Answer:
<h2>B.
r = 4cosθ</h2>
Step-by-step explanation:
Given the expression in rectangular coordinate as x²+y²−4x=0, in order to write the given expression in polar coordinates, we need to write the value of x and y as a function of (r, θ).
x = rcosθ and y = rsinθ.
Substituting the value of x and y in their polar form into the given expression we have;
x²+y²−4x=0
( rcosθ)²+( rsinθ)²-4( rcosθ) = 0
Expand the expressions in parenthesis
r²cos²θ+r²sin²θ-4rcosθ = 0
r²(cos²θ+sin²θ)-4rcosθ = 0
From trigonometry identity, cos²θ+sin²θ =1
The resulting equation becomes;
r²(1)-4rcosθ = 0
r²-4rcosθ = 0
Add 4rcosθ to both sides of the equation
r²-4rcosθ+4rcosθ = 0+4rcosθ
r² = 4rcosθ
Dividing both sides by r
r²/r = 4rcosθ/r
r = 4cosθ
<em>Hence the correct equation in polar coordinates is r = 4cosθ</em>
Step-by-step explanation:
The given inequality is :
x – 2y > 16
We need to draw the graph of this inequality.
If x = 0,
-2y>16
y>-8
If y = 0,
x>16
So, we can draw the graph as follows :
Answer:
its 15^2
Step-by-step explanation:
its 15^2 because it =225 and 13^2 is 169
give brainliest please
Sure! So this is ready as "the cube root of 125". This basically means, "what number cubed can get me 125?"
Let's go through our options.
We can rule out D, as D cubed would be unreasonably big.
We can also rule out C, because 375 cubed is easily over 10000, you know this even if you haven't computed it all, just compute the 300 cubed.
We can rule out B, too. 41 squared is already over 125, therefore it can't be the answer.
Therefore our answer is A, 5. We can check that by cubing 5, and that indeed gets us 125.
Hope this helps!
EXPLANATION
We can apply the Obtuse Triangle Theorem to check if the the measures of the sides appropiately represents the measures of the triangle:
Perimeter = 40 cm = 4x + 2x + 2 + x + 3
Adding like terms:
40 = 7x + 5
Subtracting 5 to both sides:
40 - 5 = 7x
Adding like terms:
35 = 7x
Dividing both sides by 7:
35/7 = x
Simplifying:
Then, by the obtuse triangle theorem, the following relationship should be fulfilled.