4x^2-3x+4y^2+4z^2=0
here we shall proceed as follows:
x=ρcosθsinφ
y=ρsinθsinφ
z=ρcosφ
thus
4x^2-3x+4y^2+4z^2=
4(ρcosθsinφ)^2-3(ρcosθsinφ)+4(ρsinθsinφ)^2+4(ρcosφ)
but
ρ=1/4cosθsinφ
hence we shall have:
4x^2-3x+4y^2+4z^2
=1/4cosθsinθ(cosθ(4-3sinφ))+4sin^2(φ)
Step-by-step explanation:
Since sides AD and BC are parallel,
Angle BAD + Angle ADC = 180° (co-interior angles)
Therefore Angle ADC = 180° - 115° = 65°.
Answer:
the buses are at 50 mph and 60 mph
the distance = 210 miles
Step-by-step explanation:
r1 = rate of bus 1
r2 = rate of bus 2
r1 = r2+10 (The speed of one of the buses was 10 mph greater than the speed of the other bus.)
distance = rate * time
for bus 1
distance = r1 * (3.5)
bus 2 was 1/6 short of making it to town b
so it made it 5/6 of the way
5/6 distance = r2 * (3.5)
multiply by 6/5 on each side
distance = 6/5 * r2 * 3.5
set the distances equal
r1 * (3.5)=6/5 * r2 * 3.5
r1 = 6/5 * r2
now set rate 1 equal (r1 = r2+10)
6/5 * r2 = r2 + 10
solve for r2 by subtracting r2 on each side
6/5 r2 -r2 = 10
6/5 = 1.2
1.2 r2 - r2 =10
.2 r2 = 10
divide by .2 on each side
r2 = 10/.2
r2 = 50
r1 = 50 + 10
r1 =60
to find the distance
distance = r1 * (3.5)
distance = 60 * 3.5
distance = 210 miles
Step-by-step explanation:
The scientific notation:
where 1 ≤ <em>a</em> < 10, and <em>k</em> ∈ Z (integer).
