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3 years ago

A line passes through the point (4,9) and has a slope of -3/2 What is the y-intercept?

Mathematics

2 answers:

mina [271]3 years ago

6 0

The y intercept is 15 Hope this helps

Yuki888 [10]3 years ago

4 0

The y-intercept is positive 15

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Help ASAP please!!! First answer and correct answer gets the brain...

EleoNora [17]

Answer:

U = 40/3

Hope this helps :-)

3 0

3 years ago

I need help with this

melomori [17]

N'(-2,2)
G'(-2,-4)
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5 0

3 years ago

Combine the like terms to create an equivelant expression -3x-6(-1)

Elza [17]

Answer:

-3x+6

Step-by-step explanation:

-3x-6(-1)

First, multiplying two negatives equals a positive: (-)×(-)=(+). Then,

multiply the numbers -3×+6×1 = -3x+6. Then, you got the answer and the answer is -3x+6

3 0

3 years ago

Solve the equation 0.5p -3.45= -1.2.

ddd [48]

Answer:

4.5

Step-by-step explanation:

0.5p-3.45=-1.2

add 3.45 to -1.2 which would leave the equation to look like

0.5p=2.25

then divide 0.5 to both sides cancelling 0.5 leaving the answer to be

p=4.5

6 0

3 years ago

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