Hey! i’ll give brainliest please help

zavuch27 [327]

An example of a negative mixed fraction: -5 1/2. Because slash is both signs for ... fractional exponents: 16 ^ 1/2 • adding fractions and mixed numbers: 8/5 + 6 2/7

8 0

2 years ago

Read 2 more answers

Can someone help me with this

Jobisdone [24]

9+7=16 and 5+8=13
Hope i helped, please leave a thanks! Thanks!

5 0

3 years ago

b) How many ways can you deal cards (from a deck of 52) to 4 people when each player gets 7 cards. 2 hidden and 5 visible. Assum

Mariana [72]

Answer:

${52 \choose 7}{45 \choose 7}{38 \choose 7}{31 \choose 7}{7 \choose 2}^4$

Step-by-step explanation:

The dealing of the cards can be seen in the following way:

We first have to choose which 7 cards are going to be dealt to the 1st player. So we have to pick 7 cards out of the 52 available cards. This can be done in ${52 \choose 7}$ ways. Now that we have chosen which 7 cards are going to be dealt to the 1st player, we have to choose which 2 of them are going to be the hidden ones. So we have to pick 2 cards out of the 7 cards to be the hidden ones. This can be done in ${7 \choose 2}$ ways. At this point we now know which cards are being dealt to the 1st player, and which ones are hidden for him. Then we have to choose which 7 cards to deal to the 2nd player, out of the remaining ones. So we have to pick 7 out of 52-7=45. (since 7 have been already dealt to the 1st player). This can be done in ${45 \choose 7}$ . Then again, we have to pick which 2 are going to be the hidden cards for this 2nd player. So we have to pick 2 out of the 7. This can be done in ${7 \choose 2}$ ways. Then we continue with the 3rd player. We have to choose 7 cards out of the remaining ones, which at this point are 52-7-7=38. This can be done in ${38 \choose 7}$ . And again, we have to choose which ones are the hidden ones, which can be chosen in ${7 \choose 2}$ ways. Finally, for the last player, we choose 7 out of the remaining cards, which are 52-7-7-7=31. This can be done in ${31 \choose 7}$ ways. And choosing which ones are the hidden ones for this player can be done in ${7 \choose 2}$ ways. At the end, we should multiply all our available choices on each step, to get the total choices or total ways to deal the cards to our 4 players (since dealing the cards is a process of several steps with many choices on each step).

${52 \choose 7}{7 \choose 2}{45 \choose 7}{7 \choose 2}{38 \choose 7}{7 \choose 2}{31 \choose 7}{7 \choose 2}={52 \choose 7}{45 \choose 7}{38 \choose 7}{31 \choose 7}{7 \choose 2}^4$

7 0

3 years ago

Pls helppp <br> 1/2•7•24=

Kazeer [188]

Answer:

84

Step-by-step explanation:

0.5 · 7 · 24

0.5 · 168

84

3 0

3 years ago

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Question 14 of 25What is the complete factorization of the polynomial below?x³+x² + 4x+4O A. (x-1)(x+2)(x+2)B. (x + 1)(x + 2)(x

mote1985 [20]

GIVEN:

We are given the following polynomial;

$x^3+x^2+4x+4$

Required;

We are required to factorize this polynomial completely.

Step-by-step explanation;

To factorize this polynomial, we start by grouping;

$(x^3+x^2)+(4x+4)$

We now take the common factor in each group;

$\begin{gathered} x^2(x+1)+4(x+1) \\ \\ (x^2+4)(x+1) \end{gathered}$

Next, we factorize the first parenthesis. To do this we set the equation equal to zero and solve for x as follows;

$\begin{gathered} x^2+4=0 \\ \\ Subtract\text{ }4\text{ }from\text{ }both\text{ }sides: \\ \\ x^2=-4 \\ \\ Take\text{ }the\text{ }square\text{ }root\text{ }of\text{ }both\text{ }sides: \\ \\ x=\pm\sqrt{-4} \\ \\ x=(\pm\sqrt{-1}\times\sqrt{4}) \\ \\ x=\pm2i \end{gathered}$

Therefore, the factors of the other parenthesis are;

$(x+2i)(x-2i)$

Therefore, the complete factorization of the polynomial is;

ANSWER:

$(x+1)(x+2i)(x-2i)$

Option C is the correct answer.

4 0

1 year ago