What digit appears in the units place in the number obtained when 2^320 is multiplied out?

Answers

Answer 1

Answer: Let's start with 2^x and see if we can find a recurring pattern to help us find the answer

2^1 = 2
2^2 = 4
2^3 = 8
2^4 = (1)6
2^5 = (3)2
2^6 = (6)4
2^7 = (12)8
2^8 = (25)6

If you notice, the pattern is 2,4,8,6 in the units place. Every four consecutive x's, the cycle repeats. So at 2^4, 2^8, 2^12 (4096), and 2^16(65336), the units place will all be 6, since x is always divisible by 4 here. When x= 320, we know that 320 is divisible by 4. This means that for 2^320, the units place will also be 6.

Hope this helps, even though it's rather vague!

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Factoring the expression 20a^3b^3 – 24a^5b^2 + 4a^3b^2 gives a new expression of the formUa^xb^y (Wa^2 + Vb+ z), where U > 0.

What is the value of U?

What is the value of W?

What is the value of V?

What is the value of Z?

What is the value of x?

What is the value of y?

Answers

Answer:

The value of U = 4

The value of W = -6

The value of V = 5

The value of Z = 1

The value of x = 3

The value of y =2

Step-by-step explanation:

Given:

20a³b³ – 24a⁵b² + 4a³b²

Solution:

Lets first factorize this expression:

20a³b³ – 24a⁵b² + 4a³b²

= −24a⁵b² + 20a³b³ +4a³b²

Taking 4a³b² common

= 4a³b² (−6a² + 5b + 1) ___ (1)

Now you can see this expression takes the form:

Ua^xb^y (Wa^2 + Vb+ z), where U > 0.

$Ua^(x)b^(y) (Wa^(2) + Vb + z)$

Now you can clearly see from (1)

The value of U is 4

The value of W is -6

The value of V is 5

The value of Z is 1

The value of x is 3

The value of y is 2

However if we take minus common from above equation:

20a³b³ – 24a⁵b² + 4a³b²

= −24a⁵b² + 20a³b³ +4a³b²

Taking 4a³b² common

= - 4a³b² (6a² - 5b - 1) ___ (1)

In this case the value of U is - 4, W is 6 , V is -5, Z is -1 , x is 3 and b is 2

What is the 7th term of the geometric sequence where a1 = -625 and a2 = 125

Answers

Since the sequence is geometric, there is some constant $r$ such that the sequence is recursively given by

$a_n=ra_(n-1)$

By this definition, you can recursively substitute into the right hand side the definition for $a_(n-1),a_(n-2),\ldots$ to find an explicit formula for the $n$ th term.

$a_n=ra_(n-1)=r^2a_(n-2)=r^3a_(n-3)=\cdots=r^(n-1)a_1=-625r^(n-1)$

You know the second term, which means you can find $r$ :

$a_2=-625r^(2-1)\implies125=-625r\implies r=-\frac15$

So, the 7th term of the sequence is

$a_7=-625\left(-\frac15\right)^(7-1)=-\frac1{25}$

Answer:

-0.04 is the answer

Step-by-step explanation:

The correct vocabulary term for a two term polynomial

Answers

binomial
bi=2
nomial=name
2 names

2 terms

binomial

How do you solve a quadratic equation?

Answers

Oh that's easy, all you have to do is use the quadratic equation :)
ax^2+bx+c
A would be the number squared, b would be the number with just an x and c would be the single number :) Look at the attachment and you can see how to set it all up.
Hope this helps and could you mark me as the brainliest?

if in ax^2+bx+c=0 form then
use quadratic formula
x= $\frac{-b+ \sqrt{b^(2)-4ac} }{2a}$ or $\frac{-b- \sqrt{b^(2)-4ac} }{2a}$

input and subsitute and that is the answer

Dan finishes a painting every three weeks. How many paintings will he finish a year?

Answers

Answer:

Approximately 17

Step-by-step explanation:

See added image

Carmen rolls a cube with sides labeled A, B, C, D, E, and F.What is the number of possible outcomes?

Answers

The cube may come to rest with any one of its sides upright. So there are six (6) possible outcomes of a single roll. Technically, there is also some finite probability that the cube will come to rest standing on one of its edges, or on one corner. These probabilities are so small that these additional possible outcomes are almost always ignored, and not counted.

there are six outcomes. so , you would put 1 over 6 for all of them and add them together, which results in 6 possibilities.

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