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3 years ago

Help me please why is school hard i-

Mathematics

2 answers:

Lelechka [254]3 years ago

8 0

Answer:

12 1/4

Step-by-step explanation:

You would first add the fractions which will get you 5/4. then you will add the whole numbers which will get you 11. Then there is a extra number in 5/4 so you can tranfer it to the whole number which will become... 12 and then you will be left with 12 1/4

blagie [28]3 years ago

6 0

The answer is 12 1/4. If you add the whole numbers you get 11. Then you add 3/4 and 2/4, which gets you 5/4. 5/4 can then be simplified into 1 1/4. Finally you add 1 1/4 to 11 and you get 12 1/4. Hope this helps :)

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The value of all of the quarters and dimes in a

Inessa05 [86]

Answer:

Step-by-step explanation:

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3 years ago

Consider functions f and g below.

Katyanochek1 [597]

Answer:

g(x) approaches negative infinity

Step-by-step explanation:

g(x) = -x² + 2x + 4

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I can't solve for f(x) because you didnt give me f(x)

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3 years ago

Read 2 more answers

How many terms of the arithmetic sequence {1,22,43,64,85,…} will give a sum of 2332? Show all steps including the formulas used

MA_775_DIABLO [31]

There's a slight problem with your question, but we'll get to that...

Consecutive terms of the sequence are separated by a fixed difference of 21 (22 = 1 + 21, 43 = 22 + 21, 64 = 43 + 21, and so on), so the n-th term of the sequence, a (n), is given recursively by

• a (1) = 1

• a (n) = a (n - 1) + 21 … … … for n > 1

We can find the explicit rule for the sequence by iterative substitution:

a (2) = a (1) + 21

a (3) = a (2) + 21 = (a (1) + 21) + 21 = a (1) + 2×21

a (4) = a (3) + 21 = (a (1) + 2×21) + 21 = a (1) + 3×21

and so on, with the general pattern

a (n) = a (1) + 21 (n - 1) = 21n - 20

Now, we're told that the sum of some number N of terms in this sequence is 2332. In other words, the N-th partial sum of the sequence is

a (1) + a (2) + a (3) + … + a (N - 1) + a (N) = 2332

or more compactly,

$\displaystyle\sum_{n=1}^N a(n) = 2332$

It's important to note that N must be some positive integer.

Replace a (n) by the explicit rule:

$\displaystyle\sum_{n=1}^N (21n-20) = 2332$

Expand the sum on the left as

$\displaystyle 21 \sum_{n=1}^N n-20\sum_{n=1}^N1 = 2332$

and recall the formulas,

$\displaystyle\sum_{k=1}^n1=\underbrace{1+1+\cdots+1}_{n\text{ times}}=n$

$\displaystyle\sum_{k=1}^nk=1+2+3+\cdots+n=\frac{n(n+1)}2$

So the sum of the first N terms of a (n) is such that

21 × N (N + 1)/2 - 20N = 2332

Solve for N :

21 (N ² + N) - 40N = 4664

21 N ² - 19 N - 4664 = 0

Now for the problem I mentioned at the start: this polynomial has no rational roots, and instead

N = (19 ± √392,137)/42 ≈ -14.45 or 15.36

so there is no positive integer N for which the first N terms of the sum add up to 2332.

4 0

2 years ago

How many different perfect cubes are among the positive actors of 2021^2021

9966 [12]

Answer:

hope this helps :D

Step-by-step explanation:

Perfect cube factors:

If a number is a perfect cube, then the power of the prime factors should be divisible by 3.

Example 1:Find the number of factors of293655118 that are perfect cube?

Solution: If a number is a perfect cube, then the power of the prime factors should be divisible by 3. Hence perfect cube factors must have

2(0 or 3 or 6or 9)—– 4 factors

3(0 or 3 or 6) —– 3 factors

5(0 or 3)——- 2 factors

11(0 or 3 or 6 )— 3 factors

Hence, the total number of factors which are perfect cube 4x3x2x3=72

Perfect square and perfect cube

If a number is both perfect square and perfect cube then the powers of prime factors must be divisible by 6.

Example 2: How many factors of 293655118 are both perfect square and perfect cube?

Solution: If a number is both perfect square and perfect cube then the powers of prime factors must be divisible by 6.Hence both perfect square and perfect cube must have

2(0 or 6)—– 2 factors

3(0 or 6) —– 2 factors

5(0)——- 1 factor

11(0 or 6)— 2 factors

Hence total number of such factors are 2x2x1x2=8

Example 3: How many factors of293655118are either perfect squares or perfect cubes but not both?

Solution:

Let A denotes set of numbers, which are perfect squares.

If a number is a perfect square, then the power of the prime factors should be divisible by 2. Hence perfect square factors must have

2(0 or 2 or 4 or 6 or 8)—– 5 factors

3(0 or 2 or 4 or 6) —– 4 factors

5(0 or 2or 4 )——- 3 factors

11(0 or 2or 4 or6 or 8 )— 5 factors

Hence, the total number of factors which are perfect square i.e. n(A)=5x4x3x5=300

Let B denotes set of numbers, which are perfect cubes

If a number is a perfect cube, then the power of the prime factors should be divisible by 3. Hence perfect cube factors must have

2(0 or 3 or 6or 9)—– 4 factors

3(0 or 3 or 6) —– 3 factors

5(0 or 3)——- 2 factors

11(0 or 3 or 6 )— 3 factors

Hence, the total number of factors which are perfect cube i.e. n(B)=4x3x2x3=72

If a number is both perfect square and perfect cube then the powers of prime factors must be divisible by 6.Hence both perfect square and perfect cube must have

2(0 or 6)—– 2 factors

3(0 or 6) —– 2 factors

5(0)——- 1 factor

11(0 or 6)— 2 factors

Hence total number of such factors are i.e.n(A∩B)=2x2x1x2=8

We are asked to calculate which are either perfect square or perfect cubes i.e.

n(A U B )= n(A) + n(B) – n(A∩B)

=300+72 – 8

=364

Hence required number of factors is 364.

8 0

3 years ago

Mandy cut 34 1⁄3 feet of string to use for the tail of a kite. However, she realized that the tail was too long and cut off 6 3⁄

andrezito [222]

34 1/3- 6 3/5= 416/15

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3 years ago