Two angles are said to be supplementary if their sum is 180 degrees. So, if x and y are supplementary angles, the formula is
We know that one of the angles (say x) is 98. So, the equation updates to
If we want to solve for the other angle, we can simply subtract 98 from both sides to get
Okay, so total dogs are 49.
So we know if x=large dogs and y=small dogs, then x+y=49.
Next we are told that there are 36 MORE small dogs than large dogs. We can take that more meaning addition, and x being our value for large dogs. So y=x+36.
After that we now know what value we can plug in for y. So x+x+36=49.
We can then simplify it to 2x+36=49.
Subtracting the 36 from both sides, leaves you with 2x=13.
Followed by dividing both sides by 2, gives you x=6.5. Or 6.5 large dogs.
Now we can plug this into our formula for the small dogs (y=x+36) to give us y=6.5+36, which simplifies to y=42.5. Or 42.5 small dogs. Which is our answer.
We can double check it by adding the small and large dogs together, 6.5+42.5, which gives us 49, our total entries.
Answer:
Is this the full question?
<span>Assuming that the particle is the 3rd
particle, we know that it’s location must be beyond q2; it cannot be between q1
and q2 since both fields point the similar way in the between region (due to
attraction). Choosing an arbitrary value of 1 for L, we get </span>
<span>
k q1 / d^2 = - k q2 / (d-1)^2 </span>
Rearranging to calculate for d:
<span> (d-1)^2/d^2 = -q2/q1 = 0.4 </span><span>
<span> d^2-2d+1 = 0.4d^2 </span>
0.6d^2-2d+1 = 0
d = 2.72075922005613
d = 0.612574113277207 </span>
<span>
We pick the value that is > q2 hence,</span>
d = 2.72075922005613*L
<span>d = 2.72*L</span>
the answer is
x=8y/3(1-8y)
y=3x/8(1+3x)
hope this helped l used cymath to solve this problem.
good day : )
