For each value of y, determine whether it is a solution to -2y+75-5. Is it a solution?

Mathematics

1 answer:

ser-zykov [4K]2 years ago

3 0

Answer:

• No

• Yes

• No

Step-by-step explanation:

To determine if the 4 given values of y are solutions to the inequality, start by solving the inequality. Solving an inequality is just like that of an equation, except that the direction of the sign changes when the inequality is divided by a negative number.

-2y +7≤ -5

Subtract 7 on both sides:

-2y≤ -5 -7

-2y≤ -12

Divide by -2 on both sides:

y≥ 6

This means that the solution can be 6 or greater than 6.

-10 and 3 are smaller than 6 and are not a solutions, while 7 and 6 satisfies the inequality and are therefore solutions.

_______

Alternatively, we can also substitute each value of y into the inequality and check if the value is less than or equal to -5.

Here's an example to check if -10 is a solution.

-2y +7≤ -5

When y= -10,

-2y +7

= -2(-10) +7

= 20 +7

= 27

Since 27 is greater than 5, it is <u>not</u> a solution to the inequality.

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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).

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