Help me please.on number 2

Mathematics

1 answer:

Alex_Xolod [135]4 years ago

4 0

I'm pretty sure it's "c" if 1 page for kayla is equal to 4/5 a page for molly 1/4 for Kayla is equal to 1/5 a page for molly. 1 and 1/4 is the answer.

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Integration by Parts Evaluate e-2x cos(2x) dx.

kifflom [539]

Let

$I = \displaystyle \int e^{-2x} \cos(2x) \, dx[/]texIntegrate by parts:[tex]\displaystyle \int u \, dv = uv - \int v \, du$

with

$u = e^{-2x} \implies du = -2 e^{-2x} \, dx \\\\ dv = \cos(2x) \, dx \implies v = \dfrac12 \sin(2x)$

Then

$\displaystyle I = \frac12 e^{-2x} \sin(2x) + \int e^{-2x} \sin(2x) \, dx + C$

Integrate by parts again, this time with

$u = e^{-2x} \implies du = -2 e^{-2x} \, dx \\\\ dv = \sin(2x) \, dx \implies v = -\dfrac12 \cos(2x)$

so that

$\displaystyle I = \frac12 e^{-2x} \sin(2x) - \frac12 e^{-2x} \cos(2x) - \int e^{-2x} \cos(2x) \, dx + C\\\\ \implies I = \frac{\sin(2x)-\cos(2x)}{2e^{2x}} - I + C \\\\ \implies 2I = \frac{\sin(2x) - \cos(2x)}{2e^{2x}} + C \\\\ \implies I = \boxed{\frac{\sin(2x) - \cos(2x)}{4e^{2x}} + C}$

6 0

2 years ago

Which is the polynomial function of lowest degree with rational real coefficients, a leading coefficient of 3 and roots StartRoo

anastassius [24]

Answer:

$a) f(x)=3x^{3}-6x^{2}-15x+30$

Step-by-step explanation:

1) In this question we've been given "a", the leading coefficient. and two roots:

$x_{1}=\sqrt{5}\:x_{2}=2$

2) There's a theorem, called the Irrational Theorem Root that states:

If one root is in this form $x'=\sqrt{a}+b$ then its conjugate $x''=\sqrt{a}-b$ . is also a root of this polynomial.

Therefore

$x_3=-\sqrt{5}$

3) So, applying this Theorem we can rewrite the equation, by factoring. Remembering that x is the root. Since the question wants it in this expanded form then:

$f(x)=3(x-\sqrt{5})(x+\sqrt{5})(x-2)\Rightarrow 3x^{3}-6x^{2}-15x+30$