All categories

3 years ago

An archeology dig site is shaped like a square. The archeologists make a map of the

Mathematics

2 answers:

vlabodo [156]3 years ago

8 0

Anwser: 25

step by step explanation:

ale4655 [162]3 years ago

6 0

Answer:

Step-by-step explanation:

The vertices B and D are (7,3) and (2,-2). So the area is 5*5 or 25

You might be interested in

WILL MARK BRAINLIEST + 10 POINTS.

Irina-Kira [14]

It is <span>B) x-axis only
if i am wrong sorry

</span>

5 0

3 years ago

_384<br> What is the square root of 384

sp2606 [1]

Answer:

19.5959179423

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

a wildlife manager determines that the function y=200x1.07* represents the deer population at a state park x years after the pop

Kobotan [32]

Answer:

Table:

0,200

4, 262

8,344

12, 450

Population after 12 years is 450 deer

Initial population is 200 deer.

Step-by-step explanation:

You take the x in the table and plug it in as an exponent so you will need a calculator or multiple the 1.07 by that many times so 1.07 12 times multipled by itself. and the initial is 0 years so anything to the exponent of 0 is 1 so that show i got 200 for the x=0

8 0

3 years ago

What is the defination of the focus when referring to a parabola?

Len [333]

Answer:

Focus of a Parabola. A parabola is set of all points in a plane which are an equal distance away from a given point and given line. The point is called the focus of the parabola and the line is called the directrix. The focus lies on the axis of symmetry of the parabola.

Step-by-step explanation:

6 0

3 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

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