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Kamila [148]
3 years ago
6

An entire pumkin pie (16 Slices ) has 3,840 calories. If the pie is cut into 16 slices how many calories are in each slice?

Mathematics
2 answers:
Nookie1986 [14]3 years ago
7 0

Answer:

240

Step-by-step explanation:

Levart [38]3 years ago
7 0

Answer:

240 calories

Step-by-step explanation:

3,840 divided by 16 is 240 :)

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Write a number in which the digit 3 is ten times the value of the digit 3 in 9.431
IgorC [24]

Answer:

Step-by-step explanation:

One number would be

9.31

The 3 in this number is 3/10

The 3 in the given number is 3/100

3/10 is 10 times bigger than 3/100

3/10 = 0.3

3/100 = 0.03

0.03 * x = 0.3                   Divide both sides by 0.03

0.03/0.03 x = 0.3/0.03

x = 10

6 0
3 years ago
Which of these is equivalent to 1 over x to the power of 5 ? 1 fifth x negative x to the power of 5 x to the power of 5 x to the
Kisachek [45]

Answer:

x^-5 = x to the power of negative 5

Step-by-step explanation:

Which of these is equivalent to 1 over x to the power of 5 ?

Mathematically this is expressed as

(1/x)⁵

We have a rule when it comes to expressing power

(1/a)^b = a^-b

Hence, applying this rule to our question

(1/x)⁵ = (1/x)^5

= x^-5

This is written in words as:

x to the power of negative 5

5 0
3 years ago
Find the remainder of (x4 – 2) ÷ (x + 1).
prisoha [69]
Please write   (x4 – 2) ÷ (x + 1)  as  <span>(x^4 – 2) ÷ (x + 1).

We can find the remainder using synth. div. as follows:

      _________________
-1  /  1    0    0    0    -2
              -1    1   -1    1
     ------------------------------
         1   -1    1   -1    -1

The remainder is -1.</span>
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
4 years ago
Which function does this graph correspond to?
tiny-mole [99]
ANSWER

The function that the given graph corresponds to is
f(x) = (x - 2)(x - 1)(x + 1)(x  + 2)



EXPLANATION

The given graph in the attachment has x-intercepts at

x =  - 2 ,x=-1,x=1  \:  \: and   \: \: x=2


These are the solutions of the polynomial function represented by the given graph.




We can write all these solutions as factors to obtain,

x  + 2  ,x + 1 ,x - 1   \:  \:and \:  \: x - 2


The product of all these factors gives us the function whose graph was drawn.


This implies that,

f(x) = (x - 2)(x - 1)(x + 1)(x  + 2)


Therefore the correct answer is option C.
7 0
3 years ago
Read 2 more answers
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