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Nataly [62]
3 years ago
7

Who ever answers 3 and 4 gets 10 points

Mathematics
2 answers:
mel-nik [20]3 years ago
8 0

Answer:3 and 4

Step-by-step explanation:

Hitman42 [59]3 years ago
5 0
3) clear/specific

4) part of speech
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Find the projection of u = <–6, –7> onto v = <1, 1> a. <-13/2,-13/2> b. <39,91/2> c. <-13/1764,-13/17
Alexandra [31]
<h2>Answer:</h2>

a. <-13/2,-13/2>

<h2>Step-by-step explanation:</h2>

The projection of a vector u onto another vector v is given by;

proj_vu = (\frac{u.v}{|v|^2})v               ----------------(i)

Where;

u.v is the dot product of vectors u and v

|v| is the magnitude of vector v

Given:

u = <-6, -7>

v = <1, 1>

These can be re-written in unit vector notation as;

u = -6i -7j

v = i + j

<em>Now;</em>

<em>Let's find the following</em>

(i) u . v

u . v = (-6i - 7j) . (i + j)

u . v = (-6i) (1i) + (-7j)(1j)          [Remember that, i.i = j.j = 1]

u . v = -6 -7 = -13

(ii) |v|

|v| = \sqrt{(1)^2 + (1)^2}

|v| = \sqrt{2}

<em>Substitute these values into equation (i) as follows;</em>

proj_vu = [\frac{-13}{(\sqrt{2}) ^2}][i + j]

proj_vu = \frac{-13}{2} [i + j]

This can be re-written as;

proj_vu = \frac{-13}{2}i + \frac{-13}{2}j

proj_vu =

5 0
3 years ago
At one university, the students are given z-scores at the end of the each semester, rather than the traditional gpas. the mean a
vovikov84 [41]
 <span>a) 
2=(x-2.7)/.5 
(2)(.5) = x-2.7 
1=x-2.7 
x=3.7 

-2.5=(x-2.7)/.5 
(-2.5)(.5) = x-2.7 
-1.25 =x-2.7 
x=2.7-1.25 = 1.45 

b) 
95% of the students' score lie within 2 standard deviation of the mean. Therefore, the top 2.5% corresponds to a z value of 2. 
The limits are (2, infinity) --- top 2.5 %</span>
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4 years ago
I need help I can't find which one it is
lawyer [7]
The real number has to be 0
8 0
3 years ago
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The surface of a hill is modeled by z = 100 − 4 x 2 − 2 y 2 . When a group of hikers reach the point (-3,-2,56) it begins to sno
m_a_m_a [10]

Answer:

(-24, -8)

Step-by-step explanation:

Let us recall that when we have a function f

\large f:\mathbb{R}^2\rightarrow \mathbb{R}\\f(x,y)=z

<em>if the gradient of f at a given point (x,y) exists, then the gradient of f at this point (x,y) gives the direction of maximum rate of increasing and minus the gradient of f at this point gives the direction of maximum rate of decreasing</em>. That is

\large \nabla f=(\frac{\partial f}{\partial x},\frac{\partial f}{\partial y})

at the point (x,y) gives the direction of maximum rate of increasing

\large -\nabla f

at the point (x,y) gives the direction of maximum rate of decreasing

In this case we have

\large f(x,y)=100-4x^2-2y^2

and we want to find the direction of fastest speed of decreasing at the point (-3,-2)

\large \nabla f(x,y)=(-8x,-4y) \Rightarrow -\nabla f=(8x,4y)

at the point (-3,-2) minus the gradient equals

\large -\nabla f(-3,-2)=(-24,-8)

hence the vector (-24,-8) points in the direction with the greatest rate of decreasing, and they should start their descent in that direction.

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