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den301095 [7]
4 years ago
9

Will mark bianleast 10 ponits

Mathematics
1 answer:
Marrrta [24]4 years ago
7 0

Answer:

24 in

Step-by-step explanation:

12 sides times 2in = 24in

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Number 3 is a tenth 0.3
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3 years ago
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Is 16/3 rational or irrational?
Rudik [331]
16/3 is rational simply because it is a fraction made up of whole numbers.

Irrational numbers cannot be expressed as fractions made up of whole numbers.
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Dada la ecuacion 25x2 + 4y2 = 100, determina las coordenadas de los vertices, focos, las longitudes de los respectivos ejes mayo
Likurg_2 [28]

Answer:

The given equation is

25x^{2} +4y^{2}=100

Which represents an elipse.

To find its elements, we need to divide the equation by 100

\frac{25x^{2} +4y^{2} }{100} =\frac{100}{100} \\\frac{x^{2} }{4} +\frac{y^{2} }{25} =1

Where a^{2} =25 and b^{2}=4. Remember that the greatest denominator is a, and the least is b. So, we extract the square root on each equation.

a=5 and b=2.

In a elipse, we have a major axis and a minor axis. In this case, the major axis is vertical and the minor axis is horizontal, that means this is a vertical elipse.

The length of the major axis is 2a=2(5)=10.

The length of the minor axis is 2b=2(2)=4.

The vertices are (0,5);(0,-5) and (2,0);(-2,0).

Now, the main parameters of an elipse are related by

a^{2}=b^{2} +c^{2}, which we are gonna use to find c, the parameter of the focus.

c=\sqrt{a^{2}-b^{2} }=\sqrt{25-4}=\sqrt{21}

So, the coordinates of each focus are (0,\sqrt{21}) and (0,-\sqrt{21})

The eccentricity of a elipse is defined

e=\frac{c}{a}=\frac{\sqrt{21} }{5}  \approx 0.92

The latus rectum is defined

L=\frac{2b^{2} }{a}=\frac{2(4)}{5} =\frac{8}{5} \approx 1.6

Finally, the graph of the elipse is attached.

7 0
3 years ago
State the domain of the following relation by clicking on the symbols to make the given answer correct.
alekssr [168]

Answer:

  {x: x ∈ ℝ, x ≥ 0}

Step-by-step explanation:

The relation is only defined for non-negative values of x, so that is what the domain consists of: real numbers greater than or equal to zero.

6 0
3 years ago
Plz help!
beks73 [17]

First, you need to find the derivative of this function.  This is done by multiplying the exponent of the variable by the coefficient, and then reducing the exponent by 1.  

f'(x)=3x^2-3

Now, set this function equal to 0 to find x-values of the relative max and min.

0=3x^2-3

0=3(x^2-1)

0=3(x+1)(x-1)

x=-1, 1

To determine which is the max and which is the min, plug in values to f'(x) that are greater than and less than each.  We will use -2, 0, 2.

f'(-2)=3(-2)^2-3=3(4)-3=12-3=9

f'(0)=3(0)^2-3=3(0)-3=0-3=-3

f'(2)=3(2)^2=3(4)-3=12-3=9

We examine the sign changes to determine whether it is a max or a min.  If the sign goes from + to -, then it is a maximum.  If it goes from - to +, it is a minimum.  Therefore, x=-1 is a relative maximum and x=1 is a relative miminum.

To determine the values of the relative max and min, plug in the x-values to f(x).

f(-1)=(-1)^3-3(-1)+1=-1+3+1=3

f(1)=(1)^3-3(1)+1=1-3+1=-1

Hope this helps!!

7 0
3 years ago
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