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solmaris [256]
2 years ago
11

What is the slope of the line passing through the points (0, 4) and (6, 13)

Mathematics
1 answer:
skelet666 [1.2K]2 years ago
3 0

Answer:

The slope is \frac{3}{2}

Step-by-step explanation:

Hi there!

We are given the points (0, 4) and (6, 13)

We want to find the slope of the line containing these two points

Slope (m) is the steepness of the line. When calculating from two points, you use the equation \frac{y_2-y_1}{x_2-x_1}, where (x_1, y_1) and (x_2, y_2) are points

Although we have what we need to find the slope, let's label the values of the points to avoid confusion and mistakes.

x_1=0\\y_1=4\\x_2=6\\y_2=13

Substitute these values into the equation

m=\frac{y_2-y_1}{x_2-x_1}

m=\frac{13-4}{6-0}

Subtract

m=\frac{9}{6}

Simplify

m=\frac{3}{2}

The slope of the line is 3/2

Hope this helps!

See more on calculating slopes from 2 points here: brainly.com/question/26738734

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(x +y)^5<br> Complete the polynomial operation
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Answer:

Please check the explanation!

Step-by-step explanation:

Given the polynomial

\left(x+y\right)^5

\mathrm{Apply\:binomial\:theorem}:\quad \left(a+b\right)^n=\sum _{i=0}^n\binom{n}{i}a^{\left(n-i\right)}b^i

a=x,\:\:b=y

=\sum _{i=0}^5\binom{5}{i}x^{\left(5-i\right)}y^i

so expanding summation

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

solving

\frac{5!}{0!\left(5-0\right)!}x^5y^0

=1\cdot \frac{5!}{0!\left(5-0\right)!}x^5

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also solving

=\frac{5!}{1!\left(5-1\right)!}x^4y

=\frac{5}{1!}x^4y

=\frac{5}{1!}x^4y

=\frac{5x^4y}{1}

=\frac{5x^4y}{1}

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similarly, the result of the remaining terms can be solved such as

\frac{5!}{2!\left(5-2\right)!}x^3y^2=10x^3y^2

\frac{5!}{3!\left(5-3\right)!}x^2y^3=10x^2y^3

\frac{5!}{4!\left(5-4\right)!}x^1y^4=5xy^4

\frac{5!}{5!\left(5-5\right)!}x^0y^5=y^5

so substituting all the solved results in the expression

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

=x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+y^5

Therefore,

\left(x\:+y\right)^5=x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+y^5

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