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3 years ago

Prove that if an integer is a perfect square and a perfect cube, simultaneously, then it is either of the form 7k or 7k+1

1 answer:

5 0

If an integer is both a square and a cube, it can be of the form:
(a3)^2
Now,
since a cube can be of the form 7k or 7k+-1(thanks to FoolForMath),
we write
a^3=7k
and get the no to be
49k^2
, which is in the form of 7 times something
49k^2=7×(7k^2)
Now put
a^3=7k+−1 Square it
and you'll get a number in the form of (7times something +1)

You might be interested in

How many steel rods are in a staircase frame with 12 steps?

kolbaska11 [484]

180 steel rods to be exact

3 0

3 years ago

The average of 5 quiz grades is 10. When the lowest grade is dropped and the new average is calculated, it turns out to be 11. W

V125BC [204]

Answer:

The score of the dropped grade is 6

Step-by-step explanation:

To find the average of a set of numbers you need to add all the numbers and divide by how many numbers there are. The problem wants you to solve it in reverse. Let put this into an algebraic equation with a, b, c, d, and e being the variables of the test scores

(a+b+c+d+e)/5=10

Then we can multiply both sides by 5 and get

a+b+c+d+e=50

Lets assume that c is the lowest test score. To calculate the average of that we get

a+b+d+e/4=11

Doing the same thing, we know that a+b+d+e=44

Now compare the two:

a+b+c+d+e=50

a+b+d+e=44

We can now know that c=50-44=6

So, 6 is the score of the dropped quiz. Hope that helped!

4 0

3 years ago

Y = x + 2 y = 3x - 4

exis [7]

Answer:

x=3; y=5

Step-by-step explanation:

y=x+2

y=3x-4

so x+2=3x-4

2x=6

x=3

because y=x+2

x=3

so y=3+2

8 0

3 years ago

Which equation has the solutions x=1+/- 5?

nikitadnepr [17]

Answer:

$x^{2} -2x-4=0$

Solving Steps

Use the quadratic formula

$x=\frac{-(-2)(+/-)\sqrt{(-2)^2-4*1*(-4)} }{2*1}$

Multiply / Remove the parentheses / Evaluate

$x=\frac{2(+/-)\sqrt{4+16} }{2}$

Calculate

$x=\frac{2(+/-)\sqrt{20} }{2}$

Separate the solutions

$x=\frac{2(+/-)2\sqrt{5} }{2} \\x=\frac{2-2\sqrt{5} }2}$

Simplify

$x=1+\sqrt{5} \\x=1-\sqrt{5}$

4 0

3 years ago

Every day your friend commutes to school on the subway at 9 AM. If the subway is on time, she will stop for a $3 coffee on the w

Shtirlitz [24]

Answer:

1.02% probability of spending 0 dollars on coffee over the course of a five day week

7.68% probability of spending 3 dollars on coffee over the course of a five day week

23.04% probability of spending 6 dollars on coffee over the course of a five day week

34.56% probability of spending 9 dollars on coffee over the course of a five day week

25.92% probability of spending 12 dollars on coffee over the course of a five day week

7.78% probability of spending 12 dollars on coffee over the course of a five day week

Step-by-step explanation:

For each day, there are only two possible outcomes. Either the subway is on time, or it is not. Each day, the probability of the train being on time is independent from other days. So we use the binomial probability distribution to solve this problem.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

In this problem we have that:

The probability that the subway is delayed is 40%. 100-40 = 60% of the train being on time, so $p = 0.6$

The week has 5 days, so $n = 5$

She spends 3 dollars on coffee each day the train is on time.

Probabability that she spends 0 dollars on coffee:

This is the probability of the train being late all 5 days, so it is P(X = 0).

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{5,0}.(0.6)^{0}.(0.4)^{5} = 0.0102$

1.02% probability of spending 0 dollars on coffee over the course of a five day week

Probabability that she spends 3 dollars on coffee:

This is the probability of the train being late for 4 days and on time for 1, so it is P(X = 1).

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 1) = C_{5,1}.(0.6)^{1}.(0.4)^{4} = 0.0768$

7.68% probability of spending 3 dollars on coffee over the course of a five day week

Probabability that she spends 6 dollars on coffee:

This is the probability of the train being late for 3 days and on time for 2, so it is P(X = 2).

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 2) = C_{5,2}.(0.6)^{2}.(0.4)^{3} = 0.2304$

23.04% probability of spending 6 dollars on coffee over the course of a five day week

Probabability that she spends 9 dollars on coffee:

This is the probability of the train being late for 2 days and on time for 3, so it is P(X = 3).

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 3) = C_{5,3}.(0.6)^{3}.(0.4)^{2} = 0.3456$

34.56% probability of spending 9 dollars on coffee over the course of a five day week

Probabability that she spends 12 dollars on coffee:

This is the probability of the train being late for 1 day and on time for 4, so it is P(X = 4).

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 4) = C_{5,4}.(0.6)^{4}.(0.4)^{1} = 0.2592$

25.92% probability of spending 12 dollars on coffee over the course of a five day week

Probabability that she spends 15 dollars on coffee:

Probability that the subway is on time all days of the week, so $P(X = 5)$ .

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 5) = C_{5,5}.(0.6)^{5}.(0.4)^{0} = 0.0778$

7.78% probability of spending 12 dollars on coffee over the course of a five day week

8 0

3 years ago