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2 years ago

FOR A BRAINLIST PLS HURRY

Mathematics

1 answer:

insens350 [35]2 years ago

8 0

Answer:

y = x

Step-by-step explanation:

line has a slope 1 and passes through the origin . So the equation is y = x

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The vectors x and y are unit vectors that make and angle of 30 degrees with each other. Calculate the value of | 2x + y | and th

ICE Princess25 [194]

Answer: the value of | 2x + y | = 1.39

the direction of 2x + y = 21°

Step-by-step explanation:

Given that the vectors x and y are unit vectors that make and angle of 30 degrees with each other.

Let assume that unit vector x is along the positive x-axis, and unit vector y is at +30°.

Therefore

2x-y=

2< 1,0 > - < cos(30),sin(30) >

=<2-(√2)/2, 0-1/2>

and

|2x-y|=√((2-(√2)/2)² + (-1/2)^2)

=1.3862

Therefore the value of | 2x + y | = 1.39

=atan((-1/2)÷(2-(√2)/2))

=atan(1.29289/-0.5)

=-0.369 radians

= -21.143°

Therefore the direction of 2x + y is

21°

5 0

2 years ago

PLEASE HELP AND HELP ME UNDERSTAND THIS PROBLEM

Neporo4naja [7]

Answer:

Step-by-step explanation:

Because It it the most accurate

I hope this helps! :)(

3 0

2 years ago

Use lagrange multipliers to find the point on the plane x â 2y + 3z = 6 that is closest to the point (0, 2, 4).

Arisa [49]

The distance between a point $(x,y,z)$ on the given plane and the point (0, 2, 4) is

$\sqrt{f(x,y,z)}=\sqrt{x^2+(y-2)^2+(z-4)^2}$

but since $\sqrt{f(x,y,z)}$ and $f(x,y,z)$ share critical points, we can instead consider the problem of optimizing $f(x,y,z)$ subject to $x-2y+3z=6$ .

The Lagrangian is

$L(x,y,z,\lambda)=x^2+(y-2)^2+(z-4)^2+\lambda(x-2y+3z-6)$

with partial derivatives (set equal to 0)

$L_x=2x+\lambda=0\implies x=-\dfrac\lambda2$
$L_y=2(y-2)-2\lambda=0\implies y=2+\lambda$
$L_z=2(z-4)+3\lambda=0\implies z=4-\dfrac{3\lambda}2$
$L_\lambda=x-2y+3z-6=0\implies x-2y+3z=6$

Solve for $\lambda$ :

$x-2y+3z=-\dfrac\lambda2-2(2+\lambda)+3\left(4-\dfrac{3\lambda}2\right)=6$
$\implies2=7\lambda\implies\lambda=\dfrac27$

which gives the critical point

$x=-\dfrac17,y=\dfrac{16}7,z=\dfrac{25}7$

We can confirm that this is a minimum by checking the Hessian matrix of $f(x,y,z)$ :

$\mathbf H(x,y,z)=\begin{bmatrix}f_{xx}&f_{xy}&f_{xz}\\f_{yx}&f_{yy}&f_{yz}\\f_{zx}&f_{zy}&f_{zz}\end{bmatrix}=\begin{bmatrix}2&0&0\\0&2&0\\0&0&2\end{bmatrix}$

$\mathbf H$ is positive definite (we see its determinant and the determinants of its leading principal minors are positive), which indicates that there is a minimum at this critical point.

At this point, we get a distance from (0, 2, 4) of

$\sqrt{f\left(-\dfrac17,\dfrac{16}7,\dfrac{25}7\right)}=\sqrt{\dfrac27}$

8 0

2 years ago

The following is a set of hypotheses, some information from one or more samples, and a standard error from a randomization distr

Tanya [424]

Complete Question

The following is a set of hypotheses, some information from one or more samples, and a standard error from a randomization distribution. Test $H_o : p = 0.28$ vs $H a : p < 0.28$ when the sample has n 900 , and $\^ p = 0.216$ with $SE= 0.01$ . Find the value of the standardized z -test statistic. Round your answer to two decimal places.