Answer:
Therefore the latitude closest to the equator is 6.49°(approx).
Step-by-step explanation:
Given the path of a solar ellipse is modeled by
![f(t)=0.00223t^2-0.558t+41.395](https://tex.z-dn.net/?f=f%28t%29%3D0.00223t%5E2-0.558t%2B41.395)
where f(t) is the latitude in degrees south of the equator at t minutes.
We know that ,
The minimum or maximum value of a function y(t) =at²+bt+c
is when ![t=-\frac{b}{2a}](https://tex.z-dn.net/?f=t%3D-%5Cfrac%7Bb%7D%7B2a%7D)
In this case a= 0.00223, b= - 0.558 and c= 41.395.
The minimum value of f(t)
when
=125.11
Therefore the latitude closest to the equator is
f(125.11)=0.00223(125.11)²-0.558×125.11+41.395
≈6.49
This is distributive property and to do distributive property, you distribute what is outside of the parenthesis to the numbers inside the parenthesis.
How-to-do:
2(8x-7)
2(8x) - 2(7)
16x-14
1/2 times 3/10 equals to 3/20
Answer:
Solving
we get 2g+3
Option B is correct.
Step-by-step explanation:
We need to solve ![(4g^2-9) \div (2g-3)](https://tex.z-dn.net/?f=%284g%5E2-9%29%20%5Cdiv%20%282g-3%29)
We know that ![a^2-b^2=(a-b)(a+b)](https://tex.z-dn.net/?f=a%5E2-b%5E2%3D%28a-b%29%28a%2Bb%29)
So,
can be written as ![(2g)^2-(3)^2](https://tex.z-dn.net/?f=%282g%29%5E2-%283%29%5E2)
So, using above formula we will get:
![4g^2-9\\=(2g^2)-(3)^2 \\Using \ formula a^2-b^2 = (a-b)(a+b)\\=(2g-3)(2g+3)](https://tex.z-dn.net/?f=4g%5E2-9%5C%5C%3D%282g%5E2%29-%283%29%5E2%20%5C%5CUsing%20%5C%20formula%20a%5E2-b%5E2%20%3D%20%28a-b%29%28a%2Bb%29%5C%5C%3D%282g-3%29%282g%2B3%29)
So, numerator will become
![\frac{(2g-3)(2g+3)}{(2g-3)} \\Simplifying:\\=2g+3](https://tex.z-dn.net/?f=%5Cfrac%7B%282g-3%29%282g%2B3%29%7D%7B%282g-3%29%7D%20%5C%5CSimplifying%3A%5C%5C%3D2g%2B3)
So, solving
we get 2g+3
Option B is correct.
Answer:
Option 2 and Option 5
Step-by-step explanation: