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2 years ago

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What is the length of the rectangular plot of land shown? Use pencil and paper. How are the lengths of the legs of a right tria

ngle related to the lengths of the sides of a rectangle?
The length of the rectangular plot of land is ___ ft?

1 answer:

IrinaVladis [17]2 years ago

6 0

Answer:256960ft

Step-by-step explanation: LENGTHXWIDTH . So you would want to multiply 440x584 and that is 256960 So it is 256960ft I think.

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The wall below is 90% covered with graffiti. The entire wall is 228 ft2. How much square footage does the graffiti cover?

Rama09 [41]

205.2 square feet

Convert 90% to a decimal. You get .9 Multiply 228 by .9 to find what 90% of 228 is. You should get 205.2.

6 0

3 years ago

Brian needs to start planning for an upcoming move. He is choosing two different size boxes to pack the majority of his stuff. T

GenaCL600 [577]

Solution:

Given,

Volume of the-

First cube box=1728 in³

Second cube box=13824in³

We know that volume of cube =a³

So for the first box

a³= 1728 in³

a= 12 in

Now, surface area of the first box

6a²= 6×(12)²

= 864 in²

and for the second box

b³= 13824 in³

b= 24 in

Now, the surface area of the second box

6b²= 6×(24)²

= 3456 in²

Now, the ratio of the surface area of first box to the second box will be

864:3456

=1:4

The surface area of the larger of two boxes is

3456 in²

5 0

2 years ago

Eleven thirds times seven thirds times five thirds

STatiana [176]

Answer:

three hundred and eight five twenty sevenths

4 0

2 years ago

Two arithmetic means between 3 and 24 are -

defon

The value of two arithmetic means which are inserted between 3 and 24 are 24/9 and 75/9.

<h3>What is arithmetic mean?</h3>

Arithmetic mean is the mean or average which is equal to the ratio of sum of all the group numbers to the total numbers.

The two arithmetic means between 3 and 24 are has to be inserted.

3, A₂, A₃, 24

All the four numbers are in arithmetic progression. The nth terms of AM can be found using the following formula:

t(n)=a(n-1)d

Here, d is the common difference a is the first terms and n is the total term. The first term, a=3 and t₄=24. Thus, the common difference is;

$t(n)=a(n-1)d\\t(4)=3(4-1)d\\24=3(3)d\\24=9d\\d=\dfrac{24}{9}$

The second and 3rd term are:

$A₂=3+\dfrac{24}{9}=\dfrac{51}{9}\\ A₃=\dfrac{51}{9}+\dfrac{24}{9}=\dfrac{75}{9}$

Thus, the value of two arithmetic means which are inserted between 3 and 24 are 24/9 and 75/9.

Learn more about the arithmetic mean here;

brainly.com/question/14831274

#SPJ1

8 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

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

And the magnitude of the derivative of unit tangent vector is

$||\boldsymbol{T}'||=2\sqrt{\cos ^2\left(x\right)+\sin ^2\left(x\right)} =2$

The curvature is

$\kappa=\frac{||d\boldsymbol{T}/dt||}{ds/dt}=\frac{2}{2} =1$

The tangential component of acceleration is given by the formula

$a_{\boldsymbol{T}}=\frac{d^2s}{dt^2}$

We know that $\frac{ds}{dt}=|| \boldsymbol{r}'(t)}||$ and $||\boldsymbol{r}'(t)}||=2$

$\frac{d}{dt}\left(2\right)\: = 0$ so

$a_{\boldsymbol{T}}=0$

The normal component of acceleration is given by the formula

$a_{\boldsymbol{N}}=\kappa (\frac{ds}{dt})^2$

We know that $\kappa=1$ and $\frac{ds}{dt}=2$ so

$a_{\boldsymbol{N}}=1 (2)^2=4$

3 0

3 years ago