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Alex_Xolod [135]
3 years ago
9

Create a system of linear equations with one solution. In your final answer, include the system of equations and the graphs of t

he lines.
Mathematics
1 answer:
GrogVix [38]3 years ago
8 0

Answer:

<u>System of equations</u>

x + y = 5

y - 2x = -4

(x,y) = (3,2) --- solution

See attachment for graph

Step-by-step explanation:

Solving (a): Linear equations with 1 solution

The only condition to this is that, the system must have 1 solution.

Other than that, there is no other condition.

A linear equation is represented as:

ax + by= c

For the equation to have 1 solution;

a_1 * b_2 \ne a_2 * b_1

And example of such equation is:

x + y = 5

y - 2x = -4

Solving (b): The graph

See attachment for graph

The solution is the point of intersection of both lines of the graph. So, we have:

(x,y) = (3,2)

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4 years ago
Prove that $5^{3^n} + 1$ is divisible by $3^{n + 1}$ for all nonnegative integers $n.$
Viktor [21]

When n=0, we have

5^{3^0} + 1 = 5^1 + 1 = 6

3^{0 + 1} = 3^1 = 3

and of course 3 | 6. ("3 divides 6", in case the notation is unfamiliar.)

Suppose this is true for n=k, that

3^{k + 1} \mid 5^{3^k} + 1

Now for n=k+1, we have

5^{3^{k+1}} + 1 = 5^{3^k \times 3} + 1 \\\\ ~~~~~~~~~~~~~ = \left(5^{3^k}\right)^3 + 1^3 \\\\ ~~~~~~~~~~~~~ = \left(5^{3^k} + 1\right) \left(\left(5^{3^k}\right)^2 - 5^{3^k} + 1\right)

so we know the left side is at least divisible by 3^{k+1} by our assumption.

It remains to show that

3 \mid \left(5^{3^k}\right)^2 - 5^{3^k} + 1

which is easily done with Fermat's little theorem. It says

a^p \equiv a \pmod p

where p is prime and a is any integer. Then for any positive integer x,

5^3 \equiv 5 \pmod 3 \implies (5^3)^x \equiv 5^x \pmod 3

Furthermore,

5^{3^k} \equiv 5^{3\times3^{k-1}} \equiv \left(5^{3^{k-1}}\right)^3 \equiv 5^{3^{k-1}} \pmod 3

which goes all the way down to

5^{3^k} \equiv 5 \pmod 3

So, we find that

\left(5^{3^k}\right)^2 - 5^{3^k} + 1 \equiv 5^2 - 5 + 1 \equiv 21 \equiv 0 \pmod3

QED

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2 years ago
Find the volume of the cone of with a height of 6cm and a diameter of the base or 6cm
Ainat [17]

Answer:

The volume of cone v= 113.04 cm³.

Step-by-step explanation:

Formula used to find volume of cone is:

v= \pi * r^2 * \frac{h}{3}

where r is the radius and h is the height of cone.

In the given question we are given:

height h= 6 cm

diameter d= 6 cm

We know radius r = d/2 = 6/2 = 3 cm

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So, the volume of cone v= 113.04 cm³ given height h= 6cm and radius r= 3cm

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3 years ago
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