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3 years ago

6

Line segments AD and BE intersect at C, and triangles ABC and DEC are formed. They have the following characteristics: ∠ACB and

∠DCE are vertical angles ∠B ≅ ∠E BC ≅ EC Which congruence theorem can be used to prove △ABC ≅ △DEC? HL ASA SSS SAS

1 answer:

xenn [34]3 years ago

8 0

Answer: The answer is ASA, ed2020

Step-by-step explanation:

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Solve the equation. StartFraction dx Over dt EndFraction equals StartFraction 1 Over x e Superscript t plus 7 x EndFraction An i

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Answer:

$\frac{1}{49}(7x-1)e^{7x}+e^{-t}=C$

Step-by-step explanation:

We are given that

$\frac{dx}{dt}=\frac{1}{xe^{t+7x}}$

We have to find the implicit function

Using separation variable method

$\frac{dx}{dt}=\frac{1}{xe^t\cdot e^{7x}}$

By using property $x^a\cdot x^y=x^{a+y}$

$xe^{7x}dx=e^{-t}dt$

By using property $\frac{1}{x^a}=x^{-a}$

Taking integration on both sides

$\int xe^{7x}dx=\int e^{-t}dt$

Parts integration method

$\int u\cdot v dx=u\int vdx-\int (\frac{du}{dx}\int vdx)dx$

By parts integration method

$x\int e^{7x}dx-\int (\frac{dx}{dx}\int e^{7x}dx)dx=-e^{-t}+C$

Using formula $\int e^{ax} dx=\frac{e^{ax}}{a}+C$

$\frac{xe^{7x}}{7}-\frac{1}{7}\int e^{7x}dx=-e^{-t}+C$

$\frac{xe^{7x}}{7}-\frac{1}{49}e^{7x}+e^{-t}=C$

$\frac{1}{49}(7x-1)e^{7x}+e^{-t}=C$

We are given that

$F(x,t)=C$

$F(x,t)=\frac{1}{49}(7x-1)e^{7x}+e^{-t}=C$

8 0

3 years ago