Express the polynomial as a product of linear factors f(x)=2x^3+4x^2-2x-4

1 answer:

5 0

$f(x)=2x^3+4x^2-2x-4=2(x^3+2x^2-x-2)\\\\=2[x^2(x+2)-1(x+2)]=2(x+2)(x^2-1)=2(x+2)(x^2-1^2)\\\\\boxed{f(x)=2(x+2)(x-1)(x+1)}\\\\Used\ (a-b)(a+b)=a^2-b^2$

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Two gardens are being fenced in. Both gardens will require the same amount of fencing (both gardens have the same perimeter in

Tresset [83]

Answer:

Dimensions of the square garden = 9 meters by 9 meters

Dimensions of the triangular garden = 12 meters

Step-by-step explanation:

Let the measure of a side of the square garden = x meters

Perimeter of the square = 4x meters

Since each side of the triangle is 3 meters longer than each side of the square, side of the triangle will measure = (x + 3)

Perimeter of the equilateral triangle = 3(x + 3) meters

Perimeter of the square = perimeter of equilateral triangle

4x = 3(x + 3)

4x = 3x + 9

4x - 3x = 9

x = 9 meters

Dimensions of the square garden = 9 meters by 9 meters

Dimensions of the triangular garden = 12 meters

5 0

3 years ago

Determine whether each ordered pair is a solution to the inequality x+y<−1.

polet [3.4K]

Answer:

Choice A. (10, -1)
Choice B. (-8, 9)
Choice D. (6, -3)

Step-by-step explanation:

If we plug the coordinates of point A into the inequality, then we get,
x+y > -2

10 + (-1) > -2
9 > -2
That last inequality is a true statement since 9 is to the right of -2 on the number line. That means (10,-1) is a solution. Choice A is one of the answers.
Choices B and D are also answers for similar reasons.
Something like choice C is not a solution because
x+y > -2
-1+(-9) > -2
-10 > -2
Which is false.
You should find that choice E is false as well.

If you graphed the inequality and all of the points mentioned (see below), then you can visually confirm the answers. Notice how points A, B and D are in the blue shaded region which is the solution set.
The point E on the boundary does not count as a solution. This is due to the lack of "or equal to" portion of the inequality sign. That visually shows point E is not a solution. Point C isn't a solution either as it's nowhere near the blue shaded region.