A raiz da equação 2x - 1 = - 3, é: ( U = Z)
1 answer:
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We start with 
and wish to write it as 
First, pull 2 out from the first two terms:
Let’s look at what is in parenthesis. In the final form this needs to be a perfect square. Right now we have
and we can obtain -10x by adding -5x and -5x. That is, we can build the following perfect square: 
The “problem” with what we just did is that we added to what was given. Let’s put the expression together. We have
and when we multiply that out it does not give us what we started with. It gives us 
So you see our expression is not right. It should have a -53 but instead has a -3. So to correct it we need to subtract another 50.
We do this as follows:
which gives us the final expression we seek:
If you multiply this out you will get the exact expression we were given. This means that:
a = 2
d = -5
e = -103
We are asked for the sum of a, d and e which is 2 + (-5) + (-103) = -106
First, pull 2 out from the first two terms:
Let’s look at what is in parenthesis. In the final form this needs to be a perfect square. Right now we have
The “problem” with what we just did is that we added to what was given. Let’s put the expression together. We have
So you see our expression is not right. It should have a -53 but instead has a -3. So to correct it we need to subtract another 50.
We do this as follows:
If you multiply this out you will get the exact expression we were given. This means that:
a = 2
d = -5
e = -103
We are asked for the sum of a, d and e which is 2 + (-5) + (-103) = -106
Answer:
s = 22.5 m
Step-by-step explanation:
the equation for the speed change of a coach moving along a straight section of the road and starting braking at a speed of 20 m / s has the form v (t) = 25-5t. Using integral calculus, determine the coach's braking distance.
v (t) = 25 - 5 t
at t = 0 , v = 20 m/s
Let the distance is s.
Let at t = t, the v = 20
So,
20 = 25 - 5 t
t = 1 s
So, s = 25 x 1 - 2.5 x 1 = 22.5 m