Answer:
x stays x in both equations so x stands for x
Step-by-step explanation:
y=3x+6
you can put this into a graphing caculator or draw it yourself.
6 is the y value when x equals 0. and the 3 represents the slope so then it is solved for any y value
Answer:
aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaahhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaahhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaahhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaahhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh
Answer:
a. attached graph; zero real: 2
b. p(x) = (x - 2)(x + 3 + 3i)(x + 3 - 3i)
c. the solutions are 2, -3-3i and -3+3i
Step-by-step explanation:
p(x) = x³ + 4x² + 6x - 36
a. Through the graph, we can see that 2 is a real zero of the polynomial p. We can also use the Rational Roots Test.
p(2) = 2³ + 4.2² + 6.2 - 36 = 8 + 16 + 12 - 36 = 0
b. Now, we can use Briott-Ruffini to find the other roots and write p as a product of linear factors.
2 | 1 4 6 -36
1 6 18 0
x² + 6x + 18 = 0
Δ = 6² - 4.1.18 = 36 - 72 = -36 = 36i²
√Δ = 6i
x = -6±6i/2 = 2(-3±3i)/2
x' = -3-3i
x" = -3+3i
p(x) = (x - 2)(x + 3 + 3i)(x + 3 - 3i)
c. the solutions are 2, -3-3i and -3+3i
The solution (-4,2) satisfies for the system of linear equations 3x + 13y = 14; 6x + 11y = -2
<u>Step-by-step explanation:</u>
Step 1:
Given detail is the solution of the equations (-4, 2) ie, x= - 4 and y = 2
This implies that this solution should satisfy the given linear equations.
Step 2:
Substitute values of x and y in the equations and verify whether the right hand side equals the left hand side.
System 1 Eq(1) ⇒ LHS = 3(-4) + 13 (2) = -12 + 26 = 14 = RHS
System 1 Eq(2) ⇒ LHS = 6(-4) + 11(2) = -24 + 22 = -2 = RHS
Therefore, the first system of linear equations satisfy the condition.
