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3 years ago

Find the domain of ( (a+b/b) − (a/a+b))÷( (a+b/a) − (b/a+b) )

Mathematics

2 answers:

Alexandra [31]3 years ago

7 0

Answer:

Hence the domain is given as,b is such that b is a member of all real numbers,except b=0,a=0, a=-b

Step-by-step explanation:

The domain refers to values for which the expression is defined.

This implies that, the denominators are not equal to zero.

$(\frac{a + b}{b} - \frac{a}{a + b}) \div ( \frac{a + b}{a} - \frac{b}{a + b})$

$(\frac{(a + b)(a + b)-ab}{b(a + b)})\div(\frac{(a + b)(a + b)-ab}{a(a + b)})$

$\frac{(a + b)(a + b)-ab}{b(a + b)} \times\frac{a(a + b)}{(a + b)(a + b)-ab}$

$\implies \frac{a}{b}$

Hence the domain is given as,b is such that b is a member of all real numbers,except b=0,a=0, and a=-b

lora16 [44]3 years ago

6 0

Answer:a/b

Step-by-step explanation:

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<span>Carl has 3 bags in total. One backpack weighs 4 kg and the rest two checking bags have the equal weight. The total weight of 3 bags is given to be 35 kg.

Let the weight of each checking bag is w kg. So we can write:

2 x (Weight of a checking bag) + Weight of Backpack = 35

Using the values, we get:

2w+ 4 = 35

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3 years ago

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interpret r(t) as the position of a moving object at time t. Find the curvature of the path and determine thetangential and norm

Igoryamba

Answer:

The curvature is $\kappa=1$

The tangential component of acceleration is $a_{\boldsymbol{T}}=0$

The normal component of acceleration is $a_{\boldsymbol{N}}=1 (2)^2=4$

Step-by-step explanation:

To find the curvature of the path we are going to use this formula:

$\kappa=\frac{||d\boldsymbol{T}/dt||}{ds/dt}$

where

$\boldsymbol{T}}$ is the unit tangent vector.

$\frac{ds}{dt}=|| \boldsymbol{r}'(t)}||$ is the speed of the object

We need to find $\boldsymbol{r}'(t)$ , we know that $\boldsymbol{r}(t)=cos \:2t \:\boldsymbol{i}+sin \:2t \:\boldsymbol{j}+ \:\boldsymbol{k}$ so

$\boldsymbol{r}'(t)=\frac{d}{dt}\left(cos\left(2t\right)\right)\:\boldsymbol{i}+\frac{d}{dt}\left(sin\left(2t\right)\right)\:\boldsymbol{j}+\frac{d}{dt}\left(1)\right\:\boldsymbol{k}\\\boldsymbol{r}'(t)=-2\sin \left(2t\right)\boldsymbol{i}+2\cos \left(2t\right)\boldsymbol{j}$

Next , we find the magnitude of derivative of the position vector

$|| \boldsymbol{r}'(t)}||=\sqrt{(-2\sin \left(2t\right))^2+(2\cos \left(2t\right))^2} \\|| \boldsymbol{r}'(t)}||=\sqrt{2^2\sin ^2\left(2t\right)+2^2\cos ^2\left(2t\right)}\\|| \boldsymbol{r}'(t)}||=\sqrt{4\left(\sin ^2\left(2t\right)+\cos ^2\left(2t\right)\right)}\\|| \boldsymbol{r}'(t)}||=\sqrt{4}\sqrt{\sin ^2\left(2t\right)+\cos ^2\left(2t\right)}\\\\\mathrm{Use\:the\:following\:identity}:\quad \cos ^2\left(x\right)+\sin ^2\left(x\right)=1\\\\|| \boldsymbol{r}'(t)}||=2\sqrt{1}=2$

The unit tangent vector is defined by

$\boldsymbol{T}}=\frac{\boldsymbol{r}'(t)}{||\boldsymbol{r}'(t)||}$

$\boldsymbol{T}}=\frac{-2\sin \left(2t\right)\boldsymbol{i}+2\cos \left(2t\right)\boldsymbol{j}}{2} =\sin \left(2t\right)+\cos \left(2t\right)$

We need to find the derivative of unit tangent vector

$\boldsymbol{T}'=\frac{d}{dt}(\sin \left(2t\right)\boldsymbol{i}+\cos \left(2t\right)\boldsymbol{j}) \\\boldsymbol{T}'=-2\cdot(\sin \left(2t\right)\boldsymbol{i}+\cos \left(2t\right)\boldsymbol{j})$