Answer: (-2, 0) and (0, -2)
Step-by-step explanation:
This system is:
y + x = -2
y = (x + 1)^2 - 3
To solve this we first need to isolate one of the variables in one fo the equations, in the second equation we have already isolated the variable y, so we can just replace it in the first equation:
(x + 1)^2 - 3 + x = -2
Now we can solve this for x.
x^2 + 2*x + 1 - 3 = -2
x^2 + 2*x + 1 -3 + 2 = 0
x^2 + 2*x + 0 = 0
The solutions of this equation are given by the Bhaskara's formula, then the solutions are:

The two solutions are:
x = (-2 - 2)/2 = -2
In this case, we replace this value of x in the first equation and get:
y - 2 = -2
y = -2 + 2 = 4
This solution is x = -2, y = 0, or (-2, 0)
The other solution for x is:
x = (-2 + 2)/2 = 0
If we replace this in the first equation we get:
y + 0 = -2
y = -2
This solution is x = 0, y = -2, or (0, -2)
I believe the value of y is -28
also believe m
25 is 57.4
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➷
can be simplified to 2
Multiply 2 by 6 to get the simplified version for part 1
= 
can be simplified to 
Multiply 2 by 3 to get the simplified version:
= 
It would be:

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Answer:
0.6443
Step-by-step explanation:
Given the that:
P(Z < 0.37) ; using a standard normal distribution :
The probability can be obtained using several methods :
Using a standard normal distribution table ;
P(Z < 0.37) = obtain the value at the intersection where 0.3 is on the vertical axis and 0.07 on the horizontal axis of the Z distribution table
Hence,
P(Z < 0.37) = 0.6443