We can do it through an geometric series formation.
- Where x is any number of cards and coin ratio of increase ment is 6.
So equation
25/3 v72 2.4 35/4 i think
I'm assuming you meant 62p=610. If this is the case, then p = 610/62, or 9.84.
This does not match any of the four answer choices.
I'd suggest you ensure that you have copied down this problem completely and accurately.
Answer:
19 degrees.
Step-by-step explanation:
First solve the angles in the top triangle:
FHG = 180 - 135 = 45 (angles on a straight line = 180)
FGH = 180 - 105 = 75 (angles on a straight line = 180)
Using these we get HFG = 180 - 45 - 75 = 60 (angles in a triangle = 180)
Then because HFG is directly opposite DFC on the crossed lines, we know they are equal: DFC = 60
FDC = 180 - 119 = 61 (anges on a straight line = 180)
Now if we label ABC (the target angle) as x then we known ACB = 180 - 40 - x.
With this we know FCD = 180 - ACB = 180 - (180 - 40 - x) = 40 + x (angles on a straight line).
Finally using the FDC triangle we have 40 + x + 60 + 61 = 180 so x = 180 - 40 - 60 - 61 = 19 degrees.
Answer:
The water is 208.22ft from the tug boat
Step-by-step explanation:
The governing equation is ![y=-0.0055x^2 +1.15x + 5](https://tex.z-dn.net/?f=y%3D-0.0055x%5E2%20%2B1.15x%20%2B%205)
y is the height above the ocean
x is the distance from the tugboat
if y= 6ft, the equation will now become
![6=-0.0055x^2 +1.15x + 5](https://tex.z-dn.net/?f=6%3D-0.0055x%5E2%20%2B1.15x%20%2B%205)
we can arrange this properly to form a quadratic equation by grouping like terms.
![-0.0055x^2 +1.15x -1=0](https://tex.z-dn.net/?f=-0.0055x%5E2%20%2B1.15x%20-1%3D0)
solving quadratically we have two values of x as 208.22ft and 0.873 ft.
We can take a more realistic value as a solution to our problems:
208.22ft