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Komok [63]
2 years ago
4

Tess created a Pythagorean triple of (24, 32, 40) by multiplying the known Pythagorean triple 3, 4, and 5 by 8. Is Tess correct?

Can a Pythagorean triple be created using multiples of a known Pythagorean triple.
Yes, as long as each number in the Pythagorean triple is multiplied by the same whole number.
Yes, as long as each number in the Pythagorean triple is multiplied by a different whole number.
No, Pythagorean triples only exist with small numbers.
No, multiplying a known Pythagorean triple by a whole number will not create a Pythagorean triple.
Mathematics
2 answers:
Serhud [2]2 years ago
8 0

Answer:

Yes, as long as each number in the Pythagorean Triple is multiplied by the same whole number.

General Formulas and Concepts:

<u>Pre-Algebra</u>

  • Equality Property

<u>Trigonometry</u>

[Right Triangles Only] Pythagorean Theorem: a² + b² = c²

  • a is a leg
  • b is another leg
  • c is the hypotenuse<u> </u>

Step-by-step explanation:

In order for it to <em>remain</em> a Pythagorean triple, we must follow the equality rules. To make it the exact same expression, <em>all</em> parts of an equation must be multiplied or divided by the <em>same</em> number. If this is done, then it would also <em>remain</em> a Pythagorean triple, and hence we arrive at our answer.

WITCHER [35]2 years ago
4 0

Answer:

Hello,

Answer A

Yes, as long as each number in the Pythagorean triple is multiplied by the same whole number.

Step-by-step explanation:

(a,b,c) is a Pythagorien triple  : c²=a²+b²

and k is a naturel.

(ka)^2+(kb)^2\\\\=k^2*a^2+k^2*b^2\\\\=k^2(a^2+b^2)\\\\=k^2c^2\\\\=(kc)^2\\

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Viktor [21]

Answer:

  1. 6\dfrac{2}{3} ounces of raisin.
  2. 9\dfrac{1}{3} ounces of almond.

Step-by-step explanation:

Hint- This problem can be solved using unitary method.

A recipe for trail miss uses 7 ounces of almonds with 5 ounces of raisins.

If we mix almond and raisin in equal proportion we get 7+5=12 ounces.

We know that 16 ounces make 1 pond.

Hence, in a mixture of 12 ounces we have 5 ounce of raisin,

so in a mixture of 1 ounces we have \dfrac{5}{12} ounce of raisin,

so in a mixture of 16 ounces we will have \dfrac{5}{12}\times 16=\dfrac{20}{3}=6\dfrac{2}{3} ounces of raisin.

Therefore, the amount of almond is =16-\dfrac{20}{3}=\dfrac{28}{3}=9\dfrac{1}{3} ounces.


3 0
2 years ago
Please help! Which system of inequalities represented by the graph?
MatroZZZ [7]

Answer: B

x + 2y \leq 4 \textrm{ and } 3x + y \leq 4

Step-by-step explanation:

Insert the values of x and y for any point in the heavily shaded region and see if it satisfies both inequalities

We can see that the origin (0,0) is inside the shaded region so use these values because they are the easiest to compute

Plug in x = 0 and y = 0 into the first inequality
0 + 2.0 = 0 which is greater than 5.

So we can eliminate both A and D since they both contain incorrect inequalities

Now, test the second inequality with x = 0 and y = 0

For B we get

3.0 + 0 = 0 which is indeed less than 4

C is not applicable since 0 is not greater-than-equal to 4

Hence B is the correct solution

This can be verified using a graphing tool (see image)

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This extreme value problem has a solution with both a maximum value and a minimum value. Use Lagrange multipliers to find the ex
olga nikolaevna [1]

f(x_1,\ldots,x_n)=x_1+\cdots+x_n=\displaystyle\sum_{i=1}^nx_i

{x_1}^2+\cdots+{x_n}^2=\displaystyle\sum_{i=1}^n{x_i}^2=4

The Lagrangian is

L(x_1,\ldots,x_n,\lambda)=\displaystyle\sum_{i=1}^nx_i+\lambda\left(\sum_{i=1}^n{x_i}^2-4\right)

with partial derivatives (all set equal to 0)

L_{x_i}=1+2\lambda x_i=0\implies x_i=-\dfrac1{2\lambda}

for 1\le i\le n, and

L_\lambda=\displaystyle\sum_{i=1}^n{x_i}^2-4=0

Substituting each x_i into the second sum gives

\displaystyle\sum_{i=1}^n\left(-\frac1{2\lambda}\right)^2=4\implies\dfrac n{4\lambda^2}=4\implies\lambda=\pm\frac{\sqrt n}4

Then we get two critical points,

x_i=-\dfrac1{2\frac{\sqrt n}4}=-\dfrac2{\sqrt n}

or

x_i=-\dfrac1{2\left(-\frac{\sqrt n}4\right)}=\dfrac2{\sqrt n}

At these points we get a value of f(x_1,\cdots,x_n)=\pm2\sqrt n, i.e. a maximum value of 2\sqrt n and a minimum value of -2\sqrt n.

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