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Elanso [62]
2 years ago
8

Find a vector function, r(t), that represents the curve of intersection of the two surfaces. The paraboloid z = 6x2 + y2 and the

parabolic cylinder y = 5x2.
Mathematics
1 answer:
LenKa [72]2 years ago
6 0

Answer:

\mathbf{r^{\to} (t) = ti^{\to} + 5t^2^{\to} j+(6t^2 + 25t^4) k ^{\to}}

Step-by-step explanation:

Given that:

The paraboloid surface z = 6x² + y² and the parabolic cylinder y = 5x²

Let assume that:

x = t

then from y = 5x², we have:

y = 5t²

Now replace y = 5t² and x = t into z = 6x² + y²

z = 6t² + (5t²)²

z = 6t² + 25t⁴

Hence, the curve of intersection is illustrated by the set of equations:

x = t, y = 5t², and z = 6t² + 25t⁴

As a vector equation:

\mathbf{r^{\to} (t) = ti^{\to} + 5t^2^{\to} j+(6t^2 + 25t^4) k ^{\to}}

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irina [24]

Answer:

32

Step-by-step explanation:

m-án=16, when n = 8

8 0
3 years ago
Write a sine function that has a midline of 2, an amplitude of 4 and a period of 11.
svetoff [14.1K]

Answer:

y = 4 sin(2π/11 x) + 2

Step-by-step explanation:

y = A sin(2π/T x + B) + C

where A is the amplitude,

T is the period,

B is the phase shift,

and C is the midline.

A = 4, T = 11, and C = 2.  We'll assume B = 0.

y = 4 sin(2π/11 x) + 2

3 0
3 years ago
How many ways can you choose a manager and assistant from 7-person task force? *​
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We are interested in selecting 2 people from the 7-person task force.

n = 7
r = 2

nPr = n! / (n-r)!

7!
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(7-2)!

=
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5!

7×6×5×4×3×2×1
------------------------
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5 0
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Solve for x. 1/15+x=3/10
ValentinkaMS [17]

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x=7/30

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7 0
3 years ago
Read 2 more answers
What system of inequalities is represented by the graph?
yarga [219]
As shown in the figure, we have two straight line. One of them has a negative slope and the other has a positive one. In two dimensions, the equation for non-vertical lines is often given in the slope-intercept form by:

y=mx+b

being m the slope of the line and <span>b the y-intercept of it.

On the other hand, if x = 0 then y = b.

First of all we will order the equations above without </span>inequalities<span> like this:

A. </span>y = 5x-1, y = 3x+4
<span>B. </span>y = 5x-1, y = -3x+4
C. y = 5x+1, y = -3x+4
D. y = 5x-1, y = -3x-4<span>

 As shown in the figure b = -1 for one straight and b = 4 for the second one. This values take place when x = 0. So, we discard C and D, because if x = 0, then:
</span>
For C, b = 1 and b = 4
For D, b = -1 and b = -4

Let's analyze A and B. So:

For A, m = 5 and m = 3
For B, m = 5 and m = -3

Therefore, we discard A because of the statement above.

Finally the answer is B. So, the inequalities are:  

(1) y\ \textless \ 5x-1
(2) 3x+y \geq 4

Let's prove this answer. We will take the point (2, 0) that is in the region in gray. So, substituting this point in the inequalities, we have:

(1) 0\ \textless \ 9
(2) 6 \geq 4

In fact, this is true.


7 0
3 years ago
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