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const2013 [10]
3 years ago
8

Symbol for taken away

Mathematics
2 answers:
daser333 [38]3 years ago
6 0

Hi there!


In math terms, when we think of the words "Taken away", we immediately think of subtraction/subtracting. So, the symbol that you're looking for looks sort of like a dash.


- is what it looks like.


Hope this helps!

Message me if you need anything else! I'd be happy to help you! :D

Ganezh [65]3 years ago
3 0
The symbol you are looking for is called the Subtracting sign. Here is what it looks like in an attachment.
Here are words for what it is called:
Subtract
Minus

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Mathematical induction, prove the following two statements are true
adelina 88 [10]
Prove:
1+2\left(\frac12\right)+3\left(\frac12\right)^{2}+...+n\left(\frac12\right)^{n-1}=4-\dfrac{n+2}{2^{n-1}}
____________________________________________

Base Step: For n=1:
n\left(\frac12\right)^{n-1}=1\left(\frac12\right)^{0}=1
and
4-\dfrac{n+2}{2^{n-1}}=4-3=1
--------------------------------------------------------------------------

Induction Hypothesis: Assume true for n=k. Meaning:
1+2\left(\frac12\right)+3\left(\frac12\right)^{2}+...+k\left(\frac12\right)^{k-1}=4-\dfrac{k+2}{2^{k-1}}
assumed to be true.

--------------------------------------------------------------------------

Induction Step: For n=k+1:
1+2\left(\frac12\right)+3\left(\frac12\right)^{2}+...+k\left(\frac12\right)^{k-1}+(k+1)\left(\frac12\right)^{k}

by our Induction Hypothesis, we can replace every term in this summation (except the last term) with the right hand side of our assumption.
=4-\dfrac{k+2}{2^{k-1}}+(k+1)\left(\frac12\right)^{k}

From here, think about what you are trying to end up with.
For n=k+1, we WANT the formula to look like this:
1+2\left(\frac12\right)+...+k\left(\frac12\right)^{k-1}+(k+1)\left(\frac12\right)^{k}=4-\dfrac{(k+1)+2}{2^{(k+1)-1}}

That thing on the right hand side is what we're trying to end up with. So we need to do some clever Algebra.

Combine the (k+1) and 1/2, put the 2 in the bottom,
=4-\dfrac{k+2}{2^{k-1}}+\dfrac{(k+1)}{2^{k}}

We want to end up with a 2^k as our final denominator, so our middle term is missing a power of 2. Let's multiply top and bottom by 2,
=4+\dfrac{-2(k+2)}{2^{k}}+\dfrac{(k+1)}{2^{k}}

Distribute the -2 and combine the fractions together,
=4+\dfrac{-2k-4+(k+1)}{2^{k}}

Combine like-terms,
=4+\dfrac{-k-3}{2^{k}}

pull the negative back out,
=4-\dfrac{k+3}{2^{k}}

And ta-da! We've done it!
We can break apart the +3 into +1 and +2,
and the +0 in the bottom can be written as -1 and +1,
=4-\dfrac{(k+1)+2}{2^{(k-1)+1}}
3 0
3 years ago
How to make 9.37% into decimal​
Sophie [7]

Answer: divide by 100 is .0937

Step-by-step explanation:

4 0
3 years ago
Need HELP ASAP/ Cant get this wrong!!
Pavlova-9 [17]

Answer:

B

Step-by-step explanation:

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6 0
3 years ago
27.) Find the sum of the length of line segments EF and CD
Lelechka [254]

Answer:

5√10

Step-by-step explanation:

The distance between two points A(x_1,y_1) and B(x_2,y_2) on the coordinate plane is given by:

|AB|=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2 }

From the image attached the coordinates of the points are E(-1, -1), F(8, -4), C(-7, -4) and D(-1, 2).

Hence the lengths of the line segment EF and CD are:

|EF|=\sqrt{(8-(-1))^2+(-4-(-1))^2} =\sqrt{90}=3\sqrt{10} \\\\CD=\sqrt{(-1-(-7))^2+(-2-(-4))^2}=\sqrt{40}  =2\sqrt{10} \\\\Therefore:\\\\|EF|+|CD|=3\sqrt{10} +2\sqrt{10} =5\sqrt{10}

6 0
3 years ago
URGENT!<br><br> need help<br><br><br> ...
GenaCL600 [577]

Answer:

92°

Step-by-step explanation:

If ΔABC ≈ ΔPQR then ∡A = ∡P

∡P is 92° so ∡A is 92°

8 0
3 years ago
Read 2 more answers
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