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3 years ago

5

the perimeter of a scalene triangle is 14.5 cm. The longest side is twice that of the short side. Which equation can be used to

find the side lengths if the longest side measures 6.2 cm?

1 answer:

Lynna [10]3 years ago

7 0

Keenly snakelike smells an kelsmeml salesmen smell. Skkensens.

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Find all the missing measurement

enot [183]

<h3>Answer: 21</h3>

Explanation:

Since line L is parallel to line RA, this means the pieces formed by line L cutting into the triangle are proportional.

That means we can set up the proportion below to solve for d

14/d = 24/36

14*36 = d*24

504 = 24d

24d = 504

d = 504/24

d = 21

Other equations are possible to set up, but you should get the same result regardless.

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3 years ago

The package of cheese shown below cost

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It would cost about $4.86!

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3 years ago

If the differential equation t 2y 00 − 2y 0 + (3 + t)y = 0 has y1(t) and y2(t) as a fundamental set of solutions and if W[y1, y2

podryga [215]

In the ODE, solve for $y''$ :

$t^2y''-2y'+(3+t)y=0\implies y''=\dfrac{2y'+(3+t)y}{t^2}$

The Wronskian is then

$W=\begin{vmatrix}y_1&y_2\\{y_1}'&{y_2}'\end{vmatrix}=y_1{y_2}'-{y_1}'y_2$

Differentiating the Wronskian gives

$W'=({y_1}'{y_2}'+y_1{y_2}'')-({y_1}''y_2+{y_1}'{y_2}')=y_1{y_2}''-{y_1}''y_2$

Substitute $y_1,y_2$ into the equation for $y''$ , then substitute ${y_1}'',{y_2}''$ into $W'$ :

$W'=y_1\dfrac{2{y_2}'+(3+t)y_2}{t^2}-y_2\dfrac{2{y_1}'+(3+t)y_1}{t^2}$

$\implies W'=\dfrac{2W}{t^2}$

which is another separable ODE; we have

$\dfrac{\mathrm dW}W=\dfrac2{t^2}\,\mathrm dt\implies \ln|W|=-\dfrac2t+C\implies W=Ce^{-2/t}$

Given that $W(y_1,y_2)(2)=3$ , we find

$3=Ce^{-2/2}\implies C=3e$

so that

$W(y_1,y_2)(t)=3e^{1-2/t}$

and so

$W(y_1,y_2)(6)=\boxed{3e^{2/3}}$

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3 years ago

A ball is dropped from the top of a building. The distance, y, in meters above the ground for the ball after t seconds can be mo

Harlamova29_29 [7]

The y-intercept of the equation is 100. It means that the ball is being dipped from 100 meters. An x-intercept would be when the ball would reach the ground. I believe after 98 seconds it would reach the ground but I’m not sure.

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3 years ago

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Marat540 [252]

Answer:

njbrdjbbgdbjbfjfj

Step-by-step explanation:

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3 years ago