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iVinArrow [24]
2 years ago
6

Question 3 and 4 please!

Mathematics
1 answer:
d1i1m1o1n [39]2 years ago
7 0

Question 3:

  1. f(x)=(\frac{2}{3})x^2+3x-2\\⇒standard form
  2. f(x)=3x^2-8x+5(3x-8)⇒intercept form
  3. f(x)=4(x-5)^2+20⇒ vertex form
  4. f(x)=-(3x-8)(3x+5)⇒none of the above

Question 4:Explain in detail how to solve the following function by completing the square:

x^2+8x-9=0

1. move the term to left side (done)

x^2+8x-9=0

2. use the quadratic formula

x=\frac{-b+\sqrt{b^2-4ac} }{2a}

A=1

B=8

C=-9

x=\frac{-(8)+\sqrt{(8)^2-4(1)(-9)} }{2(1)}

x=1

now check your answer.

1^2+8(1)-9=0

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Please help me!!!!!​
denpristay [2]

Answer:  see proof below

<u>Step-by-step explanation:</u>

Given: A + B + C = π               → A = π - (B + C)

                                               → B = π - (A + C)

                                               → C = π - (A + B)

Use Sum to Product Identity: sin A - sin B = 2 cos [(A + B)/2] · sin [(A - B)/2]

Use the following Cofunction Identity: cos (π/2 - A) = sin A

<u>Proof LHS → RHS:</u>

LHS:                        sin A - sin B + sin C

                             = (sin A - sin B) + sin C

\text{Sum to Product:}\quad 2\cos \bigg(\dfrac{A+B}{2}\bigg)\cdot \sin \bigg(\dfrac{A-B}{2}\bigg)+2\cos \bigg(\dfrac{C}2{}\bigg)\cdot \sin \bigg(\dfrac{C}{2}\bigg)

\text{Given:}\qquad 2\cos \bigg(\dfrac{\pi -(B+C)}{2}+\dfrac{B}{2}}\bigg)\cdot \sin \bigg(\dfrac{A-B}{2}\bigg)+2\cos \bigg(\dfrac{C}2{}\bigg)\cdot \sin \bigg(\dfrac{C}{2}\bigg)\\\\\\.\qquad \qquad =2\cos \bigg(\dfrac{\pi -C}{2}\bigg)\cdot \sin \bigg(\dfrac{A-B}{2}\bigg)+2\cos \bigg(\dfrac{C}2{}\bigg)\cdot \sin \bigg(\dfrac{C}{2}\bigg)

.\qquad \qquad =2\cos \bigg(\dfrac{\pi}{2} -\dfrac{C}{2}\bigg)\cdot \sin \bigg(\dfrac{A-B}{2}\bigg)+2\cos \bigg(\dfrac{C}2{}\bigg)\cdot \sin \bigg(\dfrac{C}{2}\bigg)

\text{Cofunction:} \qquad 2\sin \bigg(\dfrac{C}{2}\bigg)\cdot \sin \bigg(\dfrac{A-B}{2}\bigg)+2\cos \bigg(\dfrac{C}2{}\bigg)\cdot \sin \bigg(\dfrac{C}{2}\bigg)

\text{Factor:}\qquad 2\sin \bigg(\dfrac{C}{2}\bigg)\bigg[ \sin \bigg(\dfrac{A-B}{2}\bigg)+\cos \bigg(\dfrac{C}{2}\bigg)\bigg]

\text{Given:}\qquad 2\sin \bigg(\dfrac{C}{2}\bigg)\bigg[ \sin \bigg(\dfrac{A-B}{2}\bigg)+\cos \bigg(\dfrac{\pi -(A+B)}{2}\bigg)\bigg]\\\\\\.\qquad \qquad =2\sin \bigg(\dfrac{C}{2}\bigg)\bigg[ \sin \bigg(\dfrac{A-B}{2}\bigg)+\cos \bigg(\dfrac{\pi}{2} -\dfrac{(A+B)}{2}\bigg)\bigg]

\text{Cofunction:}\qquad 2\sin \bigg(\dfrac{C}{2}\bigg)\bigg[ \sin \bigg(\dfrac{A-B}{2}\bigg)+\sin \bigg(\dfrac{A+B}{2}\bigg)\bigg]

\text{Sum to Product:}\qquad 2\sin \bigg(\dfrac{C}{2}\bigg)\bigg[ 2\sin \bigg(\dfrac{A}{2}\bigg)\cdot \cos \bigg(\dfrac{B}{2}\bigg)\bigg]\\\\\\.\qquad \qquad \qquad \qquad =4\sin \bigg(\dfrac{A}{2}\bigg)\cdot \cos \bigg(\dfrac{B}{2}\bigg)\cdot \sin \bigg(\dfrac{C}{2}\bigg)

\text{LHS = RHS:}\quad 4\sin \bigg(\dfrac{A}{2}\bigg)\cdot \cos \bigg(\dfrac{B}{2}\bigg)\cdot \sin \bigg(\dfrac{C}{2}\bigg)=4\sin \bigg(\dfrac{A}{2}\bigg)\cdot \cos \bigg(\dfrac{B}{2}\bigg)\cdot \sin \bigg(\dfrac{C}{2}\bigg)\quad \checkmark

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2 years ago
What is the measure of x?
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Step-by-step explanation:

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2 years ago
Which equation is equivalent to 2x + 6y = 12?
mixer [17]

Answer:

Option C, y = -1/3x + 2

Step-by-step explanation:

2x + 6y = 12

<u>Step 1:  Solve for y</u>

2x + 6y - 2x = 12 - 2x

6y / 6 = (12 - 2x) / 6

y = 2 - 1/3x

Answer:  Option C, y = -1/3x + 2

6 0
3 years ago
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Can a hermaphrodite have a baby with them self..............
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no a hermaphrodite can not have a baby with themselves that is not possible

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