How to solve and what is the answer

Mathematics

1 answer:

TiliK225 [7]3 years ago

7 0

Two ways:
1) guess factors(trial and error)
2) use quadratic formula.
If you use this method then a = -3, b = -6 and c = -1

x = -b +/- [sqrt(b^2 -4ac)/2a]
substituting a, b, and c into our equation we get:
x = - (-6)+/- [sqrt ((-6)^2) - 4(-3)(-1))/2 (-3)]
x = + 6 +/- [sqrt (36 -4 (3)/-6)] if I didnt make a mistake in my signs
x = + 6 +/- [sqrt (36 -12)/-6)]
x = 6 +/- [sqrt (24)/-6] but sqrt 6 x sqrt of 4 = sqrt 24 hence
x = 6 +/- [ sqrt 6 x sqrt 4 /-6] that is:
x = 6 +/- [sqrt 6 x 2 /-6 ]
so x = 6 + [sqrt 6 x 2/-6] and x = 6 - [sqrt 6 x 2/-6]

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Can someone tell me if I’m right?

mestny [16]

You forgot to expand and simplify it... you only did step one, carry on your doing great 4 marks doesn’t come easy :)

8 0

3 years ago

(1 point) (a) Find the point Q that is a distance 0.1 from the point P=(6,6) in the direction of v=⟨−1,1⟩. Give five decimal pla

natima [27]

Answer:

following are the solution to the given points:

Step-by-step explanation:

In point a:

$\vec{v} = -\vec{1 i} +\vec{1j}\\\\|\vec{v}| = \sqrt{-1^2+1^2}$

$=\sqrt{1+1}\\\\=\sqrt{2}$

calculating unit vector:

$\frac{\vec{v}}{|\vec{v}|} = \frac{-1i+1j}{\sqrt{2}}$

the point Q is at a distance h from P(6,6) Here, h=0.1

$a=-6+O.1 \times \frac{-1}{\sqrt{2}}\\\\= 5.92928 \\\\b= 6+O.1 \times \frac{-1}{\sqrt{2}} \\\\= 6.07071$

the value of Q= (5.92928 ,6.07071 )

In point b:

Calculating the directional derivative of $f (x, y) = \sqrt{x+3y}$ at P in the direction of $\vec{v}$

$f_{PQ} (P) =\fracx{f(Q)-f(P)}{h}\\\\$

$=\frac{f(5.92928 ,6.07071)-f(6,6)}{0.1}\\\\=\frac{\sqrt{(5.92928+ 3 \times 6.07071)}-\sqrt{(6+ 3\times 6)}}{0.1}\\\\= \frac{0.197651557}{0.1}\\\\= 1.97651557$

$\vec{v} = 1.97651557$

In point C:

Computing the directional derivative using the partial derivatives of f.

$f_x(x,y)= \frac{1}{2 \sqrt{x+3y}}\\\\ f_x (6,6)= \frac{1}{2 \sqrt{22}}\\\\f_x(x,y)= \frac{1}{\sqrt{x+3y}}\\\\ f_x (6,6)= \frac{1}{\sqrt{22}}\\\\f_{(PQ)}(P)= (f_x \vec{i} + f_y \vec{j}) \cdot \frac{\vec{v}}{|\vec{v}|}\\\\= (\frac{1}{2 \sqrt{22}}\vec{i} + \frac{1}{\sqrt{22}} \vec{j}) \cdot \frac{-1}{\sqrt{2}}\vec{i} + \frac{1}{\sqrt{2}} \vec{j}$