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2 years ago

HELP!!!! geometry/trig pls help me thank u

Mathematics

1 answer:

patriot [66]2 years ago

7 0

Answer:

the slide it's self is 10 feet long

Step-by-step explanation:

it says the height is 8ft off the ground which means that the slide being at a 45 degree angle it has to be at least 2 ft taller than the ladder

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8. Why is it NOT useful to view science as a set of facts?

WITCHER [35]

Answer:
(assuming this isn't actually for math?) but one reason is bc science shouldn't be always taken as fact. science can be wrong.

Explanation:
not all but many scientists are skeptical and won't accept anything without proof, and even when things are accepted as right they can still be wrong later. example: humans once believed the planets revolved around the earth and that the earth was flat (even, some people still believe the earth is flat today....) but both those things are generally unaccepted today.

science shouldn't be viewed as fact always, because accepting everything as fact will lead to misassumptions and false information. it's like we should never take things for granted

3 0

2 years ago

Read 2 more answers

A parabola has vertex (2, 3) and contains the point (0, 0). find an equation that represents this parabola.

AfilCa [17]

First write it in vertex form :-

y= a(x - 2)^2 + 3 where a is some constant.

We can find the value of a by substituting the point (0.0) into the equation:-

0 = a((-2)^2 + 3

4a = -3

a = -3/4

so our equation becomes y = (-3/4)(x - 2)^2 + 3

7 0

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

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