3
x
+
2
y
>
24
3
x
+
2
y
>
24
Solve for
y
y
.
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y
>
−
3
x
2
+
12
y
>
-
3
x
2
+
12
Use the slope-intercept form to find the slope and y-intercept.
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Slope:
−
3
2
-
3
2
Y-Intercept:
12
12
Graph a dashed line, then shade the area above the boundary line since
y
y
is greater than
−
3
x
2
+
12
-
3
x
2
+
12
.
y
>
−
3
x
2
+
12
y
>
-
3
x
2
+
12
Answer:
9.72
Step-by-step explanation:
s1 = 10.6383 ; s2 = 5.21289
x1 = 147.583 ; x2 = 136.417
n1 = 12 ; n2 = 12
df1 = n1 - 1 = 12 - 1 = 11
df2 = n2 - 1 = 12 - 1 = 11
The test statistic :
(x1 - x2) / sqrt[(sp²/n1 + sp²/n2)]
Pooled variance = Sp² = (df1*s1² + df2*s2²) ÷ (n1 + n2 - 2)
Sp² = ((11*10.6383) + (11*5.21289)) / 22 = 7.926
Test statistic, T* :
(147.583 - 136.417) / √(7.926 * (1/12 + 1/12))
11.166 / √(7.926 * (1/6)
11.166 / √1.321
11.166 / 1.1493476
T* = 9.7150766
Test statistic = 9.72
31cm is the side of the new square and this the largest area shape in square
THe problem is basically telling us: 
where P is the power disappated and V^2 is our voltage squared.
So, for the second example to find the power we simply have to plug k and our voltage back in, so:
