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Dvinal [7]
3 years ago
10

What is 1.2 to the tenth power

Mathematics
2 answers:
Anna [14]3 years ago
5 0

Answer:

Step-by-step explanation:

that would be written as  1.2^10, and the end result would be the same as you'd get if you use 1.2 as a factor 10 times:  1.2*1.2*1.2* .......1.2

You could use a calculator to evaluate 1.2^10:  

Typing in 1.2^10, you'll get  6.191736422.

This is not an exact answer; there are more digits following the ones shown.

Another way in which you could do this problem would be to use logs:

Let y = 1.2^10.  Then log y = 10*log 1.2, or log y = 10(0.07918) = 0.791812.

Finding the antilog, we get y = 6.19174

Leona [35]3 years ago
5 0
14 jk idkwhat it is!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
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3 years ago
A triangle has one angle that measures 23 degrees, one angle that measures 47 degrees, and one angle that measures 110 degrees.
blsea [12.9K]

Answer:

B) Scalene Triangle

Step-by-step explanation:

This must be a scalene triangle because:

If it were a right triangle, there would be a right angle so A is out

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4 0
2 years ago
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Svetradugi [14.3K]
Find the perimeter of the rectangle with points...
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5 0
3 years ago
sebuah topi berbentuk kerucut dengan panjang diameter 18 cm dan tinggi topi tersebut 20cm berapakah volume nya​
madreJ [45]

Answer:

The area of the cone is "1,695.6 cm³".

Step-by-step explanation:

The given values are:

Diameter of cone,

d = 18 cm

then,

Radius,

r = \frac{d}{2}

 = \frac{18}{2}

 = 9 \ cm

Height,

h = 20 cm

As we know,

⇒  Area of a cone = \frac{1}{3} \pi r^2 h

On substituting the given values in the above formula, we get

⇒                            = \frac{1}{3}\times 3.14\times (9)^2\times 20

⇒                            = \frac{1}{3}\times 3.14\times 81\times 20

⇒                            = 3.14\times 27\times 20

⇒                            = 1,695.6 \ cm^3

8 0
2 years ago
In to see the<br><br> 2x+y=20<br><br> 6x 5y = 12
Andre45 [30]

Answer:

1: x = 10 − \frac{y}{2}

y = 20 − 2 x

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y = \frac{25}{x}

Step-by-step explanation:

6 0
3 years ago
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