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2 years ago

1. what is the perimeter of the triangle? 2. what is the area of the triangle?

Mathematics

1 answer:

jonny [76]2 years ago

7 0

Answer:

See below. A=45 p=39.64

Step-by-step explanation:

Hi = 9 squares (h)

Gh= 14 squares. (b)

Area of a triangle = 1/2 bxh

1/2 14 x 9

7x9= 45.

Area =45 squares.

Perimiter

A= side hi (9)

B= side gh (14)

C= side gi (unknown)

A^2 + b^2 = c^2

9^2 + 14^2 = 277

C^2 = 277

Square root of 277 =c

C length =16.64 squares

A+b+c = perimeter

9+14+16.64 = 39.64 squares.

Sorry I cannot see the units of measure in the photo

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For each of the following vector fields F , decide whether it is conservative or not by computing curl F . Type in a potential f

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The key idea is that, if a vector field is conservative, then it has curl 0. Equivalently, if the curl is not 0, then the field is not conservative. But if we find that the curl is 0, that on its own doesn't mean the field is conservative.

$\mathrm{curl}\vec F=\dfrac{\partial(5x+10y)}{\partial x}-\dfrac{\partial(-6x+5y)}{\partial y}=5-5=0$

We want to find $f$ such that $\nabla f=\vec F$ . This means

$\dfrac{\partial f}{\partial x}=-6x+5y\implies f(x,y)=-3x^2+5xy+g(y)$

$\dfrac{\partial f}{\partial y}=5x+10y=5x+\dfrac{\mathrm dg}{\mathrm dy}\implies\dfrac{\mathrm dg}{\mathrm dy}=10y\implies g(y)=5y^2+C$

$\implies\boxed{f(x,y)=-3x^2+5xy+5y^2+C}$

so $\vec F$ is conservative.

$\mathrm{curl}\vec F=\left(\dfrac{\partial(-2y)}{\partial z}-\dfrac{\partial(1)}{\partial y}\right)\vec\imath+\left(\dfrac{\partial(-3x)}{\partial z}-\dfrac{\partial(1)}{\partial z}\right)\vec\jmath+\left(\dfrac{\partial(-2y)}{\partial x}-\dfrac{\partial(-3x)}{\partial y}\right)\vec k=\vec0$

Then

$\dfrac{\partial f}{\partial x}=-3x\implies f(x,y,z)=-\dfrac32x^2+g(y,z)$

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so $\vec F$ is not conservative.

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