Your answer is -2 as the axis of symmetry is at -b/2a. So if you were to expand the equation, you would get y= x^2/4 +x -6 where b is 1 and 2a is ½ . Therefore -1 / ½ gives you -2

7 0

4 years ago

The planets move around the sun in elliptical orbits with the sun at one focus. The point in the orbit at which the planet is cl

gladu [14]

Consider the attached ellipse. Let the sun be at the right focus. Then perihelion is at right vertex on the x-axis and aphelion is at the left vertex on the x-axis.

The distances:

from perihelion to the sun in terms of ellipse is a-c;
from aphelion to the sun in terms of ellipse is a+c.

Then

$\left\{\begin{array}{l}a-c=741,000,000\\a+c=817,000,000\end{array}\right.$

Add these two equations:

$2a=1,558,000,000 \\ \\a=779,000,000$

and subtract first equation from the second:

$2c=76,000,000 \\ \\c=38,000,000.$

Note that $b=\sqrt{a^2-c^2},$ thus

$b=\sqrt{779,000,000^2-38,000,000^2}=\sqrt{605,397,000,000,000,000}.$

The equation for the planet's orbit is

$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1\Rightarrow \dfrac{x^2}{606,841,000,000,000,000}+\dfrac{y^2}{605,397,000,000,000,000}=1.$

7 0

3 years ago

Principal = $8100; rate = 17%; time = 7 years What is the amount of interest earned?

miskamm [114]

<span>it must be nnnnnnnnnnnnnnnnnnnnnnnnn
4.37%

</span>

4 0

3 years ago

How do you find the derivative of 1/logx?

valentina_108 [34]

The derivative of 1/logx is With the chain rule.

1log(x)=log(x)−1 is ,= -1xlog(x)2 .

The by-product of logₐ x (log x with base a) is 1/(x ln a). Here, the thrilling issue is that we have "ln" withinside the by-product of "log x". Note that "ln" is referred to as the logarithm (or) it's miles a logarithm with base "e".

The by-product of 1/log x is -1/x(log x)^2. Note that 1/logx is the reciprocal of log.

$With the chain rule. \\ 1log(x)=log(x)−1 \\ So you follow the chain rule to the electricity after which to the log: \\ ddxlog(x)−1=−1⋅log(x) \\ \\−2⋅ddxlog(x) \\ =−1log(x)21x \\ =−1xlog(x)2 .$