9) ) + C

Then compute

g(0) = ln(3)+C
g(4) = ln(9) = ln(3^2) = 2*ln(3)+C

Therefore,

$\displaystyle \displaystyle \int \frac{1}{\sqrt{x^2+9}}dx = \ln\left(x + \sqrt{x^2+9}\right)+C\\\\\\\displaystyle \displaystyle \int \frac{1}{\sqrt{x^2+9}}dx = g(x)\\\\\\\displaystyle \displaystyle \int_{0}^{4} \frac{1}{\sqrt{x^2+9}}dx = g(4) - g(0)\\\\\\\displaystyle \displaystyle \int_{0}^{4} \frac{1}{\sqrt{x^2+9}}dx = (2\ln(3)+C)-(\ln(3)+C)\\\\\\\displaystyle \displaystyle \int_{0}^{4} \frac{1}{\sqrt{x^2+9}}dx = \ln(3)\\\\\\$

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Extra info:

Interestingly, WolframAlpha says that the result is $\sinh^{-1}\left(\frac{4}{3}\right)$ , but we can rewrite that into ln(3) because inverse hyperbolic sine is defined as