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serious [3.7K]
3 years ago
6

Which are the solutions of the quadratic equation? x^2 = –5x – 3

Mathematics
1 answer:
ollegr [7]3 years ago
3 0
<span>x=<span><span><span>−5</span>2</span>+<span><span><span><span>12</span><span>√13</span></span><span> or </span></span>x</span></span></span>=<span><span><span>−5</span>2</span>+<span><span><span>−1</span>2</span><span>√<span>13</span></span></span></span>
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Strong moderate or weak correlation r=-0.91, r=0.82, r=-0.49, r=0.26, r=0.54, r=-0.18
Mariulka [41]

So one key thing to remember here is that the direction of the correlation is irrelevant, that is it does not matter if your correlation is + or - what matters is how close that number is to 1.0.

To help you out here are the ranges of correlation strength

  • 0.70. A strong relationship
  • 0.50. A moderate relationship
  • 0.30. A weak relationship

So to start off with 0.26 and 0.18 are very small correlations so you'd call those weak correlations.

Let me know if you need help doing the other ones? It should be simple enough with the data I gave you :)


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3 years ago
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Which power is equivalent to the expression (g-12)-3?
olasank [31]

Answer: g^36

Step-by-step explanation:

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A cylinder has radius r and height h. a. How many times greater is the surface area of a cylinder when both dimensions are multi
kondor19780726 [428]
To find it directly

A = 2 pi r h

A is proportional to rh

factor 2

A is multiplied by 2 * 2 = 4

factor 3

3 * 3 =9

factor 5

5 * 5 = 25

factor 10 

10 * 10 = 100

(b) the increase in A by factor x is x^2

c(20)^2 = 400 
3 0
3 years ago
Divide: 15m^2n + 20mn/5mn
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The answer is
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Hope that helps, have a nice night or day <4
8 0
3 years ago
) Set up a double integral for calculating the flux of F=3xi+yj+zk through the part of the surface z=−5x−2y+2 above the triangle
Fynjy0 [20]

The surface (call it S) is a triangle with vertices at the points

x=0,y=0\implies z=2\implies(0,0,2)

x=0,y=2\implies z=-2\implies(0,2,-2)

x=2,y=0\implies z=-8\implies(2,0,-8)

Parameterize S by

\vec s(u,v)=(1-v)(2,0,-8)+v\bigg((1-u)(0,2,-2)+u(0,0,2)\bigg)=(2-2v,2v-2uv,-8+6v+4uv)

with 0\le u\le1 and 0\le v\le1. Take the normal vector to S to be

\vec s_v\times\vec s_u=(20v,8v,4v)

Then the flux of \vec F across S is

\displaystyle\iint_S\vec F(x,y,z)\cdot\mathrm d\vec S=\int_0^1\int_0^1\vec F(x(u,v),y(u,v),z(u,v))\cdot(\vec s_v\times\vec s_u)\,\mathrm du\,\mathrm dv

=\displaystyle\int_0^1\int_0^1(6-6v,2v-2uv,-8+6v+4uv)\cdot(20v,8v,4v)\,\mathrm du\,\mathrm dv

=\displaystyle8\int_0^1\int_0^1(11v-10v^2)\,\mathrm du\,\mathrm dv=\boxed{\frac{52}3}

8 0
4 years ago
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