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Temka [501]
3 years ago
13

How many years older will you be 1.60 billion seconds from now? (Assume a 365-day year.)

Mathematics
1 answer:
docker41 [41]3 years ago
3 0
There are 31,536,000 Seconds in one year

In order to find out how much older you get, you just needed to divide it :

1,600,000,000 / 31,536,000 

= 50,73 years older

hope this helps
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the cable of a hoist has a cross-section of 80 mm2. the hoist is used to lift a crate weighing 500 kg. what is the stress in the
Daniel [21]

Answer:

stress on the cable is 61.312mpa

If steel is used, extension will be 0.91968mm

If polypropylene is used, extension will be 0.1532mm

Step-by-step explanation: detailed explanation and calculation is shown in the image below

3 0
3 years ago
Pls help!!!!!!!!!!!!!!!!!!
pochemuha

Answer:

x = 8

Step-by-step explanation:

All sides are equal in terms of symmetry, such as the 5 and 5 opposite the angle 120 that repeats itself at the centre 3 times

6 0
2 years ago
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The augmented matrix of a consistent system of five equa- tions in seven unknowns has rank equal to three. How many parameters a
dangina [55]

Answer:

It is necessary four parameters.

Step-by-step explanation:

Since the augmented matrix of the consistent system has rank equal to three then the number of free variables is (number of unknows)-rank then there are 7-3=4 free variables. This means that are necessary four parameter to specify all solutions.

5 0
3 years ago
A lake polluted by bacteria is treated with an antibacterial chemical. Aftertdays, thenumber N of bacteria per milliliter of wat
jeyben [28]

Answer:

N(1)=50 is a minimum

N(15)=4391.7 is a maximum

Step-by-step explanation:

<u>Extrema values of functions </u>

If the first and second derivative of a function f exists, then f'(a)=0 will produce values for a called critical points. If a is a critical point and f''(a) is negative, then x=a is a local maximum, if f''(a) is positive, then x=a is a local minimum.  

We are given a function (corrected)

N(t) = 20(t^2-lnt^2)+ 30

N(t) = 20(t^2-2lnt)+ 30

(a)

First, we take its derivative

N'(t) = 20(2t-\frac{2}{t})

Solve N'(t)=0

20(2t-\frac{2}{t})=0

Simplifying

2t^2-2=0

Solving for t

t=1\ ,t=-1

Only t=1 belongs to the valid interval 1\leqslant t\leqslant 15

Taking the second derivative

N''(t) = 20(2+\frac{2}{t^2})

Which is always positive, so t=1 is a minimum

(b)

N(1)=20(1^2-2ln1)+ 30

N(1)=50 is a minimum

(c) Since no local maximum can be found, we test for the endpoints. t=1 was already determined as a minimum, we take t=15

(d)

N(15)=20(15^2-2ln15)+ 30

N(15)=4391.7 is a maximum

7 0
3 years ago
A rectangle measures 6 inches by 15 inches. If each dimension of the rectangle is dilated by a scale factor of 1/3 to create a n
Brrunno [24]
I think it is A. I could be wrong
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3 years ago
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