Answer:
see explanation
Step-by-step explanation:
a
The tangent and the radius at the point of contact form a right angle
Using Pythagoras' identity on the right triangle formed.
Let x be the distance from the centre to P, then
x² = 4² + 10² = 16 + 100 = 116 ( take the square root of both sides )
x =
≈ 10.77 cm (to 2 dec. places )
b
let the required angle be Θ, then
Using the sine or cosine ratio in the right triangle.
cosΘ =
= 
Θ =
(
) ≈ 21.8°
The correct answer is: [A]: " <span>x(y – 5) = 2 " .
</span>_____________________________________________________
Consider:
_____________________________________________________
Choice: [A]: " x(y–5) = 2 " ;
Divide each side by "x" ;
" [x(y – 5)] / x = 2/x " ;
→ y – 5 = 2/x ;
Add "5" to each side of the equation:
y – 5 + 5 = 2/x + 5 ;
→ y = 2/x + 5 ; not a line; since one cannot divide by "zero" ; there would be no "point" on the graph at "x = 0". So, this answer choice: [A] is correct.
_______________________________________________________
Choice [B]:
y -2x -18 = 0
y - 2x = 18
y = 18 + 2x ; y = 2x + 18 ; is a line.
_________________________________________________
Choice C) 3y + 12 - 6x = 5 ;
3y = 5 - 12 + 6x ;
3y = -7 + 6x ; 3y = 6x - 7 ; y = 6x/3 - 7/3 ; y = 2x - 7/3 ; is a line.
_________________________________________________
Choice: [D]:
2(y+x) = 0 ;
[2*(y+x)] / 2 = 0/2 ; y + x = 0 ; y = -x ; is a line.
_________________________________________________
Answer:
160 inches
Step-by-step explanation:
An octagon has 8 sides;
20*8=160
Answer:
Tan C = 3/4
Step-by-step explanation:
Given-
∠ A = 90°, sin C = 3 / 5
<u>METHOD - I</u>
<u><em>Sin² C + Cos² C = 1</em></u>
Cos² C = 1 - Sin² C
Cos² C = 
Cos² C = 
Cos² C = 
Cos C = 
Cos C = 
As we know that
Tan C = 
<em>Tan C =
</em>
<em>Tan C =
</em>
<u>METHOD - II</u>
Given Sin C = 
therefore,
AB ( Height ) = 3; BC ( Hypotenuse) = 5
<em>∵ ΔABC is Right triangle.</em>
<em>∴ By Pythagorean Theorem-</em>
<em>AB² + AC² = BC²</em>
<em>AC² </em><em>= </em><em>BC² </em><em>- </em><em> AB</em><em>² </em>
<em>AC² = 5² - 3²</em>
<em>AC² = 25 - 9</em>
<em>AC² = 16</em>
<em>AC ( Base) = 4</em>
<em>Since, </em>
<em>Tan C =
</em>
<em>Tan C =
</em>
<em>Hence Tan C =
</em>
<em />
Let
x-------> the number of dinner
y-------> the number of lunch
we know that
-------> equation A
------> equation B
Substitute equation B in equation A
![8[y]+5y \leq 42](https://tex.z-dn.net/?f=8%5By%5D%2B5y%20%5Cleq%2042)



so
the greatest number of lunch is 

Hence
the greatest number of dinner is 
therefore
the greatest number of meals is

<u>the answer is</u>
