All categories

3 years ago

Prove that the value of the expression (125^2+25^2)(5^2–1) is divisible by 3

Mathematics

2 answers:

Radda [10]3 years ago

5 0

Step-by-step explanation:

$A =(125^{2} + 25^{2} ) (5^{2} - 1)\\A = [(5^{3})^{2} + (5^{2})^{2} ] . (5^{2} - 1)\\A = (5^{6} + 5^{4} ). (5^{2} - 1)\\A = [5^{4} . ( 5^{2} + 5)].(5^{2} - 1) \\A = 5^{4} . (25+ 5). (5^{2} - 1)\\A = 5^{4} . 30. (5^{2} - 1)\\$

Since 30 is divisible by 3

Thus, A is divisible by 3

Good luck!

shtirl [24]3 years ago

3 0

$(125^2+25^2)(5^2-1)=\\\\=(125^2+25^2)(25-1)=\\\\=(125^2+25^2)\cdot24=\\\\=(125^2+25^2)\cdot8\cdot{\bold{\underline3}}$

125 and 25 are integer ⇒125² and 25² are integer ⇒(125²+25²) is integer

One of factors of given expression is 3 so given expression is divisible by 3

q.e.d.

You might be interested in

Find the value of sin 0 cos(90°- 0) + cos 0 sin(90°-0)

Step2247 [10]

Step-by-step explanation:

$sin \theta \: cos(90 \degree - \theta) + cos \theta \: sin(90 \degree - \theta) \\ = sin(\theta + 90 \degree - \theta) \\ = sin(90 \degree) \\ = 1$

5 0

3 years ago

Find the cost of a cd in 6years,assuming an inflation rate if 4% , if its present cost is $11.95

choli [55]

The cost of a cd after 6 year is =$ 15.12

Step-by-step explanation:

The present cost of the cd is $11.95 and inflation rate 4%

P = $11.95 , r = 4% and n = 6 years

The cost of a cd after 6 year is = $P (1+\frac{r}{100})^n$

=$ $11.95(1+\frac{4}{100})^6$

=$ 15.12

3 0

3 years ago

What is the approximate circumference of the circle shown below?

Serggg [28]

Hey, there’s no circle for me to look at unfortunately!

4 0

3 years ago

Thanks sm need an answer asap

Novosadov [1.4K]

Answer:

0.8 s

Step-by-step explanation:

Horizontal velocity

= 61cos(35) = 49.968 f/s

To reach 40 feet, it takes

40/49.968 = 0.801 = 0.8 sec

7 0

3 years ago

Suppose small aircraft arrive at a certain airport according to a Poisson process with rate a 5 8 per hour, so that the number o

timurjin [86]

Answer:

(a) P (X = 6) = 0.12214, P (X ≥ 6) = 0.8088, P (X ≥ 10) = 0.2834.

(b) The expected value of the number of small aircraft that arrive during a 90-min period is 12 and standard deviation is 3.464.

Step-by-step explanation:

Let the random variable X = number of aircraft arrive at a certain airport during 1-hour period.

The arrival rate is, λt = 8 per hour.

(a)

For t = 1 the average number of aircraft arrival is:

$\lambda t=8\times 1=8$

The probability distribution of a Poisson distribution is:

$P(X=x)=\frac{e^{-8}(8)^{x}}{x!}$

Compute the value of P (X = 6) as follows:

$P(X=6)=\frac{e^{-8}(8)^{6}}{6!}\\=\frac{0.00034\times262144}{720}\\ =0.12214$

Thus, the probability that exactly 6 small aircraft arrive during a 1-hour period is 0.12214.

Compute the value of P (X ≥ 6) as follows:

$P(X\geq 6)=1-P(X$

Thus, the probability that at least 6 small aircraft arrive during a 1-hour period is 0.8088.

Compute the value of P (X ≥ 10) as follows:

$P(X\geq 10)=1-P(X$

Thus, the probability that at least 10 small aircraft arrive during a 1-hour period is 0.2834.

(b)

For t = 90 minutes = 1.5 hour, the value of λ, the average number of aircraft arrival is:

$\lambda t=8\times 1.5=12$

The expected value of the number of small aircraft that arrive during a 90-min period is 12.

The standard deviation is:

$SD=\sqrt{\lambda t}=\sqrt{12}=3.464$

The standard deviation of the number of small aircraft that arrive during a 90-min period is 3.464.

(c)

For t = 2.5 the value of λ, the average number of aircraft arrival is:

$\lambda t=8\times 2.5=20$

Compute the value of P (X ≥ 20) as follows:

$P(X\geq 20)=1-P(X$

Thus, the probability that at least 20 small aircraft arrive during a 2.5-hour period is 0.5298.

Compute the value of P (X ≤ 10) as follows:

$P(X\leq 10)=\sum\limits^{10}_{x=0}(\frac{e^{-20}(20)^{x}}{x!})\\=0.01081\\\approx0.0108$

Thus, the probability that at most 10 small aircraft arrive during a 2.5-hour period is 0.0108.

8 0

3 years ago