Given that \(P\propto \frac{1}{\sqrt{r}}\) and p = 3 when r = 16, find the value of r when p = 3/2
48
64
72
324
Correct answer is B
\(P \propto \frac{1}{\sqrt{r}} P=3\hspace{1mm}r=16\\
P = \frac{k}{\sqrt{r}}\Rightarrow 3 = \frac{k}{\sqrt{16}}\Rightarrow \frac{3}{1} = \frac{k}{4} \\ \Rightarrow K = 12 \Rightarrow K = 12; P\sqrt{r} = k \Rightarrow \sqrt{r} = \frac{k}{p}\\
r = \frac{(k)^2}{P}=12^2 \div \frac{3}{2}=\left(\frac{12}{1}\times \frac{2}{3}\right)=8^2\\
r = 64\)
the perpendicular bisector of the line segment X Y
a line parallel to the line segment XY
a circle with XY as diameter
the line perpendicular to the line segment XY
Correct answer is A
No explanation has been provided for this answer.
Which of the following quadratic equations has \(-\frac{1}{2}\) and \(\frac{3}{4}\) as its roots?
8x2 + 11x - 3 = 0
8x2 -11x – 3 = 0
8x2 + 2x -3 = 0
8x2 - 2x – 3 = 0
Correct answer is D
X2 - (sum of the roots)x + (product of the roots) = 0
Sum of the roots \(= -\frac{1}{2}+\frac{3}{4} = \frac{-2+3}{2}=\frac{1}{4}\)
Product of the roots = \(-\frac{1}{2}\times \frac{3}{4}=\frac{-3}{8}\\
X^2-\left(\frac{1}{4}\right)x-\frac{3}{8} = 0. Taking \hspace{1mm}the\hspace{1mm} common\)
\(8x^2-2x-3=0\);
The values of x when y = 3 are approximately
-4.7 and 1.4
-4.6 and 1.5
-3.6 and 0.4
-3.6 and 1.5
Correct answer is B
No explanation has been provided for this answer.
Use the graph to answer the Question below
What are the roots of the equation x2 + 3x - 4 = 0?
1, 4
-1, -4
-1, 4
-4, 1
Correct answer is D
x2 + 3x – 4 = 0 has roots value at the curve makes touches with the x – axis. The curve contact is at x = -4 and +1.
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