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3 years ago

Please help me answer this and learn how to find the equation for line of best fit

Mathematics

1 answer:

const2013 [10]3 years ago

6 0

Answer:

y=-12/5x+62

Step-by-step explanation:

to solve this i turned two point form, 5,50 and 17 1/2,20, into slope intercept form by using the formula y-y1=(y2-y1/x2-x1)(x-x1), which when input with the data becomes y-50=(20-50/17 1/2-5)(x-5) which then becomes y-50=-12/5(x-5), then y-50=-12/5x+12, and finally y=-12/5x+62

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Solve the question 12 = 24x + 12.

Kisachek [45]

Answer:

The answer is 0.

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

Just need help with B and C <br><br><br> 30 POINTS

saw5 [17]

Answer:

b) 6/7

c) 14

Step-by-step explanation:

b) Scale Factor = Red/(Red+Blue) = 8/(8+6) = 8/14 = 6/7.

c) x = 8+6 = 14.

EXTRA TIP: If you aren't sure, think about wheter these triangles are isosceles(from the diagram). Hope this helps!

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2 years ago

G find the area of the surface over the given region. use a computer algebra system to verify your results. the sphere r(u,v) =

Svetach [21]

Presumably you should be doing this using calculus methods, namely computing the surface integral along $\mathbf r(u,v)$ .

But since $\mathbf r(u,v)$ describes a sphere, we can simply recall that the surface area of a sphere of radius $a$ is $4\pi a^2$ .

In calculus terms, we would first find an expression for the surface element, which is given by

$\displaystyle\iint_S\mathrm dS=\iint_S\left\|\frac{\partial\mathbf r}{\partial u}\times\frac{\partial\mathbf r}{\partial v}\right\|\,\mathrm du\,\mathrm dv$

$\dfrac{\partial\mathbf r}{\partial u}=a\cos u\cos v\,\mathbf i+a\cos u\sin v\,\mathbf j-a\sin u\,\mathbf k$
$\dfrac{\partial\mathbf r}{\partial v}=-a\sin u\sin v\,\mathbf i+a\sin u\cos v\,\mathbf j$
$\implies\dfrac{\partial\mathbf r}{\partial u}\times\dfrac{\partial\mathbf r}{\partial v}=a^2\sin^2u\cos v\,\mathbf i+a^2\sin^2u\sin v\,\mathbf j+a^2\sin u\cos u\,\mathbf k$
$\implies\left\|\dfrac{\partial\mathbf r}{\partial u}\times\dfrac{\partial\mathbf r}{\partial v}\right\|=a^2\sin u$

So the area of the surface is

$\displaystyle\iint_S\mathrm dS=\int_{u=0}^{u=\pi}\int_{v=0}^{v=2\pi}a^2\sin u\,\mathrm dv\,\mathrm du=2\pi a^2\int_{u=0}^{u=\pi}\sin u$
$=-2\pi a^2(\cos\pi-\cos 0)$
$=-2\pi a^2(-1-1)$
$=4\pi a^2$

as expected.

6 0

3 years ago

in the equation dy+3x=2, for what value of d is the graph of the equation parallel to the graph of x-3y=0? the y axis?

IRISSAK [1]

Answer:

-9

Step-by-step explanation:

The slope of the the solving equation is triple the slope of the given equation. Therefore, the y coefficient must also be triple of the other y coefficient.

8 0

2 years ago

If angle 0 is in quadrant I, what is the value of

Brilliant_brown [7]

Answer:

$the \: answer \: is \: c \: = - \frac{ \sqrt{14} }{5}$