Find the value of k so that the point A lies on the line given by equation: A(2,3), kx- 2y +k=0

Mathematics

1 answer:

Gekata [30.6K]3 years ago

5 0

Answer:

Step-by-step explanation:

$\text{Put the coordinates of the given point A(2, 3)}\\\text{to the equation of a line}\ kx-2y+k=0:\\\\A(2,\ 3)\to x=2,\ y=3\\\\(k)(2)-(2)(3)+k=0\\2k-6+k=0\qquad\text{add 6 to both sides}\\2k+k=6\\3k=6\qquad\text{divide both sides by 3}\\\boxed{k=2}$

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Please help me with this!

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If the flask shown in the diagram can be modeled as a combination of a sphere and a cylinder, then its volume is

$V_{flask}=V_{sphere}+V_{cylinder}.$

Use following formulas to determine volumes of sphere and cylinder:

$V_{sphere}=\dfrac{4}{3}\pi R^3,\\ \\V_{cylinder}=\pi r^2h,$

wher R is sphere's radius, r - radius of cylinder's base and h - height of cylinder.

Then

$V_{sphere}=\dfrac{4}{3}\pi R^3=\dfrac{4}{3}\pi \left(\dfrac{4.5}{2}\right)^3=\dfrac{4}{3}\pi \left(\dfrac{9}{4}\right)^3=\dfrac{243\pi}{16}\approx 47.71;$
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If both the sphere and the cylinder are dilated by a scale factor of 2, then all dimensions of the sphere and the cylinder are dilated by a scale factor of 2. So

R'=2R, r'=2r, h'=2h.

Write the new fask volume:

$V_{\text{new flask}}=V_{\text{new sphere}}+V_{\text{new cylinder}}=\dfrac{4}{3}\pi R'^3+\pi r'^2h'=\dfrac{4}{3}\pi (2R)^3+\pi (2r)^2\cdot 2h=\dfrac{4}{3}\pi 8R^3+\pi \cdot 4r^2\cdot 2h=8\left(\dfrac{4}{3}\pi R^3+\pi r^2h\right)=8V_{flask}.$