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3 years ago

The formula for the perimeter of a rectangle is P = 2(1 + w) and the formula for the area of a rectangle is A = lw, where l is

Mathematics

1 answer:

frozen [14]3 years ago

6 0

Answer:

The perimeter is 16 inches and the area is 15 square inches.

Step-by-step explanation:

Perimeter of a Rectangle is : 2(l + w)

Area of a Rectangle is l * w

l = length

w = width

Please always note that, in a given dimension for a rectangle, the bigger value is for the LENGTH while the smaller one is the WIDTH.

From the question:

l = 5

w = 4

Perimeter

= 2(l + w)

= 2 ( 5 + 3)

= 2 ( 8 )

= 16 inch

AREA

= l * w

= 5 * 3

= 15 square inches

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3 years ago

Which equation shows an example of the associative property of addition? (–7 i) 7i = –7 (i 7i) (–7 i) 7i = 7i (–7i i) 7i × (–7i

mr Goodwill [35]

The equation which shows the associative property of addition is; (–7 + i) +7i = –7 (i +7i).

<h3>What is the associative property of addition?</h3>

Associative property of addition postulates that Changing the grouping of addends does not change the sum of the numbers. As an instance, ( 5 + 3 ) + 4 = 5 + ( 3 + 4 )

Consequently, the equation which shows the associative property of addition is; (–7 + i) +7i = –7 (i +7i)

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2 years ago

Suppose after 2500 years an initial amount of 1000 grams of a radioactive substance has decayed to 75 grams. What is the half-li

krok68 [10]

Answer:

The correct answer is:

Between 600 and 700 years (B)

Step-by-step explanation:

At a constant decay rate, the half-life of a radioactive substance is the time taken for the substance to decay to half of its original mass. The formula for radioactive exponential decay is given by:

$A(t) = A_0 e^{(kt)}\\where:\\A(t) = Amount\ left\ at\ time\ (t) = 75\ grams\\A_0 = initial\ amount = 1000\ grams\\k = decay\ constant\\t = time\ of\ decay = 2500\ years$

First, let us calculate the decay constant (k)

$75 = 1000 e^{(k2500)}\\dividing\ both\ sides\ by\ 1000\\0.075 = e^{(2500k)}\\taking\ natural\ logarithm\ of\ both\ sides\\In 0.075 = In (e^{2500k})\\In 0.075 = 2500k\\k = \frac{In0.075}{2500}\\ k = \frac{-2.5903}{2500} \\k = - 0.001036$

Next, let us calculate the half-life as follows:

$\frac{1}{2} A_0 = A_0 e^{(-0.001036t)}\\Dividing\ both\ sides\ by\ A_0\\ \frac{1}{2} = e^{-0.001036t}\\taking\ natural\ logarithm\ of\ both\ sides\\In(0.5) = In (e^{-0.001036t})\\-0.6931 = -0.001036t\\t = \frac{-0.6931}{-0.001036} \\t = 669.02 years\\\therefore t\frac{1}{2} \approx 669\ years$