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mart [117]
3 years ago
7

Write an equation that

Mathematics
1 answer:
den301095 [7]3 years ago
5 0

Answer:

y = 4x + 6

Step-by-step explanation:

This equation is written in slope-intercept form where 4 is the slope and 6 is the y-intercept.

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Look at this graph:
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3 years ago
What is the similarity ratio of the smaller to the larger similar cylinders?
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\bf \qquad \qquad \textit{ratio relations}&#10;\\\\&#10;\begin{array}{ccccllll}&#10;&Sides&Area&Volume\\&#10;&-----&-----&-----\\&#10;\cfrac{\textit{similar shape}}{\textit{similar shape}}&\cfrac{s}{s}&\cfrac{s^2}{s^2}&\cfrac{s^3}{s^3}&#10;\end{array} \\\\&#10;-----------------------------\\\\&#10;\cfrac{\textit{similar shape}}{\textit{similar shape}}\qquad \cfrac{s}{s}=\cfrac{\sqrt{s^2}}{\sqrt{s^2}}=\cfrac{\sqrt[3]{s^3}}{\sqrt[3]{s^3}}\\\\&#10;-------------------------------\\\\

\bf \cfrac{smaller}{larger}\qquad \cfrac{s}{s}=\cfrac{\sqrt{48\pi }}{\sqrt{75\pi }}\implies \cfrac{s}{s}=\cfrac{\sqrt{(2^2)^2\cdot 3}}{\sqrt{5^2\cdot 3}}\implies \cfrac{s}{s}=\cfrac{4\sqrt{3}}{5\sqrt{3}}&#10;\\\\\\&#10;\cfrac{s}{s}=\cfrac{4}{5}
8 0
3 years ago
Read 2 more answers
(x +y)^5<br> Complete the polynomial operation
Vesna [10]

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

=1\cdot \:1\cdot \:x^5

=x^5

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}

=5x^4y

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

6 0
2 years ago
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