Ask question

Ask question

All categories

2 years ago

5

True or False: To factor a polynomial using the grouping method, factor out the common terms from the first two terms and then t

he last two terms in the polynomial.

1 answer:

yan [13]2 years ago

6 0

Consider the following x^4+5x^3+x^2+5x = x^3(x+5) and x(x+5) making your 2 factors (x+5) and (x^3 +x)

You might be interested in

I need help with 89 divided by 6 in remainder

mars1129 [50]

18.433333333, please mark brainlest! :)

7 0

3 years ago

Read 2 more answers

If a soccer team played 160 games and won 65% of them, how many games did they win?

Sindrei [870]

The correct answer is 104 games.

You get that number by dividing 160 with a 100, to find out how large 1% is, and then you multiply it with 65 to find out how much 65% is, and the answer is 104 games.

4 0

3 years ago

Estimate the difference. 9 1/2- 1 6/7

Sav [38]

Answer:

$= 7 \frac{9}{14} \\$

Step-by-step explanation:

$9 \frac{1}{2} - 1 \frac{6}{7} \\ \frac{19}{2} - \frac{13}{7} \\ \frac{133 - 26}{14} \\ = \frac{107}{14} \\ = 7 \frac{9}{14}$

hope this helps

brainliest appreciated

good luck! have a nice day!

6 0

3 years ago

In the first year of a 10,000 investment the interest rate was 6%. All earned interest remained invested for the second year, an

Aleks04 [339]

Answer:

13.4%

Step-by-step explanation:

First year:

$10,000*6% = $600

New balance = $10,600

Second Year:

$10,600*7% = $742

$10,600+ $742 = $11,342

Total Return:

Final Balance - Initial balance

$11,342 - $10,000 = $1,342

$10,000*x ÷ $1,342

x = $1,342/$10,000

x = 0.1342

0.134 = 13.4%

7 0

3 years ago

Does anyone know the answer to this!?

PolarNik [594]

Answer:

$|A_y|=\left|\begin{array}{cc}12&7\\ \\17&-51\end{array}\right|$

Step-by-step explanation:

You can use these Cramer's formulas to solve for x and y:

$x=\dfrac{|A_x|}{|A|},\\ \\y=\dfrac{|A_y|}{|A|},$

where

$|A|=\left|\begin{array}{cc}12&-13\\ \\17&-22\end{array}\right|\\ \\ \\|A_x|=\left|\begin{array}{cc}7&-13\\ \\-51&-22\end{array}\right|\\ \\ \\|A_y|=\left|\begin{array}{cc}12&7\\ \\17&-51\end{array}\right|$

So,

$|A_y|=\left|\begin{array}{cc}12&7\\ \\17&-51\end{array}\right|$

4 0

3 years ago