Write the equation for a line that has an initial value of 3 and 3/4 as it’s rate of change

Mathematics

1 answer:

frutty [35]2 years ago

3 0

the slope goes by several names

• average rate of change

• rate of change

• deltaY over deltaX

• Δy over Δx

• rise over run

• gradient

• constant of proportionality

however, is the same cat wearing different costumes.

initial value of 3, namely when x = 0, y = 3, so we have the point (0 , 3) and it has a rate or slope of 3/4.

$(\stackrel{x_1}{0}~,~\stackrel{y_1}{3})\qquad \qquad \stackrel{slope}{m}\implies \cfrac{3}{4} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{3}=\stackrel{m}{\cfrac{3}{4}}(x-\stackrel{x_1}{0})\implies y=\cfrac{3}{4}x+3$

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Find the area of a triangle formed by placing the vectors [3, 6] and [7, 1] tail-to-tail.

Usimov [2.4K]

Answer:

a. $A=10u^{2}$

b. View graph

c. 6.40u

Step-by-step explanation:

knowing that the triangle area is equal to base by heigh between two, then:

$A=\frac{bh}{2}=\frac{5*4}{2}=10u^{2}$

The length of the longest altitude of your triangle is:

$c^{2}=a^{2}+b^{2}=5^{2}+4^{2}=25+16=41\\ c=\sqrt{41}=6.40u$

finally it can be seen that the position of the triangle does not matter, as long as the base and heigh are maintained, the area of the triangle will be the same