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3 years ago

What the solution set of the inequality 15y - 9 < 36

Mathematics

1 answer:

Alika [10]3 years ago

6 0

15y - 9 < 36

15y - 9 + 9 < 36 + 9

15y < 45

15y ÷ 15 < 45 ÷ 15

y < 3

Hope this helps!!
~Kiwi

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F⃗ (x,y)=−yi⃗ +xj⃗ f→(x,y)=−yi→+xj→ and cc is the line segment from point p=(5,0)p=(5,0) to q=(0,2)q=(0,2). (a) find a vector pa

DerKrebs [107]

a. Parameterize $C$ by

$\vec r(t)=(1-t)(5\,\vec\imath)+t(2\,\vec\jmath)=(5-5t)\,\vec\imath+2t\,\vec\jmath$

with $0\le t\le1$ .

b/c. The line integral of $\vec F(x,y)=-y\,\vec\imath+x\,\vec\jmath$ over $C$ is

$\displaystyle\int_C\vec F(x,y)\cdot\mathrm d\vec r=\int_0^1\vec F(x(t),y(t))\cdot\frac{\mathrm d\vec r(t)}{\mathrm dt}\,\mathrm dt$

$=\displaystyle\int_0^1(-2t\,\vec\imath+(5-5t)\,\vec\jmath)\cdot(-5\,\vec\imath+2\,\vec\jmath)\,\mathrm dt$

$=\displaystyle\int_0^1(10t+(10-10t))\,\mathrm dt$

$=\displaystyle10\int_0^1\mathrm dt=\boxed{10}$

d. Notice that we can write the line integral as

$\displaystyle\int_C\vecF\cdot\mathrm d\vec r=\int_C(-y\,\mathrm dx+x\,\mathrm dy)$

By Green's theorem, the line integral is equivalent to

$\displaystyle\iint_D\left(\frac{\partial x}{\partial x}-\frac{\partial(-y)}{\partial y}\right)\,\mathrm dx\,\mathrm dy=2\iint_D\mathrm dx\,\mathrm dy$

where $D$ is the triangle bounded by $C$ , and this integral is simply twice the area of $D$ . $D$ is a right triangle with legs 2 and 5, so its area is 5 and the integral's value is 10.

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2 years ago

Hi everyone! I have a question!

Tju [1.3M]

Answer:

7π

Step-by-step explanation:

SO first find the area of the whole circle

which has a radius of 3

a=πr^2

So 3^2=9

9π

Now the two circles which are the same in area due to having the same radius

so if you plug in the 1 for radius the answer would just be π or 1π

now multiply that by 2 becuase there are 2 identical circles

π*2=2π

Now subtract

9π-2π=7π

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3 years ago

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ruslelena [56]

Answer:

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26 = 13x

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