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PSYCHO15rus [73]
3 years ago
12

Can someone help me with these questions please?

Mathematics
1 answer:
natulia [17]3 years ago
7 0
7. is (4x+9)/7=f(x) because you switch the x and y and then solve for y.

f(x)=y
y=(7x-9)/4
x=(7y-9)/4
4x=7y-9
4x+9=7y
(4x+9)/7=y
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Ernesto buys a leaf blower price that $100 shipping and handling are an additional 10% of the price how much shipping and handli
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10% of 100 is 10 so 110
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Simplify the expression (7x² - 3x) + (11x² - x - 7) A. 18x² + 2x + 7 B. 4x² - 4x - 7 C. 18x² - 2x - 7 D. 18x² - 4x - 7
IRINA_888 [86]

Answer: D.

Step-by-step explanation:

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A storage container is in the shape of a rectangular prism. The container has a length of 10 feet, a width of 8 feet, and a heig
SVETLANKA909090 [29]

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592

Step-by-step explanation:

A=2(wl+hl+hw)

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nirvana33 [79]
The answer is 22 years old.
You set p = 2d then you solve for D.  Then you do 2+p and set that equal to 3(d-5+2) then you solve.  You keep going until you get P's age which is 22 and annie's age is 6.
6 0
3 years ago
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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 <em>tangent vector</em> 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, <em>a vector in the same direction as the projection of the helix's tangent vector (0, -π, 1) onto the tangent plane </em>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
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