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3 years ago

The point (2,6) lies on the graph of the equation y = kx^2 - 3x Find the value of K

Mathematics

1 answer:

MAVERICK [17]3 years ago

4 0

Answer:

Step-by-step explanation:

y = kx² - 3x

Since the points lie on the graph we can substitute the values of x and y that's

( 2 ,6) into the equation to find k.

We have

6 = (2)²k - 3(2)

6 = 4k - 6

4k = 6 + 6

4k = 12

Divide both sides by k

That's

$\frac{4k}{4} = \frac{12}{4}$

We have the final answer as

Hope this helps you

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3 years ago

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12345 [234]

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4 years ago

Solve the following equation by completing the square. 12x^2 - 48x = -39

daser333 [38]

Answer:

$\large\boxed{\boxed{x = \begin{cases} \frac{ \sqrt{3} }{2} + 2 \\ - \frac{ \sqrt{3} }{2} + 2 \end{cases}}}$

Step-by-step explanation:

<h3>to understand this</h3><h3>you need to know about:</h3>

quadratic equation
PEMDAS

<h3>let's solve:</h3>

$\sf divide \: both \: sides \: by \: 12 : \\ \sf {x}^{2} - 4x = - \frac{13}{4}$
$\sf \: add \: { - 2}^{2} \: to \: both \: sides : \\ \sf { {x}^{2} } - 4x + ( - {2}^{2} ) = \frac{13}{4} + ( { - 2}^{2} ) \\ {x}^{2} - 4x + 4 = \frac{13}{4} + 4$
$\sf simplify \: addition : \\ \sf { {x}^{2} } - 4x + 4 = \frac{3}{4}$
$\sf use \: {a}^{2} - 2ab + {b}^{2} = (a - b {)}^{2} : \\ \sf (x - 2 {)}^{2} = \frac{3}{4}$
$\sf squre \: root \: both \: sides : \\ \sf \sqrt{(x - 2 {)}^{2} } = \pm \sqrt{ \frac{3}{4} } \\ \begin{cases} x - 2 = \frac{ \sqrt{3} }{2} \\x - 2 = - \frac{ \sqrt{3} }{2} \end{cases}$
$\sf add \: 2 \: to \: both \: sides : \\ \sf \begin{cases}x = \frac{ \sqrt{3} }{2} + 2 \\ x = - \frac{ \sqrt{3} }{2} + 2 \end{cases} \\ \therefore \: x = \begin{cases} \frac{ \sqrt{3} }{2} + 2 \\ - \frac{ \sqrt{3} }{2} + 2 \end{cases}$

7 0

3 years ago

Read 2 more answers

Using an ordered alphabet of 26 letters, how many ways are there to choose a set of six letters such that no two letters in the

mestny [16]

To find our solution, we can start off by creating a string of 27 boxes, all followed by the letters of the alphabet. Underneath the boxes, we can place 6 pairs of boxes and 15 empty boxes.The stars represent the six letters we pick. The empty boxes to the left of the stars provide the "padding" needed to ensure that no two adjacent letters are chosen. We can create this -
$\binom {21} {6}= \frac{21*20*19*18*17*16}{6*5*4*3*2*1} =21*19*17*8=54264$
Thus, the answer is that there are $\boxed{54264}$ ways to choose a set of six letters such that no two letters in the set are adjacent in the alphabet. Hope this helped and have a phenomenal New Year! <em>2018</em>

5 0

3 years ago

Vadim26 [7]

Answer:

AAS criterion is the correct answer of this question.....

5 0

3 years ago