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Romashka-Z-Leto [24]
3 years ago
9

8,706, 2812 What is the value of 7

Mathematics
1 answer:
LiRa [457]3 years ago
5 0

Answer:

7 million is the answer to this question

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Which of the following is not a Pythagorean triple?
Ierofanga [76]

Answer:

  (7, 24, 26)

Step-by-step explanation:

A Pythagorean triple must have an odd number of even numbers. The triple (7, 24, 26) is not a Pythagorean triple.

_____

<em>Additional comment</em>

For an odd integer n, a triple can be formed as ...

  (n, (n²-1)/2, (n²+1)/2)

That is, the following will be Pythagorean triples.

  • (3, 4, 5)
  • (5, 12, 13)
  • (7, 24, 25)
  • (9, 40, 41)
  • (11, 60, 61)

Another series involves even numbers and numbers separated by 2:

  (2n, n²-1, n²+1)

  • (8, 15, 17)
  • (12, 35, 37)
  • (16, 63, 65)

In this list, if n is not a multiple of 2, the triple will be a multiple of one from the odd-number series.

It is a good idea to remember a few of these, as they tend to show up in Algebra, Geometry, and Trigonometry problems.

8 0
2 years ago
Evaluate 66 - 20 - 32 1/4
BigorU [14]

Answer:

13.75

Step-by-step explanation:

66-20- 32 1/4

66-20- 32.25

46-32.25

13.75

8 0
3 years ago
Need help on this ASAP
serg [7]

Answer:

so what to fo tell in brief

if u have time

3 0
3 years ago
What is an equation that goes through the point (-7, 10) and is perpendicular to the line -2x-4y=9
algol13

Step-by-step explanation:

m1.m2= -1

(4y = -2x -9)= (y = -1x/2 -9/4)

-1/2.m2= -1

m2 = 2

formula = y - y1 = m2 ( x - x1)

y - 10 = 2 ( x - (-7))

y - 10 = 2x + 14

y = 2x + 24

7 0
3 years ago
1.Arsenic-74 is used to locate brain tumors. It has a half-life of 17.5 days. 90 mg wereused in a procedure. Write an equation t
Sophie [7]

1)\text{   }N_t\text{ = 90\lparen}\frac{1}{2})^{\frac{t}{17.5}}

2) 5.625 mg will be left

Explanation:

1) Half-life = 17.5 days

initial amount of Arsenic-74 = 90 mg

To get the equation, we will use the equation of half-life:

\begin{gathered} N_t\text{ = N}_0(\frac{1}{2})^{\frac{t}{t_{\frac{1}{2}}}} \\ where\text{ N}_t\text{ = amount remaining} \\ N_0\text{ = initial amount} \\ t_{\frac{1}{2}\text{ }}\text{ = half-life} \end{gathered}N_t\text{ = 90\lparen}\frac{1}{2})^{\frac{t}{17.5}}

2) we need to find the remaining amount of Arsenic-74 after 70 days

t = 70

\begin{gathered} N_t=\text{ 90\lparen}\frac{1}{2})^{\frac{70}{17.5}} \\ N_t\text{ = 5.625 mg} \end{gathered}

So after 70 days, 5.625 mg will be left

4 0
1 year ago
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