That's not a linear system, but you have an awesome school system for giving you this problem.
Multiply by 6xy to clear the fractions.
That's a second degree equation, also known as a conic. That one happens to be a hyperbola.
Let's clear the fractions from the second equation, multiplying out common denominator xy:
We are being asked to find the meet of two hyperbolas, so we expect two answers, a quadratic equation.
Substituting,
We have to rule out x=0 because it's in the denominator.
Answer: (44/19, 33/20)
It should be D, I did the math :)
A.)
<span>s= 30m
u = ? ( initial velocity of the object )
a = 9.81 m/s^2 ( accn of free fall )
t = 1.5 s
s = ut + 1/2 at^2
\[u = \frac{ S - 1/2 a t^2 }{ t }\]
\[u = \frac{ 30 - ( 0.5 \times 9.81 \times 1.5^2) }{ 1.5 } \]
\[u = 12.6 m/s\]
</span>
b.)
<span>s = ut + 1/2 a t^2
u = 0 ,
s = 1/2 a t^2
\[s = \frac{ 1 }{ 2 } \times a \times t ^{2}\]
\[s = \frac{ 1 }{ 2 } \times 9.81 \times \left( \frac{ 12.6 }{ 9.81 } \right)^{2}\]
\[s = 8.0917...\]
\[therfore total distance = 8.0917 + 30 = 38.0917.. = 38.1 m \] </span>
Answer:
89447218502021127475863435723286553431
Step-by-step explanation:
exact answer yeeet thanks for the points
the answer is 6.9.
When you divide a negative by a negative it is a positive
