<span>(4 · 2^5) ÷ (2^3 · 1/16 )
</span>= (2^2 · 2^5) ÷ (2^3 · 2^-4 )
= (2^7) ÷ (2^-1)
= 2^8
Answer:
The solution of system of equation is (-2,0)
Step-by-step explanation:
Given system of equation are
Equation 1 : 2x+y=(-4)
Equation 2 : y+
x=(-1)
To plot the equation of line, we need at least two points
For Equation 1 : 2x+y=(-4)
Let x=0
2x+y=(-4)
2(0)+y=(-4)
y=(-4)
Let x=1
2x+y=(-4)
2(1)+y=(-4)
y=(-6)
Therefore,
The required points for equation is (0,-4) and (1,-6)
For Equation 2 : y+
x=(-1)
Let x=0
y+
x=(-1)
y+
(0)=(-1)
y=(-1)
Let x=2
y+
x=(-1)
y+
(2)=(-1)
y=(-2)
The required points for equation is (0,-1) and (2,-2)
Now, plot the graph using this points
From the graph,
The red line is equation 1 and blue line is equation 2
Since. The point of intersection is solution of system of equations
The solution of system of equation is (-2,0)
Answer:
Step-by-step explanation:
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