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3 years ago

Which quadratic equation is equivalent to (x + 2)2 + 5(x + 2) – 6 = 0?

Mathematics

1 answer:

Anna35 [415]3 years ago

6 0

Answer:

x² + 9x + 8 = 0

Step-by-step explanation:

A quadratic equation is expressed this way: ax² + bx + c = 0

So, so find the quadratic equivalent of (x + 2)² + 5(x + 2) – 6 = 0, we just have to do the multiplications then simplify.

(x + 2)² + 5(x + 2) – 6 = (x² + 4x + 4) + (5x + 10) - 6

After merging the terms of same degree, we get:

x² + 9x + 8 = 0

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2. Starting at a fixed time, each car entering an intersection is observed to see whether it turns left (L), right (R), or goes

jasenka [17]

Answer:

Natural numbers (integers greater than zero)

X = 3, 5, 4, 4, 3

Step-by-step explanation:

The least number of cars that can be observed in this experiment is 1, if the first car turns left. On the other hand, the experiment could go on forever if no car ever turns left, thus the highest number of cars approaches infinite.

The possible values of X are integers greater than zero, which are known as the Natural numbers.

If X = number of cars observed, simply count the number of letters in each outcome for the value of X:

Outcome = RRL, AARRL, AARL, RRAL, ARL

X = 3, 5, 4, 4, 3

7 0

3 years ago

(9x +1) 45 = (x+5) 14

ipn [44]

Use the website math papa it’ll give you the answers for your problem I promise!! Have a good day :)

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3 years ago

Cos (90-theta) • cosec90-theta) =tano. How?

denis-greek [22]

Answer:

______________________________________________________

TRIGONOMETRY IDENTITIES TO BE USED IN THE QUESTION :-

For any right angled triangle with one angle α ,

$\cos (90 - \alpha ) = \sin \alpha$ or $\sin(90 - \alpha ) = \cos\alpha$

$cosec \: (90 - \alpha ) = \sec\alpha$ or $\sec(90 - \alpha ) = cosec\:\alpha$

SOME GENERAL TRIGNOMETRIC FORMULAS :-

$\sin \alpha = \frac{1}{cosec \: \alpha }$ or $cosec \: \alpha = \frac{1}{\sin \alpha }$

$\cos \alpha = \frac{1}{\sec \alpha }$ or $\sec \alpha = \frac{1}{\cos \alpha }$

______________________________________________________

Now , lets come to the question.

In a right angled triangle , let one angle be α (in place of theta) .

So , lets solve L.H.S.

$\cos (90 - \alpha ) \times cosec(90 - \alpha )$

$=> sin\alpha \times \sec\alpha$

$=> \sin\alpha \times \frac{1}{\cos\alpha }$

$=> \frac{\sin\alpha }{\cos\alpha }$

$=> \tan\alpha$ = R.H.S.

∴ L.H.S. = R.H.S. (Proved)

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3 years ago

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Afina-wow [57]

Answer:

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Step-by-step explanation:

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4 years ago

What is the perimeter of this polygon? A(2, 3) B(-4, 0) C(0,-4) D(4,0)

vovangra [49]

Answer:2,3 hope it help you

Step-by-step explanation:

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