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2 years ago

Find the length of side x. Show steps please!

Mathematics

1 answer:

yawa3891 [41]2 years ago

8 0

Answer:

$\large{x=7.4438\ cm\approx7.4\ cm}$

Step-by-step explanation:

Use tangent:

$tangent=\dfrac{opposite}{adjacent}$

We have:

$\alpha=28^o,\ opposite=x,\ adjacent=14\ cm$

$\tan28^o\approx0.5317$

Substitute:

$\dfrac{x}{14}=0.5317$ <em>multiply both sides by 14</em>

$x=7.4438\ cm$

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Which of the following is equal to square root 9•16

natulia [17]

Answer:

Letter B

9.16

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3 years ago

Let X be the number of Heads in 10 fair coin tosses. (a) Find the conditional PMF of X, given that the first two tosses both lan

inn [45]

Answer:

Follows are the solution to the given point:

Step-by-step explanation:

For option A:

In the first point let z be the number of heads which is available on the first two trails of tosses so, the equation is:

$P(X=k | z=2 ) = \begin{pmatrix} 10-2\\ k-2\end{pmatrix} (\frac{1}{2})^{k-2} (\frac{1}{2})^{10-k}$

$= \begin{pmatrix} 8\\k-2\end{pmatrix} (\frac{1}{2})^{k-2} (\frac{1}{2})^{10-k} \ \ \ \ \ \ \ \ \ \ \\\\$

$k= 2,3..........10$ For option B:

$P(X=k | X \geq 2 ) = \sum^{10}_{i=2} \begin{pmatrix} 10-i\\ k-i\end{pmatrix} (\frac{1}{2})^{k-i} (\frac{1}{2})^{10-k}$

$k= 2,3, 4.........10$

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3 years ago

Compare the values of the underline digits. 2,402 and 64,513 the value of 4 in _______________ is _______ times the values of

Ludmilka [50]

64,513; 10; 2,402. The 4 in 64,513 is essentially 4,000 whereas the 4 in 2,402 is only 400. 400(10)=4000.

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3 years ago

How do I do this greatest common factor

zaharov [31]

The greatest common factor is 5

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2 years 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

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