All categories

3 years ago

A number,Y,is 2 less than half of a number X

Mathematics

1 answer:

zhenek [66]3 years ago

8 0

Answer: y = 0.5x - 2

Step-by-step explanation: 0.5x = half of a number X

You might be interested in

The Sweet Treat sells 40 lbs of gummy bears and 30 caramel apples every day. At this rate, how many lbs of gummy bears will be s

Sliva [168]

Answer:

8 pounds of gummy bears.

Step-by-step explanation:

To find this answer I did:

40/30 = 4/3

Then,

4/3 x 6 = 24/3

This simplified is 8.

To check:

8/6 x 5 = 40/30

3 0

3 years ago

Quadrilateral ABCD is an isosceles trapezoid. If || , AB = m + 6, CD = 3m + 2, BC = 3m, and AD = 7m – 16, solve for m.

pashok25 [27]

Answer:

m=2

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

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

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