For this case we have that by definition of trigonometric relations of a rectangular triangle that, the sine of an angle is given by the opposite leg to the angle on the hypotenuse of the triangle.
That is to say:
![Sin (30) = \frac {IR} {14}](https://tex.z-dn.net/?f=Sin%20%2830%29%20%3D%20%5Cfrac%20%7BIR%7D%20%7B14%7D)
We clear the length of the leg:
![IR = 14 * sin (30)\\IR = 14 * \frac {1} {2}\\IR = 7 \ in](https://tex.z-dn.net/?f=IR%20%3D%2014%20%2A%20sin%20%2830%29%5C%5CIR%20%3D%2014%20%2A%20%5Cfrac%20%7B1%7D%20%7B2%7D%5C%5CIR%20%3D%207%20%5C%20in)
Answer:
![IR = 7 \ in](https://tex.z-dn.net/?f=IR%20%3D%207%20%5C%20in)
Given,
2x - 1/2 = 3 - x
2x + x = 3 + 1/2
3x = 7/2
3x = 3.5
x = 3.5/3
x = 1.167
When the number 951,194 is rounded off to the nearest hundred, the result is 951,200.
<h3>How to round to the nearest hundred?</h3>
When rounding off to the nearest hundred, look at the tens digit. If this digit is 5 and greater then the hundreds number should be rounded up to the nearest whole number.
If the tens digit is 4 or below, then the hundreds digit should be rounded down to the nearest whole number.
The tens digit here is 9 which means that the hundreds digit should be rounded up to the nearest whole number. 951,194 to the nearest hundred is:
= 951,200
Find out more on rounding to the nearest 100 at brainly.com/question/1705848
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Answer:
![\frac{(x-1)^2}{81}+\frac{(y-6)^2}{72}=1](https://tex.z-dn.net/?f=%5Cfrac%7B%28x-1%29%5E2%7D%7B81%7D%2B%5Cfrac%7B%28y-6%29%5E2%7D%7B72%7D%3D1)
Step-by-step explanation:
We have been given that the center of an ellipse is (1,6). One focus of the ellipse is (-2,6). One vertex of the ellipse is (10,6).
We know that standard equation of an ellipse centered at (h,k) is in form:
, where,
a = Horizontal radius
b = Vertical radius.
Since center of the ellipse is at point (1,6) and one vertex is (10,6), so the horizontal radius would be 9 (10-1) units.
Now, we will find vertical radius using formula
, where c represents focal length.
![\text{Focal length}=1-(-2)](https://tex.z-dn.net/?f=%5Ctext%7BFocal%20length%7D%3D1-%28-2%29)
![\text{Focal length}=1+2](https://tex.z-dn.net/?f=%5Ctext%7BFocal%20length%7D%3D1%2B2)
![\text{Focal length}=3](https://tex.z-dn.net/?f=%5Ctext%7BFocal%20length%7D%3D3)
Substituting given values:
![b^2=9^2-3^2](https://tex.z-dn.net/?f=b%5E2%3D9%5E2-3%5E2)
![b^2=81-9](https://tex.z-dn.net/?f=b%5E2%3D81-9)
![b^2=72](https://tex.z-dn.net/?f=b%5E2%3D72)
Upon substituting
,
and
in standard form of ellipse, we will get:
![\frac{(x-1)^2}{9^2}+\frac{(y-6)^2}{72}=1](https://tex.z-dn.net/?f=%5Cfrac%7B%28x-1%29%5E2%7D%7B9%5E2%7D%2B%5Cfrac%7B%28y-6%29%5E2%7D%7B72%7D%3D1)
![\frac{(x-1)^2}{81}+\frac{(y-6)^2}{72}=1](https://tex.z-dn.net/?f=%5Cfrac%7B%28x-1%29%5E2%7D%7B81%7D%2B%5Cfrac%7B%28y-6%29%5E2%7D%7B72%7D%3D1)
Therefore, the required equation of the ellipse in standard form would be
.
No spaces? like t h i s? or like thiiiss??? is this bugging you ????
anyway
use elimination
multiply secn equation by -1
-3x-2y=-6
add to other equation
-3x-2y=-6
<u>4x+2y=10 +</u>
1x+0y=4
x=4
sub back to solve for y
3x+2y=6
3(4)+2y=6
12+2y=6
minus 12 both sides
2y=-6
divide 2
y=-3
(4,-3)