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Kryger [21]
3 years ago
14

Identify the first step in solving the equation below. 2003-05-04-00-00_files/i0420000.jpg A. Subtract 2w from each side. B. Mul

tiply each side by w. C. Add 2w to each side. D. Divide each side by 2.
Mathematics
2 answers:
DanielleElmas [232]3 years ago
8 0
Could you rewrite the question please  thanks
inysia [295]3 years ago
4 0
Hi!
The answer to your problem is A)0.0310 and D)1.01
A significant figure is any number that is not a stand-alone zero which means any zero in the middle of a sequence of numbers or at the end of a sequence of numbers is counted as significant, for example:           these zeros are significant (1, 2, 3, and 4 is significant as well)                |  |0000123004 |  | | |these zeros are nonsignificant

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Can someone help me find y PLEASE
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If tan x =3/4, and 0
jasenka [17]

Answer: x=0.6435011

Step-by-step explanation:

Take the inverse tangent of both sides of the equation to extract  

x

from inside the tangent.

x

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(

3

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Evaluate  

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The tangent function is positive in the first and third quadrants. To find the second solution, add the reference angle from  

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Simplify the expression to find the second solution.

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x

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3.78509376

Find the period.

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π

The period of the  

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x

=

0.6435011

+

π

n

,

3.78509376

+

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n

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6 0
3 years ago
What is a three digit number thats an odd multiple of three and the product of its digits is 24 and is larger than 15 squared?
Reika [66]
The only way 3 digits can have product 24 is 
1 x 3 x 8 = 241 x 4 x 6 = 242 x 2 x 6 = 242 x 3 x 4 = 24
So the digits comprises of 1,3,8 or 1,4,6, or 2,2,6, or 2,3,4 
To be divisible by 3 the sum of the digits must be divisible by 3.
1+  3+ 8=12, 1+ 4+ 6= 11, 2 +2 + 6=10, 2 +3 + 4=9Of those sums of digits, only 12 and 9 are divisible by 3. 

So we have ruled out all but integers whose digits consist of1,3,8, and 2,3,4.

Meanwhile they must be odd they either must end in 1 or 3.
The only ones which can end in 1 are 381 and 831.

The others must end in 3. 

They must be greater than 152 which is 225. So the

First digit cannot be 1. So the only way its digits can contain of1,3,8 and close in 3 is to be 813.

The rest must contain of the digits 2,3,4, and the only way they can end in 3 is to be 243 or 423.  
So there are precisely five such three-digit integers: 381, 831, 813, 243, and 423.
7 0
3 years ago
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