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ludmilkaskok [199]

3 years ago

8

Which function models the area of a rectangle with side lengths of 2x-4 units and x+1 units?

1 answer:

Morgarella [4.7K]3 years ago

3 0

Answer:

Step-by-step explanation:

——————————————————————————————

l

l l

l l

l____________________________________________l

either the ____ line is 2x-4 or x+1

either the "l" line is the other eq the the _____ one

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Find the intervals of DECREASING only and write in below in interval<br> notation.

Grace [21]

Answer:

(-1,0)

Step-by-step explanation:

It is decreasing when the "slope" of the function is negative. That only happens on the blue curve from x = -1 to x = 0. In interval notation, that's (-1,0)

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nikdorinn [45]

Answer:

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Step-by-step explanation:

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gogolik [260]

The correct answer for the slope is 1/4

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Find the set of values of k for which the line y=kx-4 intersects the curve y=x²-2x at 2 distinct points?

Sonbull [250]

Answer:

$-6 < k < 2$

Step-by-step explanation:

Given

$y = x^2 - 2x$

$y =kx -4$

Required

Possible values of k

The general quadratic equation is:

$ax^2 + bx + c = 0$

Subtract $y = x^2 - 2x$ and $y =kx -4$

$y - y = x^2 - 2x - kx +4$

$0 = x^2 - 2x - kx +4$

Factorize:

$0 = x^2 +x(-2 - k) +4$

Rewrite as:

$x^2 +x(-2 - k) +4=0$

Compare the above equation to: $ax^2 + bx + c = 0$

$a = 1$

$b= -2-k$

$c =4$

For the equation to have two distinct solution, the following must be true:

$b^2 - 4ac > 0$

So, we have:

$(-2-k)^2 -4*1*4>0$

$(-2-k)^2 -16>0$

Expand

$4 +4k+k^2-16>0$

Rewrite as:

$k^2 + 4k - 16 + 4 >0$

$k^2 + 4k - 12 >0$

Expand

$k^2 + 6k-2k - 12 >0$

Factorize

$k(k + 6)-2(k + 6) >0$

Factor out k + 6

$(k -2)(k + 6) >0$

Split:

$k -2 > 0$ or $k + 6> 0$

So:

$k > 2$ or k $> -6$

To make the above inequality true, we set:

$k < 2$ or $k >-6$

So, the set of values of k is:

$-6 < k < 2$

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Artemon [7]

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