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Lubov Fominskaja [6]
3 years ago
5

Simplify the expressions 1.) 6(–2y) – 9 + 4(–2y + 6) 2.) 5(b – 4) + 12

Mathematics
2 answers:
navik [9.2K]3 years ago
7 0

Step-by-step explanation:

1) -12y -9 -8y +24= -20y +15

2) 5b -20 +12= 5b-8

Mademuasel [1]3 years ago
7 0

Answer:

1. -20y-3.    2. 5b-8

Step-by-step explanation:

1. -12y-9-8y+6

-20y-3

2. 5b-20+12

5b-8

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3 years ago
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Workers are loading boxes onto a truck. So far, they have loaded 48 boxes onto the truck.
svlad2 [7]
48=4% loaded on truck and they have 96% left to load
So set up a proportion and it should look like this
48 over x = 4 over 100
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6 0
3 years ago
find each comission, given the sale and the comission rate 1. $2,500, 8% 2. $ 2,00, 7.5% 3. $600, 4.5%
Iteru [2.4K]

Answer:

1. $200 2. $150 3.$27

Step-by-step explanation:

1. 2,500 x .08 = 200

2. 2,000 x .075 = 150

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8 0
4 years ago
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Manuel's portfolio includes 66 shares of Dell, 95 shares of Coca Cola, and 180 shares of Nike. If Dell pays a yearly dividend of
KengaRu [80]
We are going to take out the dividends step by step:
 Dell:
 (66) * (1.79) = 118.14 $
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 (95) * (2.62) = 248.9 $
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7 0
4 years ago
A breeder reactor converts uranium-238 into an isotope of plutonium-239 at a rate proportional to the amount of uranium-238 pres
topjm [15]

Let U(t) denote the amount of uranium-238 in the reactor at time t. As conversion to plutonium-239 occurs, the amount of uranium will decrease, so the conversion rate is negative. Because the rate is proportional to the current amount of uranium, we have

\dfrac{\mathrm dU}{\mathrm dt}=-kU

where k>0 is constant. Separating variables and integrating both sides gives

\dfrac{\mathrm dU}U=-k\,\mathrm dt\implies\ln|U|=-kt+C\implies U=Ce^{-kt}

Suppose we start some amount u. This means that at time t=0 we have U(0)=u, so that

u=Ce^{-0k}\implies C=u\implies U=ue^{-kt}

We're given that after 10 years, 99.97% of the original amount of uranium remains. This means (if t is taken to be in years) for some starting amount u,

0.9997u=ue^{-10k}\implies k=-\dfrac{\ln(0.9997)}{10}

The half-life is the time t_{1/2} it takes for the starting amount u to decay to half, 0.5u:

0.5u=ue^{-kt_{1/2}}\implies t_{1/2}=-\dfrac{\ln(0.5)}k=\dfrac{10\ln2}{\ln(0.9997)}

or about 23,101 years. Notice that it doesn't matter what the actual starting amount is, the half-life is independent of that.

7 0
3 years ago
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