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3 years ago

Write m=-3 and (3,-1) in pointe slope form

Mathematics

1 answer:

Lady_Fox [76]3 years ago

6 0

(
3
,
−
3
)
is
y
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3
x
−
3
=
1
3
.

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What is the equation of the circle with center (0, 0) that passes through the point (–4, –6)?

eimsori [14]

Answer:

Option 4: x^2+y^2 = 52

Step-by-step explanation:

Given

Centre at origin

Point on circle = (-4, -6)

The distance between origin and point on circle will be the radius of the circle

So,

$r = \sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}}\\r= \sqrt{(-4-0)^2+(-6-0)^2} \\r = \sqrt{(-4)^2+(-6)^2}\\r = \sqrt{16+36} \\r = \sqrt{52}$

As the center is at origin the standard equation will be:

$x^2+y^2 = r^2\\$

Putting the value of r

$x^2+y^2 = (\sqrt{52})^2\\x^2+y^2 = 52$

Hence, last option i.e. Option 4 is correct ..

8 0

3 years ago

Can somebody do this budget form for me?

Rama09 [41]

Answer:8by-step explanation:

4 0

3 years ago

A metal cube

Ierofanga [76]

Answer: 2.73 cm

Step-by-step explanation:

Given that :

Side (a) of cube = 4.4cm

Volume of a cube (V) = a^3

V = 4.4^3

V = 85.184cm^3

Therefore, volume of the sphere made = 85.184cm^3

Volume of sphere = 4/3 πr^3

Where r = radius

85.184 = (4/3)*(22/7)*r^3

85.184 = (88/21)*r^3

85.184 = 4.1905 * r^3

r^3 = 85.184 / 4.1905

r^3 = 20.327884

Take the cube root of both sides

r = 2.73 cm

5 0

3 years ago

The equation for the circle below is x^2+y^2=64 what is the length of the circle’s radius

FromTheMoon [43]

Answer:

Step-by-step explanation:

The standard form of an equation of a circle:

$(x-h)^2+(y-k)^2=r^2$

(h, k) - center

r - radius

We have the equation:

$x^2+y^2=64\\\\(x-0)^2+(y-0)^2=8^2$

Therefore the center is (0, 0) and the radius r = 8

3 0

3 years ago

Evaluate the line integral using the fundamental theorem of line integrals. use a computer algebra system to verify your results

mario62 [17]

$\dfrac{\partial f}{\partial x}=\cos x\sin y\implies f(x,y)=\sin x\sin y+g(y)$

$\dfrac{\partial f}{\partial y}=\sin x\cos y=\sin x\cos y+\dfrac{\mathrm dg}{\mathrm dy}$
$\implies\dfrac{\mathrm dg}{\mathrm dy}=0\implies g(y)=C$

$f(x,y)=\sin x\sin y+C$

$\displaystyle\int_{\mathcal C}\cos x\sin y\,\mathrm dx+\sin x\cos y\,\mathrm dy=\int_{\mathcal C}\mathbf f\cdot\mathrm d\mathbf r=f\left(\frac{3\pi}2,\frac\pi2\right)-f(0,-\pi)=-1$

7 0

3 years ago