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nata0808 [166]
3 years ago
5

Will make brainiest if answered correctly write the value of the expression 3^3/3^6

Mathematics
1 answer:
Anestetic [448]3 years ago
7 0

Answer:

1/27

Step-by-step explanation:

Hey!!!

Let's put it in some LaTeX:

\frac{3^3}{3^6}

<em>The viniculum in a fraction represents subtraction with the exponents:</em>

In that case, we really have the 3 - 6. That gives -3.

So we have 3^{-3}.

That equals \frac{1}{27}.

Thus, our final answer is $\boxed{\frac{1}{27} }.

Hope this helps!!! :D

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noname [10]

Answer:

8.51x10^{-3}cm

Step-by-step explanation:

In order to add or subtract numbers in scientific notation, the powers of ten must have the same exponent.  In this caes the diameter of an average human body cell is a power of -4 and the diameter of a plant cell is a power of -3.  In order to add them, we can move the decimal over for the body cell diameter:

5.1x10^{-4}=0.51x10^{-3}

Add the two values: 0.51x10^{-3}+8x10^{-3}=8.51x10^{-3}cm

7 0
3 years ago
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Answer the photo below thanks
MArishka [77]

Answer:

i think the answer is C

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Step-by-step explanation:

7 0
3 years ago
If you answer God will bless you:
VashaNatasha [74]

Answer:

600 m³

Step-by-step explanation:

Given:

  • length of swimming pool = 30 m
  • width of swimming pool = 10 m
  • depth at deep end = 3 m
  • depth at shallow end = 1 m

To find the volume of the water in the pool, find the area of a cross section (see attached image) and multiply it by the width of the pool.

<u>Area of cross section</u>

= area of rectangle - area of triangle

= (3 × 30) - [1/2 × 30 × (3 - 1)]

= 90 - 30

= 60 m²

<u>Volume of pool</u>

= area of cross section × width

= 60 × 10

= 600 m³

6 0
2 years ago
The equation of a parabola is 116(y+3)2=x+4 . What are the coordinates of the focus? (0, −3) (−8, −3) (−4, −7) (−4, 1)
Temka [501]
Did you get the answer yet
6 0
3 years ago
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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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