All categories

2 years ago

I need help what’s the answer

Mathematics

1 answer:

alexira [117]2 years ago

7 0

Answer:

The square of a binomial pattern

Step-by-step explanation:

Given

9x² - 30x + 25

Consider the factors of the product of the coefficient of the x² term and the constant term which sun to give the coefficient of the x- term.

product = 9 × 25 = 225 and sum = - 30

The factors are - 15 and - 15

Use these factors to split the x- term

9x² - 15x - 15x + 25 ( factor the first/second and third/fourth terms )

= 3x(3x - 5) - 5(3x - 5) ← factor out (3x - 5) from each term

= (3x - 5)(3x - 5)

= (3x - 5)² ← the square of a binomial pattern

You might be interested in

A bank says you can double your money in 10 years if you put $1,000 in a simple interest account. What annual interest rate does

katrin2010 [14]

The simple interest formula allows us to calculate I, which is the interest earned or charged on a loan. According to this formula, the amount of interest is given by I = Prt, where P is the principal, r is the annual interest rate in decimal form, and t is the loan period expressed in years. The rate r must be converted from a percentage into decimal form.

Then, 2,000 = 1,000 * r * 10 ;

Finally, r = 2 ÷ 10 = 20 ÷ 100 = 0.2

hope this helps you

8 0

2 years ago

What is 9,999,999 times 24

Otrada [13]

The Answer is -
<span>
239999976</span>

3 0

3 years ago

If you had 1052 toothpicks and were asked to group them in powers of 6, how many groups of each power of 6 would you have? Put t

sukhopar [10]

1052 toothpicks can be grouped into 4 groups of third power of 6 ( $6^{3}$ ), 5 groups of second power of 6 ( $6^{2}$ ), 1 group of first power of 6 ( $6^{1}$ ) and 2 groups of zeroth power of 6 ( $6^{0}$ ).

The number 1052, written as a base 6 number is 4512

Given: 1052 toothpicks

To do: The objective is to group the toothpicks in powers of 6 and to write the number 1052 as a base 6 number

First we note that, $6^{0}=1,6^{1}=6,6^{2}=36,6^{3}=216,6^{4}=1296$

This implies that $6^{4}$ exceeds 1052 and thus the highest power of 6 that the toothpicks can be grouped into is 3.

Now, $6^{3}=216$ and $216\times 5=1080, 216\times 4=864$ . This implies that $216\times 5$ exceeds 1052 and thus there can be at most 4 groups of $6^{3}$ .

Then,

$1052-4\times6^{3}$

$1052-4\times216$

$1052-864$

$188$

So, after grouping the toothpicks into 4 groups of third power of 6, there are 188 toothpicks remaining.

Now, $6^{2}=36$ and $36\times 5=180, 36\times 6=216$ . This implies that $36\times 6$ exceeds 188 and thus there can be at most 5 groups of $6^{2}$ .

Then,

$188-5\times6^{2}$

$188-5\times36$

$188-180$

$8$

So, after grouping the remaining toothpicks into 5 groups of second power of 6, there are 8 toothpicks remaining.

Now, $6^{1}=6$ and $6\times 1=6, 6\times 2=12$ . This implies that $6\times 2$ exceeds 8 and thus there can be at most 1 group of $6^{1}$ .

Then,

$8-1\times6^{1}$

$8-1\times6$

$8-6$

$2$

So, after grouping the remaining toothpicks into 1 group of first power of 6, there are 2 toothpicks remaining.

Now, $6^{0}=1$ and $1\times 2=2$ . This implies that the remaining toothpicks can be exactly grouped into 2 groups of zeroth power of 6.

This concludes the grouping.

Thus, it was obtained that 1052 toothpicks can be grouped into 4 groups of third power of 6 ( $6^{3}$ ), 5 groups of second power of 6 ( $6^{2}$ ), 1 group of first power of 6 ( $6^{1}$ ) and 2 groups of zeroth power of 6 ( $6^{0}$ ).

Then,

$1052=4\times6^{3}+5\times6^{2}+1\times6^{1}+2\times6^{0}$