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Jobisdone [24]
3 years ago
8

Can I copy and paste a pic of a problem or diagram ?

Mathematics
1 answer:
ExtremeBDS [4]3 years ago
6 0
No you can not copy and paste a picture of a problem i believe
You might be interested in
Is 3/8 bigger than 4/10
Zigmanuir [339]

Answer:

no 4/10 is bigger than 3/8

Step-by-step explanation:

5 0
3 years ago
The olympics are held every four years. In 2016, the XXXI olympiad were held in Rio de Janeiro,Brazil. Determine the year of the
Anna007 [38]

Answer:

First Olympic held in year 1896.

Step-by-step explanation:

Number of Olympiad held in Rio = XXXI = 31st

Difference of years in two consecutive Olympiads = 4 years

Number of years spent in 31 Olympiads can be calculated by,

Number of years spent = (n - 1)d

Here n = number of Olympiads held

d = difference between two consecutive Olympiads

Number of years spent till 31st Olympiads = (31 - 1)×4

                                                                      = 120 years

Therefore, 1st Olympiad held in the year = 2016 - 120

                                                                   = 1896

6 0
3 years ago
Solve for x<br>Please show steps thank you.
const2013 [10]
The absolute value of x is simply x and followed by the expression.

X + 1 + x + 1 <_ 2

2x + 2 <_ 2

2x <_ 0

X <_ 0

Basically all values that follow this inequality will most likely hold true.

For instance. -1.

-1 + 1 = 0
Absolute value of -2 = 2.

2 is equal to 2. This expression holds true.
4 0
3 years ago
PLEASE HELP ME SOLVE THIS<br><br> THANK U!!
V125BC [204]
Step 1) Draw a dashed line through the points (0,6) and (4,7). These two points are on the line y = (1/4)x+6. To find those points, you plug in x = 0 to get y = 6. Similarly, plug in x = 4 to get y = 7. The dashed line indicates that none of the points on this line are part of the solution set.

Step 2) Draw a dashed line through (0,-1) and (1,1). These two points are on the line y = 2x-1. They are found in a similar fashion as done in step 1. 

Step 3) Shade the region that is above both dashed lines. We shade above because of the "greater than" sign. This is shown in the attached image I am providing below. The red shaded region represents all of the possible points that are the solution set. Once again, any point on the dashed line is not in the solution set. 

5 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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