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m_a_m_a [10]
3 years ago
13

I WILL GIVE BRAINLIEST TO WHOEVER IS CORRECT.

Mathematics
1 answer:
lina2011 [118]3 years ago
4 0
(1/2)^2 - 6(2 - (2/3))
(1/4) - 12 + (12/3)
(3/12) - (144/12) + (48/12)
-(93/12)
<u />
or -7.75
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(√3-√2) (√3+√2) simplify this question​
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Step-by-step explanation:

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Which expression is equivalent to–4–9?<br> 4 - 9<br> -4+ (-9)<br> -4 +9<br> 4+9
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b

B -4+(-9

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because plus a negative number is just like subtracting i belive

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In the diagram, PQRT is a rhombus. STUQ and
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How do I work out PEMDAS
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4 years ago
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A yeast culture weighing 2 grams is removed from a refrigerator unit and is expected to grow at the rate of Upper W prime (t )eq
Naily [24]

Answer:

a) during the first 10 hours of growth the weight will increase 271.8% relative to the initial state ( or 2.718 g)

b) during the first 20 hours of growth the weight will increase  467.1 % relative to the initial state ( or 4.671 g)

Step-by-step explanation:

since the growing rate law is

W'(t) = 0.1 gr/hour * e^(0.1gr/hour* t)  , W'(t) [gr/hour]

and following mathematical conventions: W'(t)= dW/dt

then

dW/dt=0.1 e^(0.1t)

∫dW =∫0.1 e^(0.1t) dt

W = e^(0.1t) + C

at the beginning, (time t=0) the weight is W=2 grams .Therefore

2 g = e^(0.1 g/h*0) + C  → 2 g = 1 g + C → C = 1 g

then

W = e^(0.1t) + 1 g

at t= 10 hours

W = e^(0.1 g/h*10h) + 1 g = 3.718 g/h

therefore the weight will increase

ΔW = 3.718 g -  1 g = 2.718 g  or 271.8% relative to the initial state

for t=20 hours

W = e^(0.1 g/h*20h) + 1 g = 8.389 g/h

thus, the from t= 10 hours to t= 20 hours the weight will increase

ΔW =  8.389 g/h -  3.718 g = 4.671 g  or 467.1 %relative to the initial state

5 0
3 years ago
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