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2 years ago

The perimeter of the fig

Mathematics

2 answers:

adoni [48]2 years ago

6 0

Add all the sides 11+4+5+16+10=46 so the perimeter of this figure is 46 Hope this Helps :)

mr_godi [17]2 years ago

5 0

4+16+10+5+11+6=52yd
Hope I helped

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Mkey [24]

1. is D number 2 is A

3 0

2 years ago

Rita is starting a running program. The table shows the total number of miles she runs in different weeks. What is the equation

prisoha [69]

Answer:

The best fit line is $y = 1.671x + 4.699$

Step-by-step explanation:

We are given,

The table representing the number of miles run in different weeks is,

Week Miles Run

1 5

2 8

4 13

6 15

8 19

10 20

Using the linear regression calculator, we see that,

<h3>The equation of the line best fit for the data is $y = 1.671x + 4.699$ .</h3>

6 0

3 years ago

Read 2 more answers

Suppose that \nabla f(x,y,z) = 2xyze^{x^2}\mathbf{i} + ze^{x^2}\mathbf{j} + ye^{x^2}\mathbf{k}. if f(0,0,0) = 2, find f(1,1,1).

lesya [120]

The simplest path from (0, 0, 0) to (1, 1, 1) is a straight line, denoted $C$ , which we can parameterize by the vector-valued function,

$\mathbf r(t)=(1-t)(\mathbf i+\mathbf j+\mathbf k)$

for $0\le t\le1$ , which has differential

$\mathrm d\mathbf r=-(\mathbf i+\mathbf j+\mathbf k)\,\mathrm dt$

Then with $x(t)=y(t)=z(t)=1-t$ , we have

$\displaystyle\int_{\mathcal C}\nabla f(x,y,z)\cdot\mathrm d\mathbf r=\int_{t=0}^{t=1}\nabla f(x(t),y(t),z(t))\cdot\mathrm d\mathbf r$

$=\displaystyle\int_{t=0}^{t=1}\left(2(1-t)^3e^{(1-t)^2}\,\mathbf i+(1-t)e^{(1-t)^2}\,\mathbf j+(1-t)e^{(1-t)^2}\,\mathbf k\right)\cdot-(\mathbf i+\mathbf j+\mathbf k)\,\mathrm dt$

$\displaystyle=-2\int_{t=0}^{t=1}e^{(1-t)^2}(1-t)(t^2-2t+2)\,\mathrm dt$

Complete the square in the quadratic term of the integrand: $t^2-2t+2=(t-1)^2+1=(1-t)^2+1$ , then in the integral we substitute $u=1-t$ :

$\displaystyle=-2\int_{t=0}^{t=1}e^{(1-t)^2}(1-t)((1-t)^2+1)\,\mathrm dt$

$\displaystyle=-2\int_{u=0}^{u=1}e^{u^2}u(u^2+1)\,\mathrm du$

Make another substitution of $v=u^2$ :

$\displaystyle=-\int_{v=0}^{v=1}e^v(v+1)\,\mathrm dv$

Integrate by parts, taking

$r=v+1\implies\mathrm dr=\mathrm dv$

$\mathrm ds=e^v\,\mathrm dv\implies s=e^v$

$\displaystyle=-e^v(v+1)\bigg|_{v=0}^{v=1}+\int_{v=0}^{v=1}e^v\,\mathrm dv$

$\displaystyle=-(2e-1)+(e-1)=-e$

So, we have by the fundamental theorem of calculus that

$\displaystyle\int_C\nabla f(x,y,z)\cdot\mathrm d\mathbf r=f(1,1,1)-f(0,0,0)$

$\implies-e=f(1,1,1)-2$

$\implies f(1,1,1)=2-e$

3 0

3 years ago

Josea went out to dinner to an Indian restaurant. The total bill was $38 .she wanted to leave a 20% tip

jekas [21]

Answer:

$20?

Step-by-step explanation:

6 0

2 years ago

A 16 foot ladder rests against the side of a house and the base of the ladder is 4 feet away. How high above the ground is the t

PtichkaEL [24]

Answer:

15.5 ft

Step-by-step explanation:

The geometry of the problem can be modeled by a right triangle with hypotenuse 16 ft and one side length of 4 ft. If x represents the height of the ladder on the building, then the Pythagorean theorem tells us ...

x^2 + (4 ft)^2 = (16 ft)^2

x^2 = 240 ft^2 . . . . . . subtract 16 ft^2

x ≈ 15.5 ft . . . . . . . . . . take the square root

The top of the ladder is about 15.5 ft above the ground.

8 0

2 years ago