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3 years ago

given the equation y = 2x - 8, find the ordered-pair solution when x = 4 and the ordered-pair solution when y = 4

Mathematics

1 answer:

babunello [35]3 years ago

6 0

Answer:

When x = 4, (4, 0)

When y = 4, (6, 4)

Step-by-step explanation:

To find the ordered-pair solution when x = 4, plug 4 into the x of the equation.

y = 2x - 8

y = 2(4) - 8

y = 8 - 8

y = 0

This produces the ordered pair (4, 0).

To find the ordered-pair solution when y = 4, plug 4 into the y of the equation.

y = 2x - 8

4 = 2x - 8

12 = 2x

6 = x

x = 6

This produces the ordered pair (6, 4).

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The correct answer is option A which is the distance between A and D will be AD = $\sqrt{(0-0)^2-(b-0)^2}=\sqrt{b^2}=b$ .

<h3>What is coordinate geometry?</h3>

A coordinate plane is a 2D plane which is formed by the intersection of two perpendicular lines known as the x-axis and y-axis.

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AD = $\sqrt{(0-0)^2-(b-0)^2}=\sqrt{b^2}=b$ .

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2 years ago

What is the coefficient of x2y3 in the expansion of (2x + y)5?

Zigmanuir [339]

Option C:

The coefficient of $x^{2} y^{3}$ is 40.

Solution:

Given expression:

$(2 x+y)^{5}$

Using binomial theorem:

$(a+b)^{n}=\sum_{i=0}^{n}\left(\begin{array}{l}n \\i\end{array}\right) a^{(n-i)} b^{i}$

Here $a=2 x, b=y$

Substitute in the binomial formula, we get

$(2x+y)^5=\sum_{i=0}^{5}\left(\begin{array}{l}5 \\i\end{array}\right)(2 x)^{(5-i)} y^{i}$

Now to expand the summation, substitute i = 0, 1, 2, 3, 4 and 5.

$$=\frac{5 !}{0 !(5-0) !}(2 x)^{5} y^{0}+\frac{5 !}{1 !(5-1) !}(2 x)^{4} y^{1}+\frac{5 !}{2 !(5-2) !}(2 x)^{3} y^{2}+\frac{5 !}{3 !(5-3) !}(2 x)^{2} y^{3}$

$$+\frac{5 !}{4 !(5-4) !}(2 x)^{1} y^{4}+\frac{5 !}{5 !(5-5) !}(2 x)^{0} y^{5}$

Let us solve the term one by one.

$$\frac{5 !}{0 !(5-0) !}(2 x)^{5} y^{0}=32 x^{5}$

$$\frac{5 !}{1 !(5-1) !}(2 x)^{4} y^{1} = 80 x^{4} y$

$$\frac{5 !}{2 !(5-2) !}(2 x)^{3} y^{2}= 80 x^{3} y^{2}$

$$\frac{5 !}{3 !(5-3) !}(2 x)^{2} y^{3}= 40 x^{2} y^{3}$

$$\frac{5 !}{4 !(5-4) !}(2 x)^{1} y^{4}= 10 x y^{4}$

$$\frac{5 !}{5 !(5-5) !}(2 x)^{0} y^{5}=y^{5}$

Substitute these into the above expansion.

$(2x+y)^5=32 x^{5}+80 x^{4} y+80 x^{3} y^{2}+40 x^{2} y^{3}+10 x y^{4}+y^{5}$

The coefficient of $x^{2} y^{3}$ is 40.

Option C is the correct answer.

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