All categories

3 years ago

What pattern do you see in the powers of 4?

2 answers:

8 0

All of them are even numbers. You will not find a power of 4 that is an odd number. Also, all of the numbers have a 4 or a 6 or both. For example,
4^3= 64

4^8= 65536

I hope this helps:)

Kay [80]3 years ago

3 0

They are all EVEN numbers

You might be interested in

Charles earns $8.75 per hour at his part-time job. On Sunday, he worked 4.25 hours. What were his total earnings for the day?

Step2247 [10]

Ok so if he earns 8.75 per hour the we have to multiply 8.75*4.25=37.1875 so in total he earns 37.19

7 0

2 years ago

if 4 shirts and 3 ties cost 241. and 3 shirts and 4 ties cost 200. what is the cost of 1 shirt and 1 tie?

Inessa05 [86]

Answer:

cost of 1 tie = 11

cost of 1 shirt = 52

Step-by-step explanation:

3 0

3 years ago

The area of a pizza is 220 square centimeters .

Zanzabum

Area of pizza = pi × r^2
pi × r^2 = 220 Sq. cm
22/7 × r^2 =220
r^2 = 220 × 7/22
r^2 = 70
r = square root of 70
r = 8.36

therefore , radius of the pizza is 8.36 cm

(please do mark as brainliest)

7 0

3 years ago

Please can someone solve this

Crazy boy [7]

Answer:

200.86

Step-by-step explanation:

Using the formulas:

A=πr2

C=2πr

Solving for A:

A=C2

4π=50.242

4·π≈200.85812

I hope this helps.

4 0

3 years ago

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

3 years ago