Find the value of ....

Mathematics

1 answer:

kaheart [24]3 years ago

8 0

I don't think this is middle school...

(2^8 x 3^-5 x 6^0)^-2 x ((3^-2)/(2^3))^4 x (2^28)

(256 x (1/243))^-2 x ((1/9)/(8))^4 x 268435456

1.05349794239^-2 x (0.01388888888)^4 x 268435456

0.90101623534 x 3.72108862e-8 x 268435456

8.99999997623

9

9 is your answer.

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Your friend attempted to factor an expression as shown. Find the error in your friends work. Then factor the expression correctl

loris [4]

Answer:

Third step is incorrect. The correct factored form is (x-1)(2x-5).

Step-by-step explanation:

The given expression is

$2x^2-7x+5$

We need to find the factored form of this expression.

Step 1: Given

$2x^2-7x+5$

Step 2: Splitting the middle term method, the middle term can be written as (-5x-2x).

$2x^2-5x-2x+5$

$(2x^2-5x)+(-2x+5)$

Step 3: Taking out common factors from each parenthesis.

$x(2x-5)-1(2x-5)$

Step 4: Taking out common factors.

$(x-1)(2x-5)$

Therefore, the third step is incorrect. The correct factored form is (x-1)(2x-5).

3 0

3 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:

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