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2 years ago

A room has an area of 121 ft2 but carpeting is only sold in m2. How much carpeting is needed to carpet the room?

Mathematics

2 answers:

stepladder [879]2 years ago

8 0

The correct answer is D. 11.24 m2

kykrilka [37]2 years ago

7 0

The correct answer for the question that is being presented above is this one: "D.)11.24 m2."
First, we need to know the conversion between feet to meters.
1 meter = 3.28 feet
1 meter^2 = 10.7584 ft^2

Given the value of 121 ft^2,
= 121 ft^2 * (1 m^2 / 10.7584 ft^2)
= 11.2470 m^2

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A portfolio consisting of four stocks is expected to produce returns of minus9%, 11%, 13% and 17%, respectively, over the next f

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Answer:

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Step-by-step explanation:

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2 years ago

Lin multiplies 7/8 times a number. The product is less than 7/8. Which could be Lin’s number?

RSB [31]

Answer:

So, the number that Lin multiplied by 7/8 MUST be less than 1. This is evident, as any positive number multiplied by a number less than 1 will decrease, doesn't matter if its a negative or positive number.

Step-by-step explanation:

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2 years ago

a direct variation function contains the points ( -8, -6 ) and (12,9 ) . Which equation represents the function ?

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Answer:

y = $\frac{3}{4}$ x

Step-by-step explanation:

Given that the quantities vary directly then the equation relating them is

y = kx ← k is the constant of variation

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2 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

2 years ago