All categories

3 years ago

....... Math practice

Mathematics

2 answers:

viva [34]3 years ago

8 0

First are 1.2 Last is 4

vlada-n [284]3 years ago

6 0

First two blanks are 1.2
Last blank is 4

You might be interested in

How do you simplify 420/450?

MatroZZZ [7]

Find the GCD (or HCF) of numerator and denominator
GCD of 420 and 450 is 30Divide both the numerator and denominator by the GCD
(420/30)/(450/30)Reduced fraction: 14/15

Equivalent fractions

7 0

3 years ago

Read 2 more answers

What is the answer to 3x - 17 = 7x - 9

Gnoma [55]

Take the numbers to one side

8 0

3 years ago

-x + 2y = 6 * finding the slope and y intercept *

Aleonysh [2.5K]

Answer:

Step-by-step explanation:

3 0

3 years ago

Determine formula of the nth term 2, 6, 12 20 30,42

nalin [4]

Check the forward differences of the sequence.

If $\{a_n\} = \{2,6,12,20,30,42,\ldots\}$ , then let $\{b_n\}$ be the sequence of first-order differences of $\{a_n\}$ . That is, for n ≥ 1,

$b_n = a_{n+1} - a_n$

so that $\{b_n\} = \{4, 6, 8, 10, 12, \ldots\}$ .

Let $\{c_n\}$ be the sequence of differences of $\{b_n\}$ ,

$c_n = b_{n+1} - b_n$

and we see that this is a constant sequence, $\{c_n\} = \{2, 2, 2, 2, \ldots\}$ . In other words, $\{b_n\}$ is an arithmetic sequence with common difference between terms of 2. That is,

$2 = b_{n+1} - b_n \implies b_{n+1} = b_n + 2$

and we can solve for $b_n$ in terms of $b_1=4$ :

$b_{n+1} = b_n + 2$

$b_{n+1} = (b_{n-1}+2) + 2 = b_{n-1} + 2\times2$

$b_{n+1} = (b_{n-2}+2) + 2\times2 = b_{n-2} + 3\times2$

and so on down to

$b_{n+1} = b_1 + 2n \implies b_{n+1} = 2n + 4 \implies b_n = 2(n-1)+4 = 2(n + 1)$

We solve for $a_n$ in the same way.

$2(n+1) = a_{n+1} - a_n \implies a_{n+1} = a_n + 2(n + 1)$

Then

$a_{n+1} = (a_{n-1} + 2n) + 2(n+1) \\ ~~~~~~~= a_{n-1} + 2 ((n+1) + n)$

$a_{n+1} = (a_{n-2} + 2(n-1)) + 2((n+1)+n) \\ ~~~~~~~ = a_{n-2} + 2 ((n+1) + n + (n-1))$

$a_{n+1} = (a_{n-3} + 2(n-2)) + 2((n+1)+n+(n-1)) \\ ~~~~~~~= a_{n-3} + 2 ((n+1) + n + (n-1) + (n-2))$

and so on down to

$a_{n+1} = a_1 + 2 \displaystyle \sum_{k=2}^{n+1} k = 2 + 2 \times \frac{n(n+3)}2$

$\implies a_{n+1} = n^2 + 3n + 2 \implies \boxed{a_n = n^2 + n}$

6 0

2 years ago

How do you know a radical expression is in simplest form?

Diano4ka-milaya [45]

Answer: To know whether a radical expression is in simplest form or not you should put the numbers and letters inside the radical in terms of prime factors. Then, the radical expression is in the simplest form if all the numbers and letters inside the radical are prime factors with a power less than the index of the radical

Explanation:

Any prime factor raised to a power greater than the index of the root can be simplified and any factor raised to a power less than the index of the root cannot be simplified

For example simplify the following radical in its simplest form:

$\sqrt[5]{3645 a^8b^7c^3}$

1) Factor 3645 in its prime factors: 3645 = 3^6 * 5

2) Since the powr of 3 is 6, and 6 can be divided by the index of the root, 5, you can simplify in this way:

- 6 ÷ 5 = 1 with reminder 1, so 3^1 leaves the radical and 3^1 stays in the radical

3) since the factor 5 has power 1 it can not leave the radical

4) the power of a is 8, then:

8 ÷ 5 = 1 with reminder 3 => a^1 leaves the radical and a^3 stays inside the radical.

5) the power of b is 7, then:

7 ÷ 5 = 1 with reminder 2 => b^1 leaves the radical and b^2 stays inside the radical

6) the power of c is 3. Since 3 is less than 5 (the index of the radical) c^3 stays inside the radical.

7) the expression simplified to its simplest form is

$3ab \sqrt[5]{3.5.a^3b^2c^3}$

And you know it cannot be further simplified because all the numbers and letters inside the radical are prime factors with a power less than the index of the radical.

7 0

3 years ago

Read 2 more answers