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aksik [14]
3 years ago
12

Find all positive integers n such that n^4- 1 is divisible by 5.​

Mathematics
1 answer:
zzz [600]3 years ago
5 0

Answer:

All positive integers except multiple of 5.

Step-by-step explanation:

From the rule of divisibility, a number's last digit should be either 0 or 5 for it to be divisible by 5.

For (n^{4} - 1) to be a multiple of 5, the last digit of (n^{4} - 1) should be 0 or 5. In other words, the last digit of n^{4} should be 1 or 6.

The last digit of n^{4} is 1 if n is an odd number except those that are multiples of 5.

The last digit of n^{4} is 6 if n is an even number except those that are multiples of 5.

Therefore, the possible values of n are all positive integers except those that are multiples of 5.

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HELP ASAP !!! The table shows several points for function j(t).
max2010maxim [7]

Answer:

d on edge

Step-by-step explanation:

its just the table flipped

4 0
2 years ago
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Given log630 ≈ 1.898 and log62 ≈ 0.387, log615 =
Tom [10]

\log_630\approx1.898\\\\\log_62\approx0.387\\\\\log_615=\log_6\dfrac{30}{2}=\log_630-\log_62\approx1.898-0.387=1.511\\\\Used:\\\\\log_ab-\log_ac=\log_a\dfrac{b}{c}

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Pleasantburg has a population growth model of P(t)=at2+bt+P0 where P0 is the initial population. Suppose that the future populat
yulyashka [42]

Answer:

The population will reach 34,200 in February of 2146.

Step-by-step explanation:

Population in t years after 2012 is given by:

P(t) = 0.8t^{2} + 6t + 19000

In what month and year will the population reach 34,200?

We have to find t for which P(t) = 34200. So

P(t) = 0.8t^{2} + 6t + 19000

0.8t^{2} + 6t + 19000 = 34200

0.8t^{2} + 6t - 15200 = 0

Solving a quadratic equation:

Given a second order polynomial expressed by the following equation:

ax^{2} + bx + c, a\neq0.

This polynomial has roots x_{1}, x_{2} such that ax^{2} + bx + c = a(x - x_{1})*(x - x_{2}), given by the following formulas:

x_{1} = \frac{-b + \sqrt{\bigtriangleup}}{2*a}

x_{2} = \frac{-b - \sqrt{\bigtriangleup}}{2*a}

\bigtriangleup = b^{2} - 4ac

In this question:

0.8t^{2} + 6t - 15200 = 0

So a = 0.8, b = 6, c = -15200

Then

\bigtriangleup = 6^{2} - 4*0.8*(-15100) = 48356

t_{1} = \frac{-6 + \sqrt{48356}}{2*0.8} = 134.14

t_{2} = \frac{-6 - \sqrt{48356}}{2*0.8} = -141.64

We only take the positive value.

134 years after 2012.

.14 of an year is 0.14*365 = 51.1. The 51st day of a year happens in February.

So the population will reach 34,200 in February of 2146.

6 0
2 years ago
3x + y = 19 , and x + 3y = 1. Find the value of 2x + 2y
soldi70 [24.7K]
3x+y=19 \\
x+3y=1 \\ \\
\hbox{add the sides of the equations:} \\
3x+x+y+3y=19+1 \\
4x+4y=20 \\ \\ \hbox{divide both sides by 2:} \\
2x+2y=10

The answer is d. 10.
6 0
3 years ago
Maya spent her allowance on playing an arcade game a few times and riding the Ferris wheel more than once.
faust18 [17]

Answer:

Sample Response: To write a two-variable equation, I would first need to know how much Maya’s allowance was. Then, I would need the cost of playing the arcade game and of riding the Ferris wheel. I could let the equation be cost of playing the arcade games plus cost of riding the Ferris wheel equals the total allowance. My variables would represent the number of times Maya played the arcade game and the number of times she rode the Ferris wheel. With this equation I could solve for how many times she rode the Ferris wheel given the number of times she played the arcade game.

6 0
3 years ago
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