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3 years ago

DEF is a right triangle. A. True B. False

Mathematics

2 answers:

Vika [28.1K]3 years ago

8 0

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Sonbull [250]3 years ago

5 0

Given triangle DEF with the lengths of all sides. If you want to check whether this triangle is right, use the Pythagorean theorem. The greatest side is EF, then only this side could be the hypotenuse.

Check, whether

$EF^2=ED^2+DF^2.$

Substitute ED=25, EF=48 and DF=27:

$DF^2+DE^2=27^2+25^2=729+625=1354,\\ \\EF^2=48^2=2304.$

Since $1354\neq 2304,$ the triangle DEF is not right triangle.

Answer: correct choice is B

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2 years ago

In a previous exercise we formulated a model for learning in the form of the differential equation dp dt = k(m − p) where p(t) m

DaniilM [7]

I assume you mean

$\dfrac{dP}{dt} = k(M-P)$

ANSWER
An expression for P(t) is

$P = M - Me^{-kt}$

EXPLANATION
This is a separable differential equation. Treat M and k as constants. Then we can divide both sides by M - P to get the P term with the differential dP and multiply both sides by dt to separate dt from the P terms

$\begin{aligned} \dfrac{dP}{dt} &= k(M-P) \\ \dfrac{dP}{M-P} &= k\, dt
\end{aligned}$

Integrate both sides of the equation.

$\begin{aligned}
\int \dfrac{dP}{M-P} &= \int k\, dt \\
-\ln|M-P| &= kt + C \\
\ln|M-P| &= -kt - C\end{aligned}$

Note that for the left-hand side, u-substitution gives us

$u = M - P \implies du = -1dP \implies dP = -du$

hence why $\int \frac{dP}{M-P} \ne \ln|M - P|$

Now we use the definition of the logarithm to convert into exponential form.

The definition is

$\ln(a) = b \iff \log_e(a) = b \iff e^b = a$

so applying it here, we get

$\begin{aligned} \ln|M-P| &= -kt - C \\ |M - P| &= e^{-kt - C} \\ 
M - P &= \pm e^{-kt - C} 
 \end{aligned}$

Exponent properties can be used to address the constant C. We use $x^{a} \cdot x^{b} = x^{a+b}$ here:

$\begin{aligned}
 M - P &= \pm e^{-kt - C} \\
M - P &= \pm e^{- C - kt} \\ 
M - P &= \pm e^{- C + (- kt)} \\ 
M - P &= \pm e^{- C} \cdot e^{- kt} \\ 
M - P &= Ke^{- kt} && (\text{\footnotesize Let $K = \pm e^{-kt}$ }) \\ 
M &= Ke^{- kt} + P\\
P &= M - Ke^{- kt}
\end{aligned}$

If we assume that P(0) = 0, then set t = 0 and P = 0

$\begin{aligned} 
0 &= M - Ke^{- k\cdot 0} \\
0 &= M - K \cdot 1 \\
M &= K
 \end{aligned}
$

Substituting into our original equation, we get our final answer of

$P = M - Me^{-kt}$

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