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2 years ago

What should be added to 4x²-3x²+1 to make it a perfect square

Mathematics

2 answers:

Vlad1618 [11]2 years ago

8 0

$im \: guessing \: its \: 4x {}^{2} - 3x + 1 \\ (correct \: me \: if \: im \: wrong)$

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

$4x {}^{2} = a {}^{2} \: \: \: \: \: \: \: \: \: \: \: b {}^{2} = 1 \\ a = 2x \: \: \: \: \: \: \: \: \: \: \: \: \: \: b = 1$

$(2x - 1) {}^{2} = 4x {}^{2} -4x + 1$

$we \: need \: to \: add \: a \: - x \\ \: to \: the \: original \: equation$

<h2>Note that this problem can be approached in many ways. For instance, we can add to the b² term instead of changing the middle term...</h2>

Vinvika [58]2 years ago

3 0

You could solve this by using completing the square and simplifying

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How to solve radical 28 plus radical 64

Finger [1]

The solution to the problem is as follows:

<span>√28 = √(4 * 7) = √4 * √7 = 2√7.
</span>
<span>√64 = 8
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Therefore, √28 + √64 = <span>2√7. + 8

I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!
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8 0

2 years ago

What is the number of terms in this geometric series? 1+2+4+...+64

ioda

Answer:

Step-by-step explanation:

1,2 ,4,8,16,32,64

4 0

2 years ago

An object launched straight up at a speed of 29.4 meters per second has a height, h, in meters of h , t seconds after the object

Nostrana [21]

Answer:

h = 44.06 meters (maximum height)

the time the object takes to complete this whole path is 6 seconds, this is why the time at which the object reaches its maximum height will between 0 and 6 seconds

Step-by-step explanation:

To solve this question, we need to first recognize that this is a constant acceleration problem, specifically, it can be thought of as a projectile motion problem.

Recall, the equations of motion:

$1) v^2 - v_0^2 = 2a(s - s_0)\\2) s = v_0^2 + \frac{1}{2} at^2\\3) v = v_0 + at$

What do we already know?

$v_0 = 29.4 ms^-1$
The launch is straight up
$a = -9.81 ms^-2$ this is the gravitational acceleration $g$

$s_0 = 0 m$ , since our reference point is at s = 0, (the ground)

We can use use the Eq(1):

we know that when any object is launched up, at maximum height its velocity is going to be zero, $v = 0 ms^-2$

$v^2 - v_0^2 = 2a(s - s_0)\\0^2 - (29.4)^2 = 2(-9.81)(s- 0)\\s = 44.06 m$

this is the maximum height!

Why does t have to between zero and six?

We can answer this using a bit visualization, if you think about the second equation

$s = v_0 t - \frac{1}{2}at^2\\ s = 29.4t - 4.905t^2$

this is the equation of the whole trajectory that object makes.

and if you solve this by making $s = 0$ , you will get the times at which the object was at the ground. the times will be 0s and 5.99s.

so the amount of time the object takes to go through this whole path is 6 seconds and this why the object will only reach its maximum height in between this time interval.

hope this helps :)