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Hatshy [7]
3 years ago
9

Someone please help me thank you

Mathematics
1 answer:
serg [7]3 years ago
8 0

Answer:

6 and 3

Step-by-step explanation:

6 · 3 = 18

6 + 3 = 9

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Answer:

One-sixth times StartFraction 2 over 1 EndFraction

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3 years ago
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A manufacturing plant is capable of producing 10 tons of product per day when it runs three shifts with no breakdowns and plenty
Temka [501]

Answer: 73%

Step-by-step explanation:

The plant is capable of producing 10 tons of product in one day.

Currently however, it is only able to produce 7.3 tons per day.

Utilization is therefore:

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= 7.3/10 * 100

= 73%

3 0
3 years ago
Eric bought a new car for $24,000. Each year, the car depreciates 15% in value. To the nearest dollar, how much will the car be
Artyom0805 [142]

Answer:

$9052

Step-by-step explanation:

a= starting car price

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r= rate

t= # of years

Equation:

a(1-r)^t

Plug in the numbers.

a= 24000   r=15%=0.15   (1-r)= 1-0.15 = 0.85  

a(1-r)^t      =      24000(0.85)^6

               =    24000(0.37714951562)

                  = 9051.588375

                    = 9052

3 0
3 years ago
The? half-life of a radioactive substance is 20 years. If you start with some amount of this? substance, what fraction will rema
yarga [219]

\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}\\ t=\textit{elapsed time}\dotfill &120\\ h=\textit{half-life}\dotfill &20 \end{cases} \\\\\\ A=P\left( \frac{1}{2} \right)^{\frac{120}{20}}\implies A=P\left( \frac{1}{2} \right)^6\implies A=P\left( \cfrac{1^6}{2^6} \right)\implies A=\cfrac{1}{64}P

6 0
4 years ago
Help with this question.<br> What is cos 45*?
Gre4nikov [31]

Answer:

  • 0.7071.

Step-by-step explanation:

Now, this can be tricky.

<u>But, we can use a calculator.</u>

  • cos 45*
  • = 0.7071.

<u>According to the calculator, the answer was 0.7071.</u>

That means 0.7071 is the answer.

This image is optional. Use it for hints.

6 0
3 years ago
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