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olga2289 [7]
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.
Mathematics
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)
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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
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