Answer:
6/8: 24
4/32: 32
Step-by-step explanation:
It looks like you want to compute the double integral

over the region <em>D</em> with the unit circle <em>x</em> ² + <em>y</em> ² = 1 as its boundary.
Convert to polar coordinates, in which <em>D</em> is given by the set
<em>D</em> = {(<em>r</em>, <em>θ</em>) : 0 ≤ <em>r</em> ≤ 1 and 0 ≤ <em>θ</em> ≤ 2<em>π</em>}
and
<em>x</em> = <em>r</em> cos(<em>θ</em>)
<em>y</em> = <em>r</em> sin(<em>θ</em>)
d<em>x</em> d<em>y</em> = <em>r</em> d<em>r</em> d<em>θ</em>
Then the integral is

Answer:
4 km
Step-by-step explanation:
2/3(6) = 4 km
Answer:
AB ≈ 15.7 cm, BC ≈ 18.7 cm
Step-by-step explanation:
(1)
Using the Cosine rule in Δ ABD
AB² = 12.4² + 16.5² - (2 × 12.4 × 16.5 × cos64° )
= 153.76 + 272.25 - (409.2 cos64° )
= 426.01 - 179.38
= 246.63 ( take the square root of both sides )
AB =
≈ 15.7 cm ( to 1 dec. place )
(2)
Calculate ∠ BCD in Δ BCD
∠ BCD = 180° - (53 + 95)° ← angle sum in triangle
∠ BCD = 180° - 148° = 32°
Using the Sine rule in Δ BCD
=
=
( cross- multiply )
BC × sin32° = 12.4 × sin53° ( divide both sides by sin32° )
BC =
≈ 18.7 cm ( to 1 dec. place )
The answer is x = -3 / 2.