Finish the pattern 5 10 30 60 180

Slope is 4 and (4,6) is on the line

erik [133]

Answer:

y-6=4(x-4)

Step-by-step explanation:

y-y1=m(x-x1)

y-6=4(x-4)

5 0

3 years ago

Brian has a math quiz every six days and a science quiz every four days on February 15 he had both tests when will he have both

Valentin [98]

A common denominator of 6 and 4 is 12. That is the Least Common Denominator, the LCD. In 12 days it would be the 27th of February. There you go.

3 0

3 years ago

Read 2 more answers

A sample of marble has a volume of 6 cm3 and a density of 2.76 g/cm3. What is its mass?

Alenkasestr [34]

To solve for the mass of a substance is to simply get the product of volume and density, that is:

mass = density * volume

So calculating for mass:

mass = (2.76 g / cm^3) * 6 cm^3

mass = 16.56 g

3 0

3 years ago

Read 2 more answers

Ez20 chaarecter long

Likurg_2 [28]

Answer:

24 oz has the lowest cost per ounce.

Step-by-step explanation:

12 oz ---> $2.49

1 oz --> $2.49/12

1 oz --> $0.2075

14 oz --> $3.15

1 oz --> $3.15/14

1 oz --> $0.225

20 oz --> $5

1 oz --> $5/20

1 oz --> $0.25

24 oz --> $4.56

1 oz --> $4.56/24

1 oz --> $0.19 ..........this is the lowest cost per ounce

36 oz --> $7.2

1 oz --> $7.2/36

1 oz --> $0.2

5 0

3 years ago

Where does the helix r(t) = cos(πt), sin(πt), t intersect the paraboloid z = x2 + y2? (x, y, z) = What is the angle of intersect

Colt1911 [192]

Answer:

Intersection at (-1, 0, 1).

Angle 0.6 radians

Step-by-step explanation:

The helix r(t) = (cos(πt), sin(πt), t) intersects the paraboloid

z = x2 + y2 when the coordinates (x,y,z)=(cos(πt), sin(πt), t) of the helix satisfy the equation of the paraboloid. That is, when

$\bf (cos(\pi t), sin(\pi t), t)$

But

$\bf cos^2(\pi t)+sin^2(\pi t)=1$

so, the helix intersects the paraboloid when t=1. This is the point

(cos(π), sin(π), 1) = (-1, 0, 1)

The angle of intersection between the helix and the paraboloid is the angle between the tangent vector to the curve and the tangent plane to the paraboloid.

The tangent vector to the helix in t=1 is

r'(t) when t=1

r'(t) = (-πsin(πt), πcos(πt), 1), hence

r'(1) = (0, -π, 1)

A normal vector to the tangent plane of the surface

$\bf z=x^2+y^2$

at the point (-1, 0, 1) is given by

$\bf (\frac{\partial f}{\partial x}(-1,0),\frac{\partial f}{\partial y}(-1,0),-1)$

where

$\bf f(x,y)=x^2+y^2$

since

$\bf \frac{\partial f}{\partial x}=2x,\;\frac{\partial f}{\partial y}=2y$

so, a normal vector to the tangent plane is

(-2,0,-1)

Hence, a vector in the same direction as the projection of the helix's tangent vector (0, -π, 1) onto the tangent plane is given by

$\bf (0,-\pi,1)-((0,-\pi,1)\bullet(-2,0,-1))(-2,0,1)=(0,-\pi,1)-(-2,0,1)=(2,-\pi,0)$

The angle between the tangent vector to the curve and the tangent plane to the paraboloid equals the angle between the tangent vector to the curve and the vector we just found.

But we now

$\bf (2,-\pi,0)\bullet(0,-\pi,1)=\parallel(2,-\pi,0)\parallel\parallel(0,-\pi,1)\parallel cos\theta$

where

$\bf \theta$ = angle between the tangent vector and its projection onto the tangent plane. So

$\bf \pi^2=(\sqrt{4+\pi^2}\sqrt{\pi^2+1})cos\theta\rightarrow cos\theta=\frac{\pi^2}{\sqrt{4+\pi^2}\sqrt{\pi^2+1}}=0.8038$

and

$\bf \theta=arccos(0.8038)=0.6371\;radians$

7 0

3 years ago