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hodyreva [135]
3 years ago
7

Write the slope-intercept form for the equation of a line parallel to y = -9 and passes through the point (12, -5)

Mathematics
1 answer:
shepuryov [24]3 years ago
8 0

Answer:

Let's see what to do buddy...

_________________________________

Step-by-step explanation:

_________________________________

STEP (1) :

The slope of the line <em>y = - 9</em> is 0(zero) because it is a Horizontal line.

So the slope of the line which is parallel to <em>y</em><em> </em><em>=</em><em> </em><em>-</em><em>9</em><em> </em><em> </em> is <em>zero</em><em> </em>.

_________________________________

STEP (2) :

We have this equation to find the slope-intercept form of the linear functions :

y - y(given \: point) = (slope) \times ( \: x -x(g \: p) \: ) \\

_________________________________

STEP (3) :

Now it is time to put the slope and the given point in the equation :

<em>given</em><em> </em><em>point</em><em> </em><em>=</em><em> </em><em>(</em><em> </em><em>1</em><em>2</em><em> </em><em>,</em><em> </em><em>-</em><em>5</em><em> </em><em>)</em>

<em>slope</em><em> </em><em>=</em><em> </em><em>0</em><em> </em>

y - ( - 5) = 0 \times (x - 12) \\  \\ y + 5 = 0 \\  \\ y =  - 5

_________________________________

And we're done.

Thanks for watching buddy good luck.

♥️♥️♥️♥️♥️

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<span><span>x2 </span>= −1.51414</span>

<span><span>x3 </span>= 3.08613</span>


x1 and x2 are in the desired interval [-2, 2]

f'(x) = 3x^2 - 4x - 4

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3x^2 - 4x - 4 = 0

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According to Rolle's theorem, we have one point in between:

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4 0
3 years ago
A rancher wants to fence in an area of 1500000 square feet in a rectangular field and then divide it in half with a fence down t
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Answer:

6000 ft

Step-by-step explanation:

Let length of rectangular field=x

Breadth of rectangular field=y

Area of rectangular field=1500000 square ft

Area of rectangular field=l\times b

Area of rectangular field=x\time y

1500000=xy

y=\frac{1500000}{x}

Fencing used ,P(x)=x+x+y+y+y=2x+3y

Substitute the value of y

P(x)=2x+3(\frac{1500000}{x})

P(x)=2x+\frac{4500000}{x}

Differentiate w.r.t x

P'(x)=2-\frac{4500000}{x^2}

Using formula:\frac{dx^n}{dx}=nx^{n-1}

P'(x)=0

2-\frac{4500000}{x^2}=0

\frac{4500000}{x^2}=2

x^2=\frac{4500000}{2}=2250000

x=\sqrt{2250000}=1500

It is always positive because length is always positive.

Again differentiate w.r.t x

P''(x)=\frac{9000000}{x^3}

Substitute x=1500

P''(1500)=\frac{9000000}{(1500)^3}>0

Hence, fencing is minimum at x=1 500

Substitute x=1 500

y=\frac{1500000}{1500}=1000

Length of rectangular field=1500 ft

Breadth of rectangular field=1000 ft

Substitute the values

Shortest length of fence used=2(1500)+3(1000)=6000 ft

Hence, the shortest length of fence that the rancher can used=6000 ft

3 0
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