Answer:
Step-by-step explanation:
We have plane 1 flying SW for 4 hours at a rate of 415 mph. The distance he covers using the d = rt formula for distance, is 415(4) = 160 miles.
We also have plane 2 flying directly east (along the x-axis) for 4 hours at 327 mph. The distance he covers using the d = rt formula for distance, is 327(4) = 1308 miles. The angle in between them at this point is 135 degrees, and what we need to find is the length of the vector connecting the 2 planes. IF this was right triangle trig that distance would be the hypotenuse and we could solve for it using Pythagorean's Theorem. BUT it is NOT a right triangle, so we have to find some other means with which to solve for that length. We will use the Law of Cosines to do this.
which simplifies a bit to
If you add all of that together, you'll get
and you'll take the square root of that to get that the distance between the 2 planes after 4 hours is
2745 miles
The is 0.666666666666666666667 so i dont know which way you want to swing because if you round up it is 0.7 and 0.6 doesn't repeat infintely so your choice
1cm..............5000 cm
or 1cm.............50 metres
=> 4cm.........4*50=200 metres, (actual distance )
It’s A because (x) the number of cake equal 12 in price
Answer:
c = 7
d = 5
Step-by-step explanation:
combining coefficient a with the nth root of b: a√ⁿb = √ⁿ((a)ⁿb). This is the reverse of simplifying a radical.
8x³√²7x = √²((8x³)²)√²(7x) = √²(64x⁶)√²(7x) = √²((64x⁶)(7x)) = √²((64 × 7)(x⁶ × x)) = √²((448)(x⁷)) = √²448x⁷ →
√²448x⁷ = √²448xᶜ.
7 → c, √²448x⁷ = √²448x⁷, c = 7.
4x√³9x² = √³((4x)³)√³(9x²) = √³(64x³)√³(9x²) =
√³((64x³)(9x²)) = √³((64 × 9)(x³ × x²)) = √³((576)(x⁵)) = √³576x⁵ →
√³576x⁵ = √³576xᵈ.
5 → d, √³576x⁵ = √³576x⁵, d = 5.
