Please write an equation in slope intercept form that is parallel to the equation

Mathematics

1 answer:

gtnhenbr [62]3 years ago

4 0

Answer:

y=1/4x-1.5

Step-by-step explanation:

So, because the line is parallel, the slope is the same. Then, to get the y-int, you do b=y-mx with the point, so the equation is b=-1-(1/4)(2), then you multiply to get b=-1-(0.5), so your y-int is -1.5.

(:P)

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Find the root<br><br> X^3-125=0

Firlakuza [10]

Steps to solve:

x^3 - 125 = 0

~Add 125 to both sides

x^3 = 125

~Take cube root of both sides

x = 5

Best of Luck!

8 0

3 years ago

Read 2 more answers

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).

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