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3 years ago

Help me 3×(x-y)-7(y-x)²

Mathematics

1 answer:

forsale [732]3 years ago

6 0

Answer:

${ \tt{3(x - y) - 7 {(y - x)}^{2} }} \\ \\\dashrightarrow \: { \tt{3(x - y) - 7(y - x)(y - x)}} \\ \\ \dashrightarrow \: { \tt{3x - 3y - 7 {y}^{2} + 14xy - 7 {x}^{2} }} \\ \\ \dashrightarrow \: { \tt{ - 7( {x}^{2} + {y}^{2}) + 14xy + 3x - 3y }} \\ \\ \dashrightarrow \: { \tt{ - 7 \{ {(x + y)}^{2} - 2xy \} + 14xy + 3x - 3y}} \\ \\ \dashrightarrow \: { \tt{7 {(x + y)}^{2} + 12xy + 3x - 3y }}$

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How many planes contain each line and point?

nikitadnepr [17]

There are infinite many planes that contain each line and point

<h3>How to determine the number of planes?</h3>

The given parameters are given as:

Line KL and G
Line JI and G

As a general rule, a line and a point can be used to draw as many planes as possible

This means that there are infinite many planes possible

Hence, there are infinite many planes that contain each line and point

Read more about planes and points at:

brainly.com/question/14366932

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2 years ago

Mrs. Teabottle is quite worried about the annual Gardenia Banquet. Ms. Aggravant, Mr. Bellicose, Mrs. Colic and Dr. Dreck all de

nasty-shy [4]

Answer:

Step-by-step explanation:

To abbreviate, I will refer to each person by their initials (A,B,C,D).

We will count all the possibilities depending on C and completing each case.

C can sit on the blue, red or green tables. Suppose that C sits on the blue table. Now, A can sit on the yellow or orange tables, then there are 2 possibilities to sit A. After choosing, there remains only one choice for B, either the yellow or orange table, the one which A didn't sit.

Now, the blue, yellow and orange tables are occupied, so D can sit on the red or green tables. Hence, we have 2 choices for D. In total, we have 2×2=4 ways to sit everyone if we choose blue for C.

Suppose that C sits on the green or red tables (2 choices). Now, we split this in two cases.

If we use the yellow and blue tables for A and B, there are 2 ways of seating them (A to yellow, B to blue, or A to blue, B to yellow). For D, we have 2 choices, the red/green table (depending on C) and the orange table. Thus, in this case, we have 2×2×2=8 ways of seat the guests.

If we don't use precisely the yellow and blue tables for A and B, one of them must sit at the orange table. There are 2 ways of doing this (A in orange or B in orange). The other must sit on the blue or yellow tables (2 choices). Finally, D has only 1 choice, the table not used by C. Then we have 2×2×2=8 ways of doing this.

Therefore, considering all the cases we have 4+8+8=20 ways of seating the guests.

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3 years ago

Y'ALL! WHAT IS:<br><br>46(74 + 46) + 57(9 + 36) <br>

koban [17]

Answer:

331

Step-by-step explanation:

Add 74 + 46 + 57 = 177

Add 9 + 36 = 45

Add 45 + 46 = 91

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3 years ago

A there are 3 red pen pens, 4 blue pens, and 5 green pens in a drawer. suppose you choose a pen at random

Anastaziya [24]

Step-by-step explanation:

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2 years ago

exam p A drawer contains four pairs of socks, with each pair a different color. One sock at a time is randomly drawn from the dr

Marta_Voda [28]

Answer:

The probability that the maximum number of draws is required is 0.2286

Step-by-step explanation:

The probability that the maximum number of draws happens when you pick <em>different colors in the first four pick</em>.

Assume you picked one sock in the first draw. Its probability is 1, since you can draw any sock.

In the second draw, 7 socks left and you can draw all but the one which is the pair of the first draw. Then the probability is $\frac{6}{7}$

In the third draw, 6 socks left and you can draw one of the two pair colors which are not drawn yet. Its probability is $\frac{4}{6}$

In the forth draw, 5 socks left and only one pair color, which is not drawn. The probability of drawing one of this pair is $\frac{2}{5}$

In the fifth draw, whatever you draw, you would have one matching pair.

The probability combined is 1× $\frac{6}{7}$ × $\frac{4}{6}$ × $\frac{2}{5}$ ≈ 0.2286

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4 years ago