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Radda [10]
4 years ago
14

Six machines at a certain factory operate at the same constant rate. If four of these machines, operating simultaneously, take 2

7 hours to fill a certain production order, how many fewer hours does it take all six machines, operating simultaneously, to fill the same production order?A: 9B: 12C: 16D: 18E: 24
Mathematics
1 answer:
andreyandreev [35.5K]4 years ago
7 0

Answer:

A) 9

Step-by-step explanation:

No. of hours required: 4×27 = 108

If 6 machines working:

108÷6 = 18 hours

27 - 18 = 9 hours fewer

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FIANCE 50 HELP ASAP
Assoli18 [71]

Calculate the risk measure (beta) compared to the returns of asset and market premium.

1.4 = 4 + 9((rm^-4)

1.4 = 9rm ^-32

33.4 = 9rm

Rm = 3.71

The answer is 3.71

3 0
3 years ago
The greatest number that divide 60 and 84 without leaving a reminder ​
a_sh-v [17]

Step-by-step explanation:

HCF of 60 and 84 = 6

So the required greatest number that divides 60 and 84 without leaving remainder is 6.

6 0
3 years ago
Read 2 more answers
What two numbers add to 8 and multiply to 28
anygoal [31]
Find numbers that multiply to 28 and add them to see if they add to 8
28=
1 and 28=29 not 8
2 and 14=16 not 8
4 and 7=11 not 8
that's it'
no 2 numbers
we must use quadratic formula

x+y=8
xy=28

x+y=8
subtract x fromb oths ides
y=8-x
subsitute
x(8-x)=28
distribute
8x-x^2=28
add x^2 to both sides
8x=28+x^2
subtract 8x
x^2-8x+28=0

if you have
ax^2+bx+c=0 then x=\frac{-b+/- \sqrt{b^{2}-4ac} }{2a}
so if we have
1x^2-8+28=0 then
a=1
b=-8
c=28

x=\frac{-(-8)+/- \sqrt{(-8)^{2}-4(1)(28)} }{2(1)}
x=\frac{8+/- \sqrt{-48} }{2}
x=\frac{8+/- 4 \sqrt{-3} }{2}
x=4+/- 2 \sqrt{-3}
x=4+/- 2i \sqrt{3}
there are no real numbers that satisfy this
5 0
4 years ago
Please explain this answer ☹️
GuDViN [60]

Answer:

see explanation

Step-by-step explanation:

Note that cos315° = cos45° and sin315° = - sin45° and

cos45° = sin45° = \frac{1}{\sqrt{2} } = \frac{\sqrt{2} }{2}

Hence

12(cos315° + isin315°)

= 12(cos45° - isin45°)

= 12(\frac{\sqrt{2} }{2} - i \frac{\sqrt{2} }{2})

= 6\sqrt{2} - 6i \sqrt{2}

7 0
3 years ago
226Ra has a half-life of 1599 years. How much is left after 1000 years if the initial amount was 10 g?
Ray Of Light [21]
\bf \textit{Amount for Exponential Decay using Half-Life}\\\\
A=P\left( \frac{1}{2} \right)^{\frac{t}{h}}\qquad 
\begin{cases}
A=\textit{accumulated amount}\\
P=\textit{initial amount}\to &10\\
t=\textit{elapsed time}\to &1000\\
h=\textit{half-life}\to &1599
\end{cases}
\\\\\\
A=10\left( \frac{1}{2} \right)^{\frac{1000}{1599}}
8 0
3 years ago
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