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2 years ago

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On a coordinate grid, both point (−4, −1) and (2, 6) point are reflected across the y-axis. What are the coordinates of the refl

ected points?
A. (4, 1) and (−2, −6)
B. (−4, 1) and (−2, 6)
C. (−4, 1) and (2, −6)
D. (4, −1) and (−2, 6)

2 answers:

Alex777 [14]2 years ago

8 0

Answer:

Option D. (4, −1) and (−2, 6)

Step-by-step explanation:

we know that

The rule of the reflection of a point across the y-axis is equal to

(x,y) -----> (-x,y)

so

Applying the rule of the reflection

(−4, −1) -----> (4, −1)

(2, 6)----- (-2, 6)

Delicious77 [7]2 years ago

8 0

Answer:

Option D. (4, −1) and (−2, 6)

Step-by-step explanation:

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Read 2 more answers

For the binomial expansion of (x + y)^10, the value of k in the term 210x 6y k is a) 6 b) 4 c) 5 d) 7

Sloan [31]

Answer:

a) 6

Step-by-step explanation:

Expanding the polynomial using the formula:

$(x+y)^n=\sum_{k=0}^n \binom{n}{k} x^{n-k} y^k$

Also

$\binom{n}{k}=\frac{n!}{(n-k)!k!}$

I think you mean $210x^6y^4$

We can deduce that this term will be located somewhere in the middle. So I will calculate $k= 5; k=6 \text{ and } k =7$ .

For $k=5$

$\binom{10}{5} (y)^{10-5} (x)^{5}=\frac{10!}{(10-5)! 5!}(y)^{5} (x)^{5}= \frac{10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5! }{5! \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 } \\ =\frac{30240}{120} =252 x^{5} y^{5}$

Note that we actually don't need to do all this process. There's no necessity to calculate the binomial, just $x^{n-k} y^k$

For $k=6$

$\binom{10}{6} \left(y\right)^{10-6} \left(x\right)^{6}=\frac{10!}{(10-6)! 6!}\left(y\right)^{4} \left(x\right)^{6}=210 x^{6} y^{4}$

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