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kap26 [50]
3 years ago
5

Which number has a cube root between 7 and 8

Mathematics
2 answers:
lubasha [3.4K]3 years ago
3 0

<u>Step-by-step explanation:</u>

<u>Here </u> ,There are no options given to choose which number is a cube root between 7 & 8 . So below provided answer is way to choose which numbers have cube root between 7 & 8 .

Here  , We have to find that Which number has a cube root between 7 and 8  . Let's find out :

We know that ,

\sqrt[3]{343}  = 7\\\sqrt[3]{512}  = 8

So , the number which have cube root between 7 & 8 will surely lie in between of 343 & 512  . Suppose the numbers which which have cube root between 7 & 8  are x_1,x_2,x_3,....,x_n , So these numbers lie between 7 & 8 i.e.

⇒ 343

Therefore, all the numbers which lies between 343 and 512 or 343  , have a cube root between 7 & 8 .

alexira [117]3 years ago
3 0

<u>Step-by-step explanation:</u>

The numbers having cube roots between 7-8 are as follows -

Cube Roots - Cube root is a value of the Number such as  27=3*3*3

 here cube root of 27 is 3 as when we multiply 3 thrice we get the number.

The above question requires a number which has cube root between 7 and 8.

1.92*1.92*1.92=7.07  

1.93*1.93*1.93=7.18

1.94*1.94*1.94=7.30

1.95*1.95*1.95=7.41

1.96*1.96*1.96=7.52

1.97*1.97*1.97=7.64

1.98*1.98*1.98=7.76

1.99*1.99*1.99=7.88

the numbers which are cube roots between 7 and 8 are - 1.92,1.93,1.94,1.95,1.96,1.97,1.98,1.99.

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Chris labels every lattice point in the coordinate plane with the square of the distance from the point to the origin (a lattice
stiv31 [10]

Answer:

4 times

Step-by-step explanation:

A lattice point may be defined as the point of intersection of two grid lines or more than two grid lines that is placed in a regularly spaced points arrays. This is called a lattice point.

In the context, Chris tries to label every lattice point in a coordinate plane with its square of distance from the point to its origin. The lattice points means that the numbers are both the integers. So for number 25, Chris has to label 4 times

i.e. (55),(-5,5),(5,-5),(-5,-5)

4 0
3 years ago
Find the sum or difference. a. -121 2 + 41 2 b. -0.35 - (-0.25)
s344n2d4d5 [400]

Answer:

2

Step-by-step explanation:

The reason an infinite sum like 1 + 1/2 + 1/4 + · · · can have a definite value is that one is really looking at the sequence of numbers

1

1 + 1/2 = 3/2

1 + 1/2 + 1/4 = 7/4

1 + 1/2 + 1/4 + 1/8 = 15/8

etc.,

and this sequence of numbers (1, 3/2, 7/4, 15/8, . . . ) is converging to a limit. It is this limit which we call the "value" of the infinite sum.

How do we find this value?

If we assume it exists and just want to find what it is, let's call it S. Now

S = 1 + 1/2 + 1/4 + 1/8 + · · ·

so, if we multiply it by 1/2, we get

(1/2) S = 1/2 + 1/4 + 1/8 + 1/16 + · · ·

Now, if we subtract the second equation from the first, the 1/2, 1/4, 1/8, etc. all cancel, and we get S - (1/2)S = 1 which means S/2 = 1 and so S = 2.

This same technique can be used to find the sum of any "geometric series", that it, a series where each term is some number r times the previous term. If the first term is a, then the series is

S = a + a r + a r^2 + a r^3 + · · ·

so, multiplying both sides by r,

r S = a r + a r^2 + a r^3 + a r^4 + · · ·

and, subtracting the second equation from the first, you get S - r S = a which you can solve to get S = a/(1-r). Your example was the case a = 1, r = 1/2.

In using this technique, we have assumed that the infinite sum exists, then found the value. But we can also use it to tell whether the sum exists or not: if you look at the finite sum

S = a + a r + a r^2 + a r^3 + · · · + a r^n

then multiply by r to get

rS = a r + a r^2 + a r^3 + a r^4 + · · · + a r^(n+1)

and subtract the second from the first, the terms a r, a r^2, . . . , a r^n all cancel and you are left with S - r S = a - a r^(n+1), so

(IMAGE)

As long as |r| < 1, the term r^(n+1) will go to zero as n goes to infinity, so the finite sum S will approach a / (1-r) as n goes to infinity. Thus the value of the infinite sum is a / (1-r), and this also proves that the infinite sum exists, as long as |r| < 1.

In your example, the finite sums were

1 = 2 - 1/1

3/2 = 2 - 1/2

7/4 = 2 - 1/4

15/8 = 2 - 1/8

and so on; the nth finite sum is 2 - 1/2^n. This converges to 2 as n goes to infinity, so 2 is the value of the infinite sum.

8 0
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AleksAgata [21]
The "compound amount" formula is A = P(1+r/n)^(nt),
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Answer:

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Step-by-step explanation:

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on movies = 2/5 × 20 = $8

total spent = 2.5+8 =$ 10.5

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klasskru [66]

Answer:

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Dan : X

Harry: X+5

Regan: 2X

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6 0
2 years ago
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