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Vaselesa [24]
4 years ago
9

Which is the correct formula to calculate the volume of a sphere?

Mathematics
1 answer:
slega [8]4 years ago
3 0
The volume for a sphere is 4/3Pi(r^3), or
4/3 • 3.14 (Pi estimated) • (r • r • r).

Pi is 3.14 estimated, and r is the radius (from a curve to the midpoint of a circle).

I hope this helps!
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Need help with this!!!!
antoniya [11.8K]

Answer:

Cuatro es la respuesta también conocida como d

6 0
3 years ago
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Find the sum of the geometric sequence. 1, 1/2, 1/4, 1/8, 1/16
dezoksy [38]
I would convert them all to the denominator 16, due to the smallest fraction and all of them can be converted to it.

1 = 16/16
1/2 = 8/16
1/4 = 4/16
1/8 = 2/16
1/16 = 1/16

1+ 2 + 4 + 8 + 16 = 31

31/16 = 1 15/16 

Hope this helps!
7 0
3 years ago
Read 2 more answers
The midpoint of AB = ([?],[ ])
3241004551 [841]

Answer:

(0,0)

Step-by-step explanation:

x_{m}=\frac{-2+2}{2}=0\\y_{m}=\frac{-2+2}{2}=0\\

4 0
3 years ago
There is a line that includes the point (2, 10) and has a slope of 1. What is its equation in
rjkz [21]

Answer:

The line is y=1x+8

Step-by-step explanation:

We can use point-slope form of a line to make an equation. Point-slope form is as follows:

y-y₁=m(x-x₁)

**The variables y₁ and x₁ are where we will plug in our given coordinates and the variable m is your slope.

1. Plug in the given info:

y-10=1(x-2)

2. Distribute the 1:

y-10=1x-2

3. Add 10 to both sides:

y=1x+8

5 0
3 years ago
Which of the following is a solution to 2cos2x − cos x − 1 = 0?
Pavel [41]

Answer:

Option A is correct.

Solution for the given equation is, x = 0^{\circ}

Step-by-step explanation:

Given that : 2\cos^2x -\cos x -1 =0

Let \cos x =y

then our equation become;

2y^2-y-1= 0           .....[1]

A quadratic equation is of the form:

ax^2+bx+c =0.....[2] where a, b and c are coefficient and the solution is given by;

x = \frac{-b\pm \sqrt{b^2-4ac}}{2a}

Comparing equation [1] and [2] we get;

a = 2 b = -1 and c =-1

then;

y = \frac{-(-1)\pm \sqrt{(-1)^2-4(2)(-1)}}{2(2)}

Simplify:

y = \frac{ 1 \pm \sqrt{1+8}}{4}

or

y = \frac{ 1 \pm \sqrt{9}}{4}

y = \frac{ 1 \pm 3}{4}

or

y = \frac{1+3}{4} and y = \frac{1 -3}{4}

Simplify:

y = 1 and y = -\frac{1}{2}

Substitute y = cos x we have;

\cos x = 1

⇒x = 0^{\circ}

and

\cos x = -\frac{1}{2}

⇒ x = 120^{\circ} \text{and} x = 240^{\circ}

The solution set:  \{0^{\circ}, 120^{\circ} , 240^{\circ}\}

Therefore, the solution for the given equation  2\cos^2x -\cos x -1 =0 is, 0^{\circ}





8 0
4 years ago
Read 2 more answers
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