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4 years ago

How to solve this equation by getting a value and b value

Mathematics

1 answer:

goblinko [34]4 years ago

3 0

Answer:

a=±5

Step-by-step explanation:

25=5^2

Given that, x^2−10x+25=x^2−10x+5^2

Identity: a^2−2(ab)+b^2=(a-b)^2

Here, a=x and b=5

∴=(x−5)^2

Hope this helps

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#3 please <br> What’s the answer & how do u do it

Monica [59]

Answer:

<h2>The desired equation is y = (4/3)x - 14/3.</h2>

Step-by-step explanation:

This new line passes thru (-4, 8) and (0, 5).

As we move from (-4, 8) to (0, 5), x increases by 4 and y decreases by 3.

Thus, the slope of this line is m = rise / run = -3/4.

Thus, the slope of a line perpendicular to the given line is the negative reciprocal of -3/4: 4/3.

Thus, the equation of a line perpendicular to the given line is

y = (4/3)x + b.

This line passes thru the point (5, 2). Thus, the following is true:

2 = (4/3)(5) + b, or 2 = 20/3 + b

Mult. this 2 by 3/3, we get 6/3, so that 6/3 = 20/3 + b. b = -14/3.

<h2>The desired equation is y = (4/3)x - 14/3.</h2>

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3 years ago

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Answer:

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Step-by-step explanation:

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3 years ago

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If i = StartRoot negative 1 EndRoot, what is the value of i 3?

12345 [234]

Answer:

i³ = - i

Step-by-step explanation:

<u><em>Explanation</em></u>:-

<u><em>Step(i)</em></u>:-

Given $i = \sqrt{-1}$

we know that cos π = -1

sin π = 0

now given data

we can write $i = \sqrt{cos\pi + i sin\pi }$

$i =( {cos\pi + i sin\pi })^{\frac{1}{2} }$

<u><em>step(ii)</em></u>:-

In complex numbers nth roots of a complex number

Z = r(cosθ+i sinθ)

$Z^{\frac{n}{2} } = r^{\frac{n}{2} } (cos\pi +isin\pi )^{\frac{n}{2} }$

$Z^{\frac{n}{2} } = r^{\frac{n}{2} } (cosn\pi/2 }+isinn\pi/2 )$

Step(iii):-

now $i =( {cos\frac{\pi }{2} + i sin\frac{\pi }{2} }) }$

$i^{3} =( {cos\frac{\pi }{2} + i sin\frac{\pi }{2} })^{3} }$

$i^{3} =( {cos\frac{3\pi }{2} + i sin\frac{3\pi }{2} })}$

By using trigonometric

$cos\frac{3\pi }{2} = cos (\pi +\frac{\pi }{2} ) = -cos\frac{\pi }{2} =0$

$sin\frac{3\pi }{2} = sin (\pi +\frac{\pi }{2} ) = -sin\frac{\pi }{2} = -1$

$i^{3} =( 0+ i (-1) ) = -i$

<u><em>Final answer:-</em></u>

i³ = - i

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4 years ago