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3 years ago

3.76+(y-1/2)/2 need help noww homework

Mathematics

1 answer:

Zepler [3.9K]3 years ago

4 0

Answer:

351+50y/100

Step-by-step explanation:

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Exspress the ratio from the simplest form 72:36

Vedmedyk [2.9K]

First, find the GCF. For these two numbers, it is 36. (Because 36 times 1 is 36 and 36 times 2 is 72)
Divide both sides by this, which gives you a
2:1 ratio

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3 years ago

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A store sell 5 bundle of firewood for $ 40 .which equation could be used to find the total cost c if a customer wants to buy 3 b

dimaraw [331]

that firewood be expensive tho

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3 years ago

Shawn picked some vegetables from his garden.he picked 2.234 pounds of corn 2.205 pounds of tomatoes and 2.25 pounds of potatoes

Tatiana [17]

Answer: The correct answer is H.

Step-by-step explanation: The correct order of the weights from largest to smallest is shown in choice H. The correct order is 2.25, 2.234, 2.205.

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3 years ago

Nike and Kane were born on consecutive years and on same date. Nick is younger. The sum of their age is 11. Find the age of the

klemol [59]

Nick is 5 and Kane is 6.

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3 years ago

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Consider the following region R and the vector field F. a. Compute the two-dimensional curl of the vector field. b. Evaluate bo

Shalnov [3]

Looks like we're given

$\vec F(x,y)=\langle-x,-y\rangle$

which in three dimensions could be expressed as

$\vec F(x,y)=\langle-x,-y,0\rangle$

and this has curl

$\mathrm{curl}\vec F=\langle0_y-(-y)_z,-(0_x-(-x)_z),(-y)_x-(-x)_y\rangle=\langle0,0,0\rangle$

which confirms the two-dimensional curl is 0.

It also looks like the region $R$ is the disk $x^2+y^2\le5$ . Green's theorem says the integral of $\vec F$ along the boundary of $R$ is equal to the integral of the two-dimensional curl of $\vec F$ over the interior of $R$ :

$\displaystyle\int_{\partial R}\vec F\cdot\mathrm d\vec r=\iint_R\mathrm{curl}\vec F\,\mathrm dA$

which we know to be 0, since the curl itself is 0. To verify this, we can parameterize the boundary of $R$ by

$\vec r(t)=\langle\sqrt5\cos t,\sqrt5\sin t\rangle\implies\vec r'(t)=\langle-\sqrt5\sin t,\sqrt5\cos t\rangle$

$\implies\mathrm d\vec r=\vec r'(t)\,\mathrm dt=\sqrt5\langle-\sin t,\cos t\rangle\,\mathrm dt$

with $0\le t\le2\pi$ . Then

$\displaystyle\int_{\partial R}\vec F\cdot\mathrm d\vec r=\int_0^{2\pi}\langle-\sqrt5\cos t,-\sqrt5\sin t\rangle\cdot\langle-\sqrt5\sin t,\sqrt5\cos t\rangle\,\mathrm dt$

$=\displaystyle5\int_0^{2\pi}(\sin t\cos t-\sin t\cos t)\,\mathrm dt=0$

7 0

3 years ago

3.76+​(y-​1/​2)/​2 need help noww homework

3.76+(y-1/2)/2 need help noww homework