6×6=36. also get a calculator
Should be 7. fkshfjsdhfjkhksjh
So there are two ways.
First you could solve for y and plug it into the calculator. one equation for y1 and another for y2. Then go 2nd trace and press 5 and then press enter 3 times to find the intersection.
the second way is to solve by hand. I would suggest solve the first equation for y and plug the equation solved for y into the other equation. so y = -3x + 9 and 3x-5y=15. Then you could do 3x - 5(-3x+9) = 15. Finally solve for x and plug in x into both equations to see if you get the same y value.
The answer should be (10/3, -1)
Answer
f(t)=2cos(t)
Step-by-step explanation:
let's describe the situation first, a fly is starting 2 meters from the bulb and flies towards the bulb and when it is really close to the bulb, it flies 2 meters further away from the bulb this means that it reaches d = 0m (distance from the bulb) and then flies further 2 meters so d = -2m.
it this point the fly returns back and touches the bulb and flies away (ends it's oscillatory motion ). d = 0 again and story ends here.
here if we want to model this problem with time function , the cosine function seems the best fit with amplitude of 2, so the answer is f(t) = 2cos(t).
Now you can ask why cosine function? well if you look at the graph or the plot of the function it perfectly captures the physical situation going on here in this problem.
Domain is 0 to 2
because it is one complete cycle and the range is -2 to +2.
Answer: The climb angle is 14°
Step-by-step explanation:
We can create a rectangle triangle, where the velocity of the plane is the hypotenuse, the horiontal component (1000 ft by unit of time) is a cathetus and the vertical component is (250ft by unit of time).
We want to find the angle that is adjacent to the horizontal component, so we can use the relationship.
Tan(angle) = opposite cathetus/adjacent cathetus
Tan(angle) = 250/1000 = 1/4
Angle = Atan(1/4) = 14°
