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saw5 [17]
3 years ago
9

Hi. I need help with these questions. See image for question. Answer 14 and 13

Mathematics
1 answer:
irinina [24]3 years ago
8 0

Answer:

  • 13) 36x² + 9x - 4 = 0
  • 14) 21x² + 20x + 100 = 0

Step-by-step explanation:

13.

<h3>Given</h3>

<u>Quadratic equation</u>

  • 4x² - 3x - 4 = 0
  • With the roots of α and β
<h3>To Find </h3>
  • The quadratic equation with roots of 1/(3α) and 1/(3β)
<h3>Solution</h3>

<u>The sum and the product of the roots of the given equation:</u>

  • α + β = -b/a ⇒ α + β = -(-3)/4 = 3/4
  • αβ = c/a ⇒ αβ = -4/4 = - 1

<u>New equation is:</u>

  • (x - 1/(3α))(x - 1/(3β)) = 0
  • x² - (1/(3α) + 1/(3αβ))x + 1/(3α3β) = 0
  • x² - ((3α + 3 β)/(3α3β))x + 1/(3α3β) = 0
  • x² - ((α + β)/(3αβ))x + 1/(9αβ) = 0
  • x² - (3/4)/(3(-1))x + 1/(9(-1)) = 0
  • x² + 1/4x - 1/9 = 0
  • 36x² + 9x - 4 = 0

===================

14.

<h3>Given</h3>

<u>Quadratic equation</u>

  • 3x² +2x + 7 = 0
  • With the roots of α and β
<h3>To Find </h3>
  • The quadratic equation with roots of α + 1/β and β + 1/α
<h3>Solution</h3>

<u>The sum and the product of the roots of the given equation:</u>

  • α + β = -b/a ⇒ α + β = -2/3
  • αβ = c/a ⇒ αβ = 7/3

<u>New equation is:</u>

  • (x - (α + 1/β))(x - (β + 1/α)) = 0
  • x² - (α + 1/β + β + 1/α)x + (α + 1/β) (β + 1/α) = 0
  • x² - (α + β + (α + β)/αβ )x + αβ + 1/αβ + 2 = 0
  • x² - (-2/3 - (2/3)/(7/3))x + 7/3 + 1/(7/3) + 2 = 0
  • x² - (-2/3 - 2/7)x + 7/3 + 3/7 + 2 = 0
  • x² + (14 + 6)/21x + (49 + 9 + 42/21) = 0
  • x² + 20/21x + 100/21 = 0
  • 21x² + 20x + 100 = 0
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Step-by-step explanation:

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3 0
3 years ago
A. Evaluate ∫20 tan 2x sec^2 2x dx using the substitution u = tan 2x.
irakobra [83]

Answer:

The integral is equal to 5\sec^2(2x)+C for an arbitrary constant C.

Step-by-step explanation:

a) If u=\tan(2x) then du=2\sec^2(2x)dx so the integral becomes \int 20\tan(2x)\sec^2(2x)dx=\int 10\tan(2x) (2\sec^2(2x))dx=\int 10udu=\frac{u^2}{2}+C=10(\int udu)=10(\frac{u^2}{2}+C)=5\tan^2(2x)+C. (the constant of integration is actually 5C, but this doesn't affect the result when taking derivatives, so we still denote it by C)

b) In this case u=\sec(2x) hence du=2\tan(2x)\sec(2x)dx. We rewrite the integral as \int 20\tan(2x)\sec^2(2x)dx=\int 10\sec(2x) (2\tan(2x)\sec(2x))dx=\int 10udu=5\frac{u^2}{2}+C=5\sec^2(2x)+C.

c) We use the trigonometric identity \tan(2x)^2+1=\sec(2x)^2 is part b). The value of the integral is 5\sec^2(2x)+C=5(\tan^2(2x)+1)+C=5\tan^2(2x)+5+C=5\tan^2(2x)+C. which coincides with part a)

Note that we just replaced 5+C by C. This is because we are asked for an indefinite integral. Each value of C defines a unique antiderivative, but we are not interested in specific values of C as this integral is the family of all antiderivatives. Part a) and b) don't coincide for specific values of C (they would if we were working with a definite integral), but they do represent the same family of functions.  

3 0
3 years ago
5(2x − 12) + 21 = 2x + 41<br><br> I know the answer is 10 but can someone explain how to get that?
dsp73
5(2x-12)+21=2x+41\\\\5\cdot2x-5\cdot12+21=2x+41\\\\10x-60+21=2x+41\\\\10x-39=2x+41\qquad|+39\\\\10x-39+39=2x+41+39\\\\10x=2x+80\qquad|-2x\\\\10x-2x=2x+80-2x\\\\8x=80\qquad|:8\\\\\boxed{x=10}
6 0
3 years ago
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