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avanturin [10]
3 years ago
6

State whether the triangles are similar. If so, write a similarity statement and the postulate or theorem you used.

Mathematics
1 answer:
never [62]3 years ago
8 0

Answer:

Triangles ABC and MNO are similar by SSS

Step-by-step explanation:

we know that

If two triangles are similar, then the ratio of its corresponding sides is proportional

In this problem

If this problem the corresponding sides are

AB and MN

BC and NO

AC and MO

Verify if the corresponding sides are proportional

\frac{AB}{MN}=\frac{BC}{NO}=\frac{AC}{MO}

substitute the given values

\frac{5}{7.5}=\frac{5}{7.5}=\frac{8}{12}

0.67=0.67=0.67 ----- is true

so

When two triangles have corresponding sides with identical ratios , the triangles are similar by SSS Similarity

therefore

Triangles ABC and MNO are similar by SSS

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4 years ago
1. Use separation of variables to find the solution to the differential equation subject to the given initial condition.
andrew11 [14]

Answer:

Step-by-step explanation:

Given the differential equation dy/dx = 5y/x subject to the condition y = 4 and x = 1. Using the variable separable method of solving differential equation, we will have;

dy/dx = 5y/x

Separate the variables

dy/5y = dx/x

Integrate both sides of the expression

\frac{1}{5}\int\limits \frac{1}{y}  \, dy = \int\limits \frac{dx}{x} \\ \\\frac{1}{5}lny = lnx + C\\\\lny = 5lnx+5C\\

using the initial condition y = 4 while x = 1

ln4 = 5ln1 + 5C

ln4 = 0+5C

C = ln4/5

Substituting the value of C back into the expression;

lny = 5 lnx+5(ln4/5)\\lny = 5lnx+ln4\\lny = lnx^5+ln4\\lny = ln(4x^5)\\y = 4x^5

<em>Hence the solution to the differential equation is y = 4x⁵</em>

<em></em>

b) Given 4(du/dt) = u²

du/dt = u²/4

du/ u² = dt/4

u⁻²du = 1/4 dt

integrate both sides of the equation

\int\limit {u^{-2}} \, du  = \int\limits\frac{1}{4}  \, dt\\\\\frac{u^{-1}}{-1} = \frac{t}{4} + C\\\\\frac{-1}{u} =  \frac{t}{4} + C

Imputing the initial condition u(0) = 7 i.e when t = 0, u = 7

\frac{-1}{7} =  \frac{0}{4} + C\\\\\frac{-1}{7} =  C\\

\frac{-1}{u} =  \frac{t}{4} - \frac{1}{7}

<em>Hence the solution to the DE is </em>\frac{-1}{u} =  \frac{t}{4} - \frac{1}{7}

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3 years ago
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