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2 years ago

Guys I wouldn't be asking if I had absolutely no clue on how to solve this but I am so please please help me!!!!

Mathematics

1 answer:

nalin [4]2 years ago

8 0

Its simple, so these are two "y=mx+b" form lines that are limited by the signs on the right. Your first line f(x)=-x-4 has to be between -3 and 1 on the x-axis.

You would plot a point at (0,-4) like normal, then go down one, right one as the slope is -1. you would make the dot at 1 though unfilled as x does not equal to 1. It is less than 1. º Should look like the dot I just inserted.

The second line would be plotted, at (0,-6). You would then make a horizontal line between 1 and 5. º-------• (Should look like this)

IMPORTANT:

FILLED DOT - EQUAL TO

UNFILLED DOT - Not equal to

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Consider F and C below. F(x, y) = x4y5i + x5y4j, C: r(t) = t3 − 2t, t3 + 2t , 0 ≤ t ≤ 1 (a) Find a function f such that F = ∇f.

vovikov84 [41]

$\mathbf f(x,y)=x^4y^5\,\mathbf i+x^5y^4\,\mathbf j$

We want a scalar function $f(x,y)$ whose gradient is equivalent to the vector field. That means

$\dfrac{\partial f}{\partial x}=x^4y^5\implies f(x,y)=\dfrac{x^5y^5}5+g(y)$
$\dfrac{\partial f}{\partial y}=x^5y^4\implies x^5y^4=x^5y^4+\dfrac{\mathrm dg}{\mathrm dy}$
$\implies\dfrac{\mathrm dg}{\mathrm dy}=0\implies g(y)=C$

So (a)

$f(x,y)=\dfrac{x^5y^5}5+C$

which means (b)

$\displaystyle\int_{\mathcal C}\mathbf f\cdot\mathrm d\mathbf r=f(\mathbf r(1))-f(\mathbf r(0))=f(-1,3)-f(0,0)=-\dfrac{243}5-0=-\dfrac{243}5$

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2 years ago

The sides of a triangle are in the ratio 7:7:11. what is the measure of the largest angle and classify the triangle by its sides

zheka24 [161]

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3 years ago

Tan(– 225°) + sec(-180°)

Alecsey [184]

Ok thanks you guys 279$

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2 years ago

Think About the Process The length of a rectangle is twice the width. The area of the rectangle

natima [27]

Answer:

length of each square= $\sqrt{43}$ ≈ 6.6

the length of the rectangle = 2 $\sqrt{43}$ ≈ 13.1

width of the rectangle ≈ $\sqrt{43}$ ≈ 6.6

Step-by-step explanation:

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Answer: 8.73

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3 years ago