please answer this---
Answers
Answer:
I don't know
Step-by-step explanation:
1. State whether the following sets are empty, finite or infinite sets. In case of
(non-empty) finite sets, mention the cardinal number.
(i) (all colours of a rainbow
(ii) {x x is a prime number between 7 and 11)
(iii) {x | x is a digit in the numeral 550131527}
(iv) (x | x is a letter in the word 'SUFFICIENT'}
plz do answer fastEquationoflinemakinginterceptsonxandy−axis,
= x/a + y/b = 1
Comparing with Ax + By + C = 0,
A = 1/x
B = 1/b
C = -1
Also,
\sf Perpendicular \: Distance \: of \: line \: Ax + By + C = 0 \: from \: the \: point \: (x,y),PerpendicularDistanceoflineAx+By+C=0fromthepoint(x,y),
D = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2+b^2} }D=
A
2
+b
2
∣Ax
1
+By
1
+C∣
Where,
(x₁, y₁) = (0,0) (Given)
D = p (Given)
A = 1/x
B = 1/b
C = -1
Substituting values,
\begin{gathered}\sf D = \dfrac{|Ax_1 + By_1 + C|}{\sqrt{A^2+b^2}} \\ \\ \sf p = \dfrac{|\frac{1}{a} \times 0 + \frac{1}{b} \times 0 + (-1)|}{\sqrt{(\frac{1}{a})^2+(\frac{1}{b})^2}} \\ \\ \sf p = \dfrac{|-1|}{\sqrt{\frac{1}{a^2}+\frac{1}{b^2}}} \\ \\ \sf p = \dfrac{1}{\sqrt{\frac{1}{a^2}+\frac{1}{b^2}}}\end{gathered}
D=
A
2
+b
2
∣Ax
1
+By
1
+C∣
p=
(
a
1
)
2
+(
b
1
)
2
∣
a
1
×0+
b
1
×0+(−1)∣
p=
a
2
1
+
b
2
1
∣−1∣
p=
a
2
1
+
b
2
1
1
Squaring both sides,
\sf p^2 = \dfrac{1}{\frac{1}{a^2}+\frac{1}{b^2}}p
2
=
a
2
1
+
b
2
1
1
Taking reciprocal,
\frac{1}{p^2} = \frac{1}{a^2}+\frac{1}{b^2}
p
2
1
=
a
2
1
+
b
2
1
Hence proved.
Answer:
ok
don't no
ok
bye so
sorry this
ok