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Vaselesa [24]
3 years ago
7

What is the value of x? see attachment below

Mathematics
1 answer:
Kitty [74]3 years ago
4 0

Answer:

  x = 31

Step-by-step explanation:

The perimeter is the sum of the equal-length sides:

  63 = (x -10) +(x -10) +(x -10)

  63 = 3x -30 . . . . . . . . collect terms

  21 = x -10 . . . . . . . . . . divide by 3

  31 = x . . . . . . . . . . . . . add 10

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Researchers are monitoring two different radioactive substances. They have 300 grams of substance A which decays at a rate of 0.
Korolek [52]

Answer:

231.59 years

Step-by-step explanation:

To model this situation we are going to use the exponential decay function:

f(t)= a (1-b)^t

where f(t) is the final amount remaining after t years of decay

a is the final amount

b is the decay rate in decimal form

t is the time in years

For Substance A:

Since  we have 300 grams of the substance, a=300. To convert the decay rate to decimal form, we are going to divide the rate by 100%:

r = 0.15/100 = 0.0015. Replacing the values in our function:

f(t) = a (1-b)^t

f(t) = 300 (1-0.0015)^t

f(t) = 300 (0.9985)^t equation (1)

For Substance B:

Since we have 500 grams of the substance, a= 500. To convert the decay rate to decimal form, we are going to divide the rate by 100%:

r=0.37/100= 0.0037. Replacing the values in our function:

f(t) = a (1-b)^t

f(t)= 500 (1-0.0037)^t

f(t)=500(0.9963)^t equation (2)

Since they are trying to determine how many years it will be before the substances have an equal mass M, we can replace f(t) with M in both equations:

M=300(0.9985)^t equation (1)

M=500(0.9963)^t equation (2)

We can conclude that the system of equations that can be used to determine how long it will be before the substances have an equal mass, M, is :

{M=300(0.9985)^t

{M=500(0.9963)^t

7 0
3 years ago
Simplify the rational expression:<br> 5y+5/2 ÷ 25y-20/40y²-32y
NNADVOKAT [17]

Step-by-step explanation:

Your problem → 5y+5/2 / 25y-20/40y^2-32y

5y+5/2÷25y-20/40y^2-32y

=5⋅y+5/2÷25×y-20/40×y^2-32×y

=5y+5/2÷25×y-20/40×y^2-32y

=5y+5/2×1/25×y-20/40×y^2-32y

=5y+y/10-20/40×y^2-32y

=5y+y/10-y^2/2-32y

=5y×10+y-y^2×5-32y×10 ÷ 10

=50y+y-5y2-320y ÷ 10

= -269y-5y^2 / 10

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