The height & speed are mathematically given as
t = 0.2 s
u = 4 m/s
<h3>What is the
height & speed?</h3>
Generally, the equation for height is mathematically given as
h = 0.2gt^2
Therefore
0.2 = 0.5 * 9.8 *
0.2 = 4.9 *
t = 0.2 s
b)
In conclusion, the initial speed is
s = ut
Therefore
u = 0.8 / 0.2
u = 4 m/s
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Answer:
Step-by-step explanation:
It appears that the number of bulbs produced on the first shift is quite a bit more than the second shift. 5.5 times more, actually. That means that the first shift's output is based on the second shift's output. Second shift produced an unknown number of bulbs, so we will call that number x. First shift produced 5.5 times more, so first shift produced 5.5x. The total number of bulbs produced between the 2 shifts is given as 16,250, so
x + 5.5x = 16,250 so
6.5x = 16,250 and
x = 2500
The second shift produced 2500 bulbs, and the first shift produced 5.5(2500) = 13,750 bulbs.
To find the answer, you would divide 938 ÷ 2680, which is 0.35. Then you would convert this into a fraction, which is 35%.
Answer:
8 is the only one that will work
Step-by-step explanation:
(f o g)(x)=f(g(x)).
So this means the x will first be plug into g.
So let's check your choices.
g(6)=1/(6-6)=1/0 so 6 is not in the domain of g which means it isn't in the domain of (f o g).
g(8)=1/(8-6)=1/2 so this is a number so 8 is in the domain of g,
Let's check if 1/2 is in the domain of f.
f(1/2)=sqrt(6*1/2)=sqrt(3) so this is a number so since 1/2 is in the domain of f then 8 is in the domain of (f o g).
g(4)=1/(4-6)=1/(-2)=-1/2 so 4 is in the domain of g,
f(-1/2)=sqrt(6*-1/2)=sqrt(-3) so this is a problem because you can't square root negative numbers so -1/2 isn't in the domain of f, and therefore 4 isn't in the domain of (f o g).
g(2)=1/(2-6)=1/-4=-1/4 so 2 is in the domain of g.
f(-1/4)=sqrt(6*-1/4)=sqrt(-3/2) so again this is a problem because we can't square root negative numbers so -1/4 isn't in the domain of f, and therefore 2 isn't in the domain of (f o g).