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Mathematics

1 answer:

Ksju [112]3 years ago

6 0

Answer:

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Step-by-step explanation:

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Find the difference. (4x 2 - 6x + 8) - (3x 2 - 6x + 2) x2 + 6 7x2 - 12x - 6 x2 + 12x + 6 x2 - 12x - 6

Natasha_Volkova [10]

Answer:

x² + 6

Step-by-step explanation:

Given

(4x² - 6x + 8) - (3x² - 6x + 2) ← distribute by - 1

= 4x² - 6x + 8 - 3x² + 6x - 2 ← collect like terms

= x² + 6

3 0

2 years ago

2/3x=10 thankyou for helping me

Novay_Z [31]

(2/3)x = 10
(3/2)(2/3)x = 10(3/2)
x = 30/2 = 15

6 0

2 years ago

Read 2 more answers

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

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