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natta225 [31]
2 years ago
7

Which of the following is a factor from problem #1? 1) X^2 -3x -10

Mathematics
1 answer:
Pavlova-9 [17]2 years ago
5 0
The answer is (X+2) x (X-5)
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If 6!+7!+8!=(n)(6!), then the value of n is<br> A) 15<br> B)16<br> C)63<br> D)64<br> E)15!
gladu [14]
6!+7!+8!=(n)(6!)
calculate the individual values first:
6!=720
7!=5040
8!=40320

plug them into the equation:
720+5040+40320=n720
solve for n
5760+40320=n720
46080=n720
divide both sides by 720 to isolate n
46080/720=n720/720
64=n
6 0
3 years ago
What is the density of the oak board, it's 5.5 inches wide, 1.5 inches thick , 3 feet long and weighs 8.05 lbs? Please show work
laiz [17]

Answer:

37 - 56 lb / ft3

0.6 - 0.9 103 kg / m3

Step-by-step explanation:

The density or hardness of wood varies by species, and the value is necessary to approximate the weight of lumber by volume. In this table, the density of different species of wood is expressed as weight in pounds per cubic foot and kilograms per cubic meter. The density will vary based on the moisture content of the wood.

8 0
3 years ago
Write a vector equation of the line that passes through point P(-4,11) and is parallel to a=[-3,8]. Then write parametric equati
Assoli18 [71]

Answer:

\vec l = (-4, 11) + t\cdot (-3,8), x = -4 - 3\cdot t and y = 11 + 8\cdot t

Step-by-step explanation:

The vector equation of the line is:

\vec l = \vec P + t \cdot \vec A

\vec l = (-4, 11) + t\cdot (-3,8)

The parametric equations of the line are:

x = -4 - 3\cdot t and y = 11 + 8\cdot t

4 0
3 years ago
2 less than the product of 5 and n
Deffense [45]

The product of 5 and n is written as 5n.

Two less than this product is 5n-2

7 0
3 years ago
How would you determine the axis of symmetry in order to graph x^2 = 8y^2.
Tanya [424]
The picture illustrates the definition. The point P is a typical point on the parabola so that its distance from the directrix, PQ, is equal to its distance from F, PF. The point marked V is special. It is on the perpendicular line from F to the directix. This line is called the axis of symmetry of the parabola or simply the axis of the parabola and the point V is called the vertex of the parabola. The vertex is the point on the parabola closest to the directrix.

Finding the equation of a parabola is quite difficult but under certain cicumstances we may easily find an equation. Let's place the focus and vertex along the y axis with the vertex at the origin. Suppose the focus is at (0,p). Then the directrix, being perpendicular to the axis, is a horizontal line and it must be p units away from V. The directrix then is the line y=-p. Consider a point P with coordinates (x,y) on the parabola and let Q be the point on the directrix such that the line through PQ is perpendicular to the directrix. The distance PF is equal to the distance PQ. Rather than use the distance formula (which involves square roots) we use the square of the distance formula since it is also true that PF2 = PQ2. We get

<span>(x-0)2+(y-p)2 = (y+p)2+(x-x)2. 
x2+(y-p)2 = (y+p)2.</span>If we expand all the terms and simplify, we obtain<span>x2 = 4py.</span>

Although we implied that p was positive in deriving the formula, things work exactly the same if p were negative. That is if the focus lies on the negative y axis and the directrix lies above the x axis the equation of the parabola is

<span>x2 = 4py.</span><span>The graph of the parabola would be the reflection, across the </span>x<span> axis of the parabola in the picture above. A way to describe this is if p > 0, the parabola "opens up" and if p < 0 the parabola "opens down".</span>

Another situation in which it is easy to find the equation of a parabola is when we place the focus on the x axis, the vertex at the origin and the directrix a vertical line parallel to the y axis. In this case, the equation of the parabola comes out to be

<span>y2 = 4px</span><span>where the directrix is the verical line </span>x=-p and the focus is at (p,0). If p > 0, the parabola "opens to the right" and if p < 0 the parabola "opens to the left". The equations we have just established are known as the standard equations of a parabola. A standard equation always implies the vertex is at the origin and the focus is on one of the axes. We refer to such a parabola as a parabola in standard position.

Parabolas in standard positionIn this demonstration we show how changing the value of p changes the shape of the parabola. We also show the focus and the directrix. Initially, we have put the focus on the y axis. You can select on which axis the focus should lie. Also you may select positive or negative values of p. Initially, the values of p are positive. 

5 0
3 years ago
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