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3 years ago

How do I find the slope from a table if the value added onto each (x,y) value is different? Please help!!! I need the slope!

Mathematics

1 answer:

Bess [88]3 years ago

5 0

ANSWER:
________

Line q = 3
Line v = -1/2

FORMULA
__________

y2 - y1 / x2 - x1

EXPLANATION:
_____________

First, we need to find the slope for line a only

We need to select two points from line q. The ones I selected are (-3,18) and (2,33)

y2 = 33
y1 = 18

x2 = 2
x1 = -3

33 - 18 = 15
2 - -3 = 5

15/5 = 3

Next, we need to find the slope for line v.

The two points I picked are (0,8) and (10,3)

y2 = 8
y1 = 3

x2 = 0
x1 = 10

8 - 3 = 5
0 - 10 = -10

5/-10 = -1/2

And that’s how you get your answer :)

PUN OF THE DAY
_______________

Why did Adele cross the road? To say hello from the other side.

I hoped this helped! And have a •AMAZING• day :3

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3 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}$