Is anyone good at math investigations? Can anyone help me to do this investigation for me please?

PLEASE HELP!!! Which equation is relevant to 4^x^+^3=64

Aleksandr [31]

I believe it’s the second option 2^2x+6=2^6

6 0

3 years ago

A cleaner recommends mixing 1½ cup of cleaner for every 12 cups of water. What is the ratio of cleaner to water in simplest form

guapka [62]

We know that

1½ cup of cleaner-------> (1*2+1)/2-----> 3/2 cup of cleaner

the ratio of cleaner to water=(3/2)/12---> 3/(2*12)--> 3/24--> 1/8

the answer is
1/8

6 0

4 years ago

PLEASE HELP

Vilka [71]

9514 1404 393

Answer:

y = -x +3

m = -1

b = 3

Step-by-step explanation:

To solve the given equation for y, divide by the coefficient of y.

(-3y)/(-3) = (3x -9)/(-3)

y = -x +3

__

The slope is the x-coefficient, M = -1.

__

The y-intercept is the added constant, B = 3.

__

Both equations are graphed in the attachment. Texture has been added to the original so you can see the graphs are the same line.

4 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

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