Answer:
You should <u>carry 2 other batteries.</u>
Step-by-step explanation:
If the average life of each battery tends to last 6.9 months, and what you need is a year of battery supply, you probably need to carry two batteries.
The battery inside the flashlight has probably already spent much of its half-life.
We are told that the flashlight already has a battery inside, the batteries are discharged inside the flashlight even if the flashlight is not being used.
Which means that the average life of that battery is wearing out.
It would take 2 more batteries in addition to the one already inside the flashlight to supply the need for these for a year.
Answer:
Step-by-step explanation:
Answer:
0.65%, 9.2%, 40%, 656%
Step-by-step explanation:
see answer
<span> we know the length of the cable is 9m.
That means the magnitude of </span><span><span>r<span><span><span>AB</span></span><span></span></span></span>=9</span><span>m.
The unit vector, denoted u, is each of </span><span>r<span><span><span>AB</span></span><span></span></span></span><span> divided by the magnitude.
</span>u=<span>(<span><span>x/9</span><span></span></span>i−<span>y/9<span></span></span>j−<span>z/9<span></span></span>k<span>)
</span></span><span>We can also figure out the unit vector of F.
</span>u=(350i - 250j - 450k)/√(350² +(-250)² +(-450)²)
u=0.562i−0.401j−0.723<span>k
</span>
<span>Force F is directed from point A to B, then both unit vectors must be equal.
Therefore
</span>(x/9i−y/9j−z/9k)=0.562i−0.401j−0.723k
<span>We can now solve for each term
x/9=0.562----- > x=5.058 m
-y/9=-0.401--------- > y=3.609 m
-z/9=-0.723------- > z=6.507 m
The answer is
the coordinates of point a
(5.058,3.609,6.507)
</span>
Hi there :-)
4c+7.5a=3312.50
Let
C=x
A=500-x
4x+7.5 (500-x)=3312.50
Solve for x
4x+3750-7.5x=3312.50
4x-7.5x=3312.5-3750
-3.5x=-437.5
X=437.5/3.5
X=125 of children
A=500-125=375 of adults
Hope it helps
