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3 years ago

Find all Values of b that will make the polynomial a perfect square trinomial.

Mathematics

1 answer:

weeeeeb [17]3 years ago

8 0

Answer:

10,404/334,084

Step-by-step explanation:

Given the polynomial

289r^2 - 102r + c

We are to find the value of c that will make it a perfect square

Divide through by 289

289r²/289 - 102r/289 + c/289

Half of the coefficient of r is 1/2(102/289)

Half of the coefficient of r = 102/578

Square the result

r² = (102/578)²

r² = 10,404/334,084

Hence the required constant is 10,404/334,084

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Sin(60-theta)sin(60+theta)

ehidna [41]

Step-by-step explanation:

$\sin(60 - \theta) \sin(60 + \theta) \\ = \{ \sin(60) \cos( \theta) - \sin( \theta) \cos(60) \} \{ \sin(60) \cos( \theta) + \cos(60) \sin( \theta) \} \\ = \{ \frac{ \sqrt{3} }{2} \cos( \theta) - \frac{1}{2} \sin( \theta) \} \{ \frac{ \sqrt{3} }{2} \cos( \theta) + \frac{1}{2} \sin( \theta) \} \\ \\$

from difference of two squares:

${ \boxed{(a - b)(a + b) = ( {a}^{2} - {b}^{2} ) }}$

therefore:

$= \{ {( \frac{ \sqrt{3} }{2}) }^{2} { \cos }^{2} \theta \} - \{ {( \frac{ \sqrt{3} }{2} )}^{2} { \sin }^{2} \theta \} \\ \\ = \frac{3}{4} { \cos }^{2} \theta - \frac{3}{4} { \sin}^{2} \theta$

factorise out ¾ :

$= \frac{3}{4} ( { \cos }^{2} \theta - { \sin}^{2} \theta) \\ \\ = { \boxed{ \frac{3}{4} \cos(2 \theta) }}$

3 0

3 years ago

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Maura runs 5 miles in 1.25 hours. Ava runs 3 miles in 50 minutes or 0.83¯ hour.<br> Who runs faster?

kenny6666 [7]

Answer:

║⊕║∵°·∵∴Hello :)∴∵·°∵∴║⊕║

Your answer should be:

Maura runs faster.

Step-by-step explanation:

I hope this helped!

<em>Brainliest appreciated</em>

8 0

3 years ago

PLEASE HELP MEH (MATH)

emmainna [20.7K]

7₈/7₈, 2(-2) ˣ 2(-3), 4₂ ˣ 49-1), 5(-10)/5(-12)

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3 years ago

Factor the trinomial x^2-2x-35

Oliga [24]

Answer:

(x-7)(x+5)

Step-by-step explanation:

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3 years ago

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Find the area of the triangle formed by the origin and the points of intersection of parabolas y=− 1/3 (x−1)^2+8 and y=x^2−2x−3.

Kobotan [32]

Answer:

The area of the triangle formed by origin, and the points (4,5) and (-2,5) will be 15 sq. units.

Step-by-step explanation:

The two parabolas are $y = - \frac{1}{3}(x - 1)^{2} + 8$ and y = x² - 2x - 3.

Now, solving those two equations we will get the points of intersection.

So, $- \frac{1}{3}(x - 1)^{2} + 8 = x^{2} - 2x - 3$

⇒ - (x - 1)² + 24 = 3x² - 6x - 9

⇒ -x² + 2x - 1 + 24 = 3x² - 6x - 9

⇒ 4x² - 8x - 32 = 0

⇒ x² - 2x - 8 = 0

⇒ x² - 4x + 2x - 8 = 0

⇒ (x - 4)(x + 2) = 0

So, x = 4 or - 2.

Now, for x = 4 , y = 4² - 2(4) - 3 = 5 and the point of intersection is (4,5).

Or, for x = - 2, y = (- 2)² - 2(- 2) - 3 = 5 and the point of intersection is (-2,5).

Now, points (4,5) and (-2,5) make a straight line parallel to the x-axis at a perpendicular distance of 5 units from origin and its length is (4 - (- 2)) = 6 units.

So, the area of the triangle formed by origin, and the points (4,5) and (-2,5) will be = 0.5 × 5 × 6 = 15 sq. units. (Answer)

8 0

3 years ago