Complete question reads;
3. It's most accurate to say that hand-tied arrangements
A. are especially popular for weddings.
B. include only one or two floral materials.
C. are difficult to carry.
D. require a concealed water supply.
Answer:
<u>B. include only one or two floral materials.</u>
Explanation:
Remember, when flowers are been arranged by a florist consideration is made to allow for efficient handling. This involves using a few flower stems and constructing them using one’s hand to form the arrangement. The stems therefore would form a handle which could be held, thus only one or two floral materials are used.
The point slope form of the asymptote of the hyperbola is; y - 4 = ±(6/7)(x - 6)
- We are given the equation of the hyperbola as;
[(y + 4)²/36] - [(x - 6)²/49] = 1
- We can rewrite this equation as;
[(y + 1)²/6²] - [(x - 6)²/7²] = 1
- The general equation form of hyperbola is;
[(y - k)²/a²] - [(x - h)²/b²] = 1
- Comparing this general form to our given equation we can see that;
k = -4
h = 6
a = 6
b = 7
- Now the equation of the asymptote of a hyperbola follows the general formula;
y = ±[a(x - h)/b] + k
- Plugging in the relevant values gives us;
y = ±(6/7)(x - 6) - 4
>> y - 4 = ±(6/7)(x - 6)
Read more about hyperbola asymptotes at; brainly.com/question/12919612
Answer: Twice the previous time would be taken to reach the same speed v with the puck of mass 2m.
Explanation:
Let a Force pushes the hockey puck of mass m.
Then acceleration, a= \frac{F}{m}a=mF
From the equation of motion,
\begin{gathered}\➪ v=u+at\\ v=0+\frac{F}{m}\Delta t\end{gathered}⇒v=u+atv=0+mFΔt ......(1)
In the second case, when mass is 2m, then acceleration,
a'=\frac{F}{2m}a′=2mF
and t' is the time taken.
The final speed is v,
\begin{gathered}\➪ v=0+ a't'\\ \➪ \frac {F}{m}\Delta t=\frac{F}{2m}t'\\ \➪ t'= 2\Delta t\end{gathered}⇒v=0+a′t′⇒mFΔt=2mFt′⇒t′=2Δt using equation (1)
Hence, it would take two times the previous amount of time to push the pluck of double mass.
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