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IrinaVladis [17]
4 years ago
11

Whats 1 plus 4555555555555

Mathematics
2 answers:
ELEN [110]4 years ago
5 0
<span>4.5555556e+12   is the answer</span>
Llana [10]4 years ago
3 0
The answer would be 4555555555556
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-6 and 3 what is the midpoint?<br>​
miskamm [114]

Answer:

-1.5

Step-by-step explanation:

As you can see, to calculate the midpoint of 2 numbers, we sum them together and then divide by 2, we got:

(-6 + 3) / 2

= -3 / 2

= -1.5

Hope this help you :3

5 0
4 years ago
Suppose the diameter of a circle is 16 units. What is its circumference?
igomit [66]

Answer:

50.27

Step-by-step explanation:

7 0
4 years ago
Read 2 more answers
6n-1=-5+5n<br><img src="https://tex.z-dn.net/?f=6n%20%20-%201%20%3D%20%20-%205%20%2B%206" id="TexFormula1" title="6n - 1 = - 5
Liula [17]

6n - 1 = -5 + 5n

+1

6n = -4 + 5n

-5n

n = -4


6n - 1 = -5 + 6

+1

6n = -5 + 6 + 1

6n = 2

/6

n = 0.333

7 0
4 years ago
AB=diameter of the circle O. OC=radius. Arc AXB is an arc of the circle with centre C. Prove areas of the shaded region are =
alexandr1967 [171]
1.
Draw a circle with center C And radius CA, as shown in the attached picture.

Let the lengths of radii AO, OB, OC be R. Triangle ABC is inscribed in the circle with center O and one of its sides is a diameter, this means that the angle ACB is a right angle.
|AO|=|OC|=R, by the Pythagorean theorem |AC|=\sqrt{2}R.

these are all shown in the picture.

2.

Area of triangle ABC is 1/2 * 2R * R= R^2

3.

Let the area between arc BXA and chord AB be Y. (the yellow region).

and let G be the shaded region between arcs AB and AXB.

G=1/2(Area circle with center O)-Y
   =\frac{1}{2} \pi R^{2}-Y

To find Y:

Notice that the area of the sector ACB is 1/4 of the area of circle with center C, since m(ACB) is 1/4 of 360 degrees.

So Area of sector ACB = \frac{1}{4} \pi (\sqrt{2} R)^{2}=\frac{1}{4} \pi*2 R^{2} =\frac{1}{2} \pi R^{2}

Y =area of sector ABC-Area(triangle ABC)=\frac{1}{2} \pi R^{2}- \frac{1}{2}*2R*R=\frac{1}{2} \pi R^{2}- R^{2}


4. 

Finally,

G=\frac{1}{2} \pi R^{2}-Y=\frac{1}{2} \pi R^{2}-(\frac{1}{2} \pi R^{2}- R^{2})=R^{2}

This proves that the 2 shaded regions have equal area.
 

5 0
3 years ago
Solve the equation 18t - 8 = 1 for t
bulgar [2K]

Answer:

T=1/2

Step-by-step explanation:

18t-8=1

18t=9

T=9/18

T=1/2

6 0
4 years ago
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