arielpraczko1
10.01.2020 •
Mathematics
Solve the system of equations y = 4x + 1 y = x^2 + 2x - 2 a. (-3,-13) and (1,3) b. (-3,13) and (1,-3) c. (1,3) and (-3,13) d. (-1,-3) and (3,13)
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Ответ:
option D
(-1,-3) and (3,13)
Step-by-step explanation:
Given in the question two equations
y = 4x + 1
y = x² + 2x - 2
Equate both functions
4x + 1 = x² + 2x - 2
rearrange the x term and constant
-x² + 4x - 2x + 2 + 1 = 0
-x² + 2x + 3 = 0
factors
-x * 3x = -3x²
-x + 3x = 2x
-x² -x + 3x + 3 = 0
-x(x+1) +3(x+1) = 0
solve
(x+1)(3-x) = 0
x = -1
and
x = 3
Plug these values in equation to find y
x = -1
y = 4(-1)+ 1
y = -3
x = 3
y = 4(3)+ 1
y = 13
Ответ:
D. (-1,-3) and (3,13)
Step-by-step explanation:
The given equation is;
and
Equate both equations to get;
Split the middle term to obtain;
Factor by grouping;
x=-1,x=3
When x=-1, y=4(-1)+1=-3
(-1,-3)
When x=3, y=4(3)+1=13
(3,13)
Ответ:
V= π r² h
Step-by-step explanation: