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erastovalidia [21]
3 years ago
13

1440 tons of grain were brought to a silo over two days. On the second day, 80% of the amount of the grain delivered during the

first day was brought in. How many tons of grain were delivered to the silo on the first day?
Mathematics
1 answer:
stellarik [79]3 years ago
6 0
Answer is 800 
because if you do
1440=0.8t+t
1440=1.8t
800=t 
so the first day is 800 tons and the second day is 640 tons
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Initially, there are 40 grams of A and 50 grams of B, and for each gram of B, 2 grams of A is used. It is observed that 15 grams
hram777 [196]

Answer:

X(16)=25.71grams

Step-by-step explanation:

let X(t) denote grams of C formed in  t mins.

For X grams of C we have:

\frac{2}{3}Xg of A and \frac{1}{3}Xg of B

Amounts of A,B remaining at any given time is expressed as:

40-\frac{2}{3}Xg of A and  50-\frac{1}{3}Xg  of B

Rate at which C is formed satisfies:

\frac{dX}{dt} \infty(40-\frac{2}{3}X)(50-\frac{1}{3}X)->\frac{dX}{dt}=k(90-X)\\\therefore \frac{dX}{(90-X)^2}=kdt->\int{\frac{dX}{(90-X)^2}} \, =\int {k} \, dt  \\\therefore \frac{1}{90-X}=kt+c->90-X=\frac{1}{kt+c}\\\\X(t)=90-\frac{1}{kt+c}

Apply the initial condition,X(0)=0 ,to the expression above

0=90-\frac{1}{c} \ \ ->c=\frac{1}{90}\\\therefore\\X(t)=90-\frac{1}{kt+\frac{1}{90}} \ \ ->X(t)=90-\frac{90}{90kt+c}

Now at X(8)=15:

15=90-\frac{90}{90\times 8k+1}  \ ->75=\frac{90}{720k+1}\\k=0.0002778

Substitute  in X(t) to get

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5 0
3 years ago
Find the area of a rectangle with a length of 7 centimeters and a width of 24 centimeters.
nadezda [96]

Answer:

168 cm^2

Step-by-step explanation:

Given,

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