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2 years ago

Line m passes through C(-2, 0) and D(1, -3). What is the equation of the line m in standard form?

1 answer:

8 0

From general equation of a line;
y=mx+c
m= $\frac{change in y}{change in x}$
m= $\frac{0--3}{-2-1}$
m=-1
y=-x+c
Finding value of c;
using point (-2,0)
0=-(-2)+c
c=-2
y=-x-2

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5( x + 7) - 3 ( x - 4) = 7x + 2 what is x?

S_A_V [24]

Answer:

x=9

Step-by-step explanation:

5( x + 7) - 3 ( x - 4) = 7x + 2

Distribute

5x+35 - 3x +12 = 7x +2

2x +47 = 7x+2

Subtract 2x from each side

2x+47-2x =7x -2x+2

47 = 5x+2

Subtract 2 from each side

47-2= 5x+2-2

45= 5x

Divide by 5

45/5 =5x/5

9 =x

8 0

2 years ago

Read 2 more answers

1. Find the volume of a sphere with a radius of 10 inches. Use 3.14 for pi.

dolphi86 [110]

Answer: $1256in^{2}$

Step-by-step explanation:

The formula to find the radius of a sphere is $4(pi)r^{2}$

3.14*4*100=1256

I hope this helped

7 0

2 years ago

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A bouqet had 10 flower and sold it for 90$ which is the rate of $_ per flower

lana [24]

Answer:

$9 per flower

Step-by-step explanation:

divide cost ($90) by the number of items (10 flowers)

6 0

2 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

2 years ago

Identify the vertical asymptotes of f(x) = 2 over quantity x squared plus 3 x minus 10.

zysi [14]

The function is:
f ( x ) = 2 / ( x² + 3 x - 10 )
x² + 3 x - 10 = x² + 5 x - 2 x - 10 = x ( x + 5 ) - 2 ( x + 5 ) =
= ( x + 5 ) ( x - 2 )
x + 5 = 0 => x = - 5
x - 2 = 0 => x = 2
Answer:
The vertical asymptotes are: x = - 5 and x = 2.

3 0

2 years ago