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Aleks04 [339]
4 years ago
6

Why is it important to use estimation when solving problems

Mathematics
1 answer:
stellarik [79]4 years ago
5 0
Well, it might not be important for most problems but..
If you're dividing, and the answer is a repeat..
You can estimate.

Example - 2.678 | You can estimate it to 2.68 and use that as an answer instead of a very long answer that's repeated.

I hope this helps.
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Find the remainder when 3x^3+ 6x-1 is divided by x-3​
Sauron [17]

Answer:

93

Step-by-step explanation:

by remainder theorem p(3) is the remainder

and therefore it is 93

3 0
3 years ago
Express in exponential form log(2)16 = 4
S_A_V [24]
\bf \textit{exponential form of a logarithm}
\\\\
log_a  b=y \implies   a^y=  b\qquad\qquad 
%  exponential notation 2nd form
a^y=  b\implies log_a  b=y \\\\
-------------------------------\\\\
log_2(16)=4\qquad \qquad \implies \qquad 2^4=16
8 0
4 years ago
Please help,I wasn't there when the teacher taught this:(
nordsb [41]
The distance she rides increases by 1/4 every 1 minute.
5 0
4 years ago
Divide. and simplify 4/7÷8/3
Xelga [282]
(4/7)/(8/3)=(4/7)*(3/8)=3/14
5 0
3 years ago
Not sure how to figure out how to get b2 by itself..can anyone help me out?
Arte-miy333 [17]

As with any "solve for ..." problem, you start by looking at the operations performed on the variable of interest. Here, when you evaluate this expression according to the order of operations, you

  1. add b1
  2. multiply by (h/2)

When you want to solve for b2, you undo these operations in reverse order. To undo multiplication by a fraction, you multiply by the inverse (reciprocal) of the fraction. To undo addition, you add the opposite.

Whatever you do must be done to both sides of the equation.

Here we go ...

... (2/h)A = b1 + b2 . . . . . we undid the multiply by h/2, by multiplying by 2/h

... (2/h)A - b1 = b2 . . . . . we undid the addition of b1 by adding the opposite of b1

Then your solution is

... b2 = 2A/h - b1

If you want to, you can combine these terms over a single denominator to get

... b2 = (2A -h·b1)/h

8 0
4 years ago
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