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photoshop1234 [79]

1 year ago

15

Explain why you cannot find the intersection points of the two lines shown below. Give both an algebraic reason and a graphical

reason. y=4x+1 and y= 4x+10

1 answer:

Art [367]1 year ago

5 0

Answer:

$m_1 = m_2 = 4$

Step-by-step explanation:

Given

$y = 4x + 1$

$y = 4x + 10$

Required

Show that these two lines have no point of intersection

The general equation of a line is

$y = mx + b$

Where m represents slope

In $y = 4x + 1$

$m = 4$

Similarly; in $y = 4x + 10$

$m = 4$

Since the slope of both lines are equal;

$m_1 = m_2 = 4$

<em>This implies that the lines are parallel and parallel lines have no point of intersection</em>

<em></em>

<em>See Attachment for Graph</em>

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Answer:

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1) Tofu is healthy

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If we symbolize "For all x" by the symbol ∀ then then the propositions 1(a), 2(a) and 3(a) can be written as:

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vredina [299]

You're looking for the extreme values of $x^2+3y^2+13x$ subject to the constraint $x^2+y^2\le1$ .

The target function has partial derivatives (set equal to 0)

$\dfrac{\partial(x^2+3y^2+13x)}{\partial x}=2x+13=0\implies x=-\dfrac{13}2$

$\dfrac{\partial(x^2+3y^2+13x)}{\partial y}=6y=0\implies y=0$

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$x=\cos u$

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$T(u)$ has critical points where $T'(u)=0$ :

$T'(u)=-13\sin u+4\sin u\cos u=\sin u(4\cos u-13)=0$

$(1)\quad\sin u=0\implies u=0,u=\pi$

$(2)\quad4\cos u-13=0\implies\cos u=\dfrac{13}4$

but $|\cos u|\le1$ for all $u$ , so this case yields nothing important.

At these critical points, we have temperatures of

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