All categories

3 years ago

Solve the following equation for x: 5x + 3y = 15.

Mathematics

1 answer:

Blizzard [7]3 years ago

7 0

Answer:

x = negative three fifthsy + 3

Step-by-step explanation:

5x + 3y = 15

Subtract 3y from both the sides,

5x = 15 - 3y

Divide by 5 on both sides,

x = -3y/5 + 15/5

x = -3y/5 + 3

You might be interested in

Please help ASAP

Alisiya [41]

This is hard math is hard

5 0

3 years ago

Find - 2 2/6 + (5/6)

Bad White [126]

Answer:-3/2

Step-by-step explanation:

-2 2/6+5/6

-7/3 + 5/6

(2x-7+1x5) ➗ 6

(-14+5) ➗ 6

-9 ➗ 6

-9/6=-3/2

5 0

3 years ago

Plz help me with this

shtirl [24]

2x-7y=18 _(1)
-2x+4y=5

11y=23 , Y=23/11 sub in 1

2x-7×23/11=18

2x = 18×11 + 7×23

x = ( 198 + 161 )/2 = 359/2

5 0

4 years ago

Can somebody pls help me with this question I dont get it

alekssr [168]

Answer:

1. 195=6.50·c

2. 30

Step-by-step explanation:

195=6.5·c

Divide both sides by 6.5

30=c

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