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2 years ago

Question 1 of 4

1 answer:

6 0

OB. The digits in the millions period are 41, the digits in the thousands period are 000, and the digits in the ones period are 138.

The digits of a number can be classified into into 3 groups and each such group is called a period.

Here, we are give a number 41,000,138.

In this number, the million period consists of

ten millions - 4

and millions - 1

Hence the digits in the million periods are 41

The thousand period consists of all zeroes, hence the digits in thousand periods are 000.

The ones period contains-

hundreds - 1

tens - 3

and ones - 8

Thus, the digits in the ones period are 138

Hence, option B is the correct answer.

Learn more about place value here-

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Read more on cost per unit;

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3 0

2 years ago

What is this expand 2(2x-3)

klemol [59]

Answer:

4x - 6

Step-by-step explanation:

2 × 2x = 4x

2 × -3 = -6

4x -6

3 0

3 years ago

Read 2 more answers

Evaluate the expression below. (8 - 7)- 4 + 3(-1) 1 -7 -1 7

Masja [62]

Answer:

answer is -1 according to calculation

4 0

3 years ago

What are the next letters in the following pattern??

Debora [2.8K]

answer is OP

count the difference in letters between the first letters

a -c-f-j c is 2 away from a, f is 3 away from c, etc so the next letter in that pattern would be O

the second letter is going backwards tsrq, so P comes before q

3 0

4 years ago

Q is a degree 5 polynomial that passes through the origin, has zeros i and 7 - i, and q(-1)= -520. find the equation for q

Lunna [17]

Using the factor theorem, the equation for q is:

$q(x) = 4(x^5 - 14x^4 + 51x^3 - 14x^2 + 50x)$

The Factor Theorem states that a polynomial function with roots $x_1, x_2, ..., x_n$ is given by:

$f(x) = a(x - x_1)(x - x_2)...(x - x_n)$

In which a is the leading coefficient.

In this problem:

Passes through the origin, so x = 0 is a zero, that is $x_1 = 0$ .
x = i is a zero, and thus, it's conjugate also is, that is, x = -i, hence $x_2 = i, x_3 = -i$

x = 7 - i is a zero, and thus, it's conjugate also is, that is, x = 7 + i, hence $x_4 = 7 - i, x_3 = 7 + i$

Then, the equation is:

$q(x) = a(x - 0)(x - i)(x + i)(x - 7 + i)(x - 7 - i)$

$q(x) = ax(x^2 - i^2)((x-7)^2 - i^2)$

Considering that $i^2 = -1$

$q(x) = ax(x^2 + 1)(x^2 - 14x + 50)$

$q(x) = ax(x^4 - 14x^3 + 51x^2 - 14x + 50)$

$q(x) = a(x^5 - 14x^4 + 51x^3 - 14x^2 + 50x)$

q(-1) = -520 means that when $x = -1, q = -520$ , and this is used to find a.

$q(x) = a(x^5 - 14x^4 + 51x^3 - 14x^2 + 50x)$

$-520 = a(-1 - 14 - 51 - 14 - 50)$

$130a = 520$

$a = \frac{520}{130}$

$a = 4$

Hence, the equation is:

$q(x) = 4(x^5 - 14x^4 + 51x^3 - 14x^2 + 50x)$

A similar problem is given at brainly.com/question/24380382

4 0

2 years ago