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diamong [38]
3 years ago
15

What is the greatest perfect square of 1290

Mathematics
2 answers:
Alex73 [517]3 years ago
6 0
<span>It has no perfect square factors, unless you count 1 = 1^2. That's sort of a degenerate case we don't usually count, since every integer has that factor. We would usually say that 1290 is a square-free integer. Hope this helps!!</span>
xenn [34]3 years ago
3 0

There can't be greatest perfect Square of an Integer or Natural Number.There is only one Square of a Natural Number.

So, (1290)²

=1290×1290

<u>=(1300-10)×(1300-10)</u>

=1300×(1300-10)-10×(1300-10)

=1,690,000-13,000-13,000+100

=1,690,000-26,000+100

=1,664,000+100

=1,664,100

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Answer:

He should buy 4 boxes

Step-by-step explanation:

25 divided by 8 is 3.125, but you want to have enough so the least amount you can buy is 4.

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"Four times a number is no more than 36."
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Step-by-step explanation:

if we consider x as the number we can write this inequality:

4x ≤ 36

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3 years ago
How to solve these 3 problems
Bingel [31]

Answer:

  1. a. decay; b. growth; c. decay; d. neither
  2. r = 4; a = 1; y = 1·4^x
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Step-by-step explanation:

There are two kinds of exponential problems here.

  1. exponential functions of the form f(x) = a·b^x
  2. exponential sequences of the explicit form an = a1·r^(n-1)

The second problem gives you a table that suggests the sequence form, but it asks for the exponential function form. The third problem does something similar.

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<h3>1.</h3>

In an exponential function of the form f(x) = a·b^x, the function grows if b>1 and decays if b<1. Using this check, we can easily answer ...

  a. 0.4 < 1 . . . decay

  b. 1.3 > 1 . . . growth

  c. 1/2 < 1 . . . decay

  d. 1 = 1 . . . neither growth nor decay; the function is constant: j(x) = 1.

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<h3>2.</h3>

The value of x is given starting at 1, so we can consider this a geometric sequence. The common ratio is r = 16/4 = 4. The first term is a1 = 4, so the explicit formula for the sequence is ...

  an = 4·4^(n-1)

When this is expanded to get rid of the constant in the exponent, we have ...

  an = 4·(4^n)·(4^-1) = 1·4^n

We recognize this form as matching the functional form f(x) = a·r^x. The multiplier of the exponential factor is a=1. In summary, ...

  r = 4; a = 1; f(x) = 1·4^x

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<h3>3.</h3>

The first term of this geometric sequence is a1 = 3. The common ratio is r = 15/3 = 5. Using the explicit formula, we have ...

 a. explicit form: an = 3·5^(n-1)

Using the method of question 2 to write the functional form, we find ...

  an = 3(5^n)(5^-1) = (3/5)(5^n)

  b. functional form: f(x) = (3/5)(5^x)

  c. function family: exponential growth functions

 d. y-intercept: (3/5) . . . . read this from the f(x) form

     1st term: the first term listed in the given sequence is 3

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<em>Additional comment</em>

The "y-intercept" of a sequence is irrelevant (undefined), as the sequence term numbering starts with 1, not 0. The domain of the explicit formula is <em>natural numbers</em>, which does not include 0.

Similarly, the "first term" of a function f(x) needs further definition. Here, we've answered the question by saying the first term is f(1). There is no conventional definition of a "first term" for a continuous function.

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Answer:

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Step-by-step explanation:

Distrubute

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IRISSAK [1]
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8 0
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