(8.88×10^4) + (7.48×10^4)

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Irina18 [472]

Your answer is, Negative Correlation

It is showing negative correlation because the points are decreasing and they are all pin pointed in around the same area.

Hope this helps!

Please Mark me Brainiest!

6 0

2 years ago

What is the distance between -4.32 and 0 on the number line?

saveliy_v [14]

B looks the most reasonable

5 0

3 years ago

Read 2 more answers

PLEASE HELP I NEED HELP !!!!!!!! The graph of a line passes through the two points (-2, 1) and (2, 1). What is the equation of t

MrRissso [65]

Answer:

y - 1 = 0

Step-by-step explanation:

We know that Equation of a line is given by,

y - y₁ / x - x₁ = y₂ - y₁ / x₂ - x₁

Let A(-2,1) and B(2,1)

x₁ = -2 ; x₂ = 2 ; y₁ = 1 ; y₂ = 1

Substituting the values,

y - 1 / x - (-2) = 1 - 1 / 2 - (-2)

y - 1 / x + 2 = 0

y - 1 = 0

7 0

3 years ago

Read 2 more answers

Find an equation of the tangent plane to the given parametric surface at the specified point.

Neko [114]

Answer:

Equation of tangent plane to given parametric equation is:

$\frac{\sqrt{3}}{2}x-\frac{1}{2}y+z=\frac{\pi}{3}$

Step-by-step explanation:

Given equation

$r(u, v)=u cos (v)\hat{i}+u sin (v)\hat{j}+v\hat{k}$ ---(1)

Normal vector tangent to plane is:

$\hat{n} = \hat{r_{u}} \times \hat{r_{v}}\\r_{u}=\frac{\partial r}{\partial u}\\r_{v}=\frac{\partial r}{\partial v}$

$\frac{\partial r}{\partial u} =cos(v)\hat{i}+sin(v)\hat{j}\\\frac{\partial r}{\partial v}=-usin(v)\hat{i}+u cos(v)\hat{j}+\hat{k}$

Normal vector tangent to plane is given by:

$r_{u} \times r_{v} =det\left[\begin{array}{ccc}\hat{i}&\hat{j}&\hat{k}\\cos(v)&sin(v)&0\\-usin(v)&ucos(v)&1\end{array}\right]$

Expanding with first row

$\hat{n} = \hat{i} \begin{vmatrix} sin(v)&0\\ucos(v) &1\end{vmatrix}- \hat{j} \begin{vmatrix} cos(v)&0\\-usin(v) &1\end{vmatrix}+\hat{k} \begin{vmatrix} cos(v)&sin(v)\\-usin(v) &ucos(v)\end{vmatrix}\\\hat{n}=sin(v)\hat{i}-cos(v)\hat{j}+u(cos^{2}v+sin^{2}v)\hat{k}\\\hat{n}=sin(v)\hat{i}-cos(v)\hat{j}+u\hat{k}\\$

at u=5, v =π/3

$=\frac{\sqrt{3} }{2}\hat{i}-\frac{1}{2}\hat{j}+\hat{k}$ ---(2)

at u=5, v =π/3 (1) becomes,

$r(5, \frac{\pi}{3})=5 cos (\frac{\pi}{3})\hat{i}+5sin (\frac{\pi}{3})\hat{j}+\frac{\pi}{3}\hat{k}$

$r(5, \frac{\pi}{3})=5(\frac{1}{2})\hat{i}+5 (\frac{\sqrt{3}}{2})\hat{j}+\frac{\pi}{3}\hat{k}$

$r(5, \frac{\pi}{3})=\frac{5}{2}\hat{i}+(\frac{5\sqrt{3}}{2})\hat{j}+\frac{\pi}{3}\hat{k}$

From above eq coordinates of r₀ can be found as:

$r_{o}=(\frac{5}{2},\frac{5\sqrt{3}}{2},\frac{\pi}{3})$

From (2) coordinates of normal vector can be found as

$n=(\frac{\sqrt{3} }{2},-\frac{1}{2},1)$

Equation of tangent line can be found as:

$(\hat{r}-\hat{r_{o}}).\hat{n}=0\\((x-\frac{5}{2})\hat{i}+(y-\frac{5\sqrt{3}}{2})\hat{j}+(z-\frac{\pi}{3})\hat{k})(\frac{\sqrt{3} }{2}\hat{i}-\frac{1}{2}\hat{j}+\hat{k})=0\\\frac{\sqrt{3}}{2}x-\frac{5\sqrt{3}}{4}-\frac{1}{2}y+\frac{5\sqrt{3}}{4}+z-\frac{\pi}{3}=0\\\frac{\sqrt{3}}{2}x-\frac{1}{2}y+z=\frac{\pi}{3}$

5 0

3 years ago

PLEASE HELP

GuDViN [60]

The equation of a circle with center (h,k) and radius r is
(x-h)²+(y-k)²=r²
so
given center is (3,-4)
(x-3)²+(y-(-4))²=r²
(x-3)²+(y+4)²=r²
find r² by subsitutiong the point taht it passes through which is (6,5)

so
(6-3)²+(5+4)²=r²
3²+9²=r²
9+81=r²
90=r²
so

(x-3)²+(y+4)²=90 is da equation

4th equation

7 0

3 years ago