If (x - 3)2 + (y - 5)2 + (z - 4)2 = 0, then the value of $$\frac{{{x^2}}}{9}{\text{ + }}\frac{{{y^2}}}{{25}}{\text{ + }}\frac{{{z^2}}}{{16}}$$ is?
A. 12
B. 9
C. 3
D. 1
Answer: Option C
Solution(By Examveda Team)
$$\eqalign{ & {\left( {x - 3} \right)^2}{\text{ + }}{\left( {y - 5} \right)^2}{\text{ + }}{\left( {z - 4} \right)^2} = 0 \cr & \therefore {\left( {x - 3} \right)^2} = 0{\text{ }}x = 3 \cr & {\left( {y - 5} \right)^2} = 0{\text{ }}y = 5 \cr & {\left( {z - 4} \right)^2} = 0{\text{ }}z = 4{\text{ }} \cr & \therefore \frac{{{x^2}}}{9}{\text{ + }}\frac{{{y^2}}}{{25}}{\text{ + }}\frac{{{z^2}}}{{16}} \cr & \Rightarrow \frac{9}{9} + \frac{{25}}{{25}} + \frac{{16}}{{16}} \cr & \Rightarrow 3 \cr} $$Related Questions on Algebra
If $$p \times q = p + q + \frac{p}{q}{\text{,}}$$ then the value of 8 × 2 is?
A. 6
B. 10
C. 14
D. 16
A. $$1 + \frac{1}{{x + 4}}$$
B. x + 4
C. $$\frac{1}{x}$$
D. $$\frac{{x + 4}}{x}$$
A. $$\frac{{20}}{{27}}$$
B. $$\frac{{27}}{{20}}$$
C. $$\frac{6}{8}$$
D. $$\frac{8}{6}$$
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