All categories

3 years ago

Explain the following composite numbers into product of prime factors.5929

Mathematics

1 answer:

Alexxx [7]3 years ago

7 0

Since, the prime factors of 5929 are 7, 11. Therefore, the product of prime factors = 7 × 11 = 77.

You might be interested in

Helpppp PLEASEEE AHHHHHH

Allisa [31]

Answer:

Step-by-step explanation:

18+12-6

6 0

2 years ago

Read 2 more answers

First right answer gets brilliant. Can someone please help? It is due today!

nikitadnepr [17]

Answer:

Step-by-step explanation:

A, x=31

E, b=11

N, m=61

Y, b=15

I, a=80

G, k=-78

U, x=-13

R, u=88

O, y=13

A, w=104

H, d=-7

V, n=1/36

Sorry, don't want to give away all the answers!

Hope this helps!!!

7 0

3 years ago

6x + 3y + z = 21<br> y = 7x + 8z - 29<br> -6y - 8z = 6

kicyunya [14]

Answer:

What is the question

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

NEED HELP IN TEST Please

Elden [556K]

Answer:

did you try

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

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:

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

Therefore the half-life is between 600 and 700 years

5 0

3 years ago

Explain the following composite numbers into product of prime factors.5929​

Explain the following composite numbers into product of prime factors.5929