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Triss [41]
3 years ago
7

What is significant digit rules and how is it used?

Mathematics
1 answer:
torisob [31]3 years ago
6 0

Answer:

Non-zero digits are always significant. Any zeros between two significant digits are significant. A final zero or trailing zeros in the decimal portion ONLY are significant.

Step-by-step explanation:

  • Non-zero digits are always significant.
  • Any zeros between two significant digits are significant.
  • A final zero or trailing zeros in the decimal portion ONLY are significant.

Heres a something that might help:

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You might be interested in
__________________________________________
xz_007 [3.2K]

Answer:

3.3 x 10⁻⁴

Step-by-step explanation:

Given expression:

           \frac{2.64 x 10^{-3} }{8.0 x 10^{0} }

 To solve this problem;

    10°   = 1

So;

             \frac{2.64 x 10^{-3} }{8.0}

    =  \frac{2.64}{8}  x 10⁻³

    = 0.33 x 10⁻³

    = 3.3 x 10⁻¹  x 10⁻³

     = 3.3 x 10⁻⁴

4 0
3 years ago
Myron got his paycheck of $40 this weck. He had $11 left after he spent m dollars on new video
solniwko [45]

Answer:

I dont know the answer but all I know is that dude broke as heck

6 0
3 years ago
Read 2 more answers
Please answer question three and four if you can :)<br> Show full working out ty;)
nikklg [1K]

Step-by-step explanation:

3

Let D be the mid point of side BC, [B(2, - 1), C(5, 2)].

Therefore, by mid-point formula:

D = ( \frac{2 + 5}{2},  \:  \:  \frac{ - 1 + 2}{2} ) = ( \frac{7}{2}, \:  \:  \frac{ 1}{2} ) \\ \therefore D= (3.5, \:  \: 0.5) \\  \& \: A=(-1,\:\:4)...(given) \\\\  now \: by \: distance \: formula \\  \\ Length  \: of \:  segment  \: AD \\  =  \sqrt{( - 1 - 3.5)^{2}  +  {(4 - 0.5)}^{2} }  \\ =  \sqrt{(4.5)^{2}  +  {(3.5)}^{2} }  \\ =  \sqrt{20.25 + 12.25 }  \\  =  \sqrt{32.5}  \\    \red{ \boxed{\therefore Length  \: of \:  segment  \: AD  = 5.7 \: units}}

4 (a)

Equation of line AB[A(2, 1), B(-2, - 11)] in two point form is given as:

\frac{y-y_1}{y_1-y_2} =\frac{x-x_1}{x_1 - x_2} \\\\\therefore \frac{y-1}{1-(-11)} =\frac{x-2}{2 - (-2) } \\\\\therefore \frac{y-1}{1+11} =\frac{x-2}{2 +2} \\\\\therefore \frac{y-1}{12} =\frac{x-2}{4} \\\\\therefore \frac{y-1}{3} =\frac{x-2}{1} \\\\\therefore y-1= 3(x - 2)\\\\\therefore y= 3x - 6+1\\\\\therefore y= 3x - 5\\\\ \huge \purple {\boxed {\therefore 3x - y-5=0}} \\

is the equation of line AB.

Now we have to check whether C(4, 7) lie on line AB or not.

Let us substitute x = 4 & y = 7 on the Left hand side of equation of line AB and if it gives us 0, then C lies on the line.

LHS = 3x - y-5\\=3\times 4-7-5\\= 12-12\\=0\\= RHS

Hence, point C (4, 7) lie on the straight line AB.

4(b)

Like we did in 4(a), first find the equation of line AB and then substitute the coordinates of point C in equation and if they satisfy the equation, then all the three points lie on the straight line.

4 0
4 years ago
I NEED HELP!! ASAP PLEASE!!!
mariarad [96]

Answer:

D

Step-by-step explanation:

a:angles given do not necessarily prove that ADP and BCP is 30 they can be different

c:angles given are not in triangles

b:with regards to the square I don't think you can really prove the statement

5 0
3 years ago
How do i solve this please help
Sveta_85 [38]

Answer:

48

Step-by-step explanation:

138-90(right angled triangle)=48

3 0
3 years ago
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