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2 years ago

5x(4-2)2/(-4-(-8)) HELP PLEASE

Mathematics

2 answers:

IRISSAK [1]2 years ago

8 0

Answer:

when i evalute and simplify it i get 5x

so i hope this helps

Step-by-step explanation:

mr Goodwill [35]2 years ago

5 0

Answer:

=5x

Step-by-step explanation:

5x(4−2)(2)

−4−(−8)

=5x :)

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2 years ago

Read 2 more answers

Hey, I need help with questions 1 and 2, if anyone can help, please? thanks!

Eva8 [605]

The correct works are:

$Blue(s + h) = 2s^2 + 4sh + 2h^2 + 3$ .
$\frac{Blue(s + h) - Blue(s)}{h} = 4s + 2h$

<h3>Function Notation</h3>

The function is given as:

$Blue(s) = 2s^2 + 3$

The interpretation when Steven is asked to calculate Blue(s + h) is that:

Steven is asked to find the output of the function Blue, when the input is s + h

So, we have:

$Blue(s + h) = 2(s + h)^2 + 3$

Evaluate the exponent

$Blue(s + h) = 2(s^2 + 2sh + h^2) + 3$

Expand the bracket

$Blue(s + h) = 2s^2 + 4sh + 2h^2 + 3$

So, the correct work is:

$Blue(s + h) = 2s^2 + 4sh + 2h^2 + 3$

<h3>Simplifying Difference Quotient</h3>

In (a), we have:

$Blue(s + h) = 2s^2 + 4sh + 2h^2 + 3$

$Blue(s) = 2s^2 + 3$

The difference quotient is represented as:

$\frac{f(x + h) - f(x)}{h}$

So, we have:

$\frac{Blue(s + h) - Blue(s)}{h} = \frac{2s^2 + 4sh + 2h^2 + 3 - 2s^2 - 3}{h}$

Evaluate the like terms

$\frac{Blue(s + h) - Blue(s)}{h} = \frac{4sh + 2h^2}{h}$

Evaluate the quotient

$\frac{Blue(s + h) - Blue(s)}{h} = 4s + 2h$

Hence, the correct work is:

$\frac{Blue(s + h) - Blue(s)}{h} = 4s + 2h$

Read more about function notations at:

brainly.com/question/13136492

4 0

1 year ago

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

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