Answer:
445788
Step-by-step explanation:
OMG too difficult man!!
Thanks to my calculator to solve this hard question
Answer:
4 sides
Step-by-step explanation:
A lot of math is about pattern matching. Here, you're asked if sides are the same length (one side matches another), if there are right angles (square corners), and if the number of angles or sides is 4.
You might make the observations ...
<u>pink</u>: 2 pairs of parallel sides that are the same length (4 sides). Adjacent sides are different length.
<u>orange</u>: all 4 sides are congruent and at right angles
<u>blue</u>: 2 pairs of parallel sides that are the same length (4 sides). Adjacent sides are different length.
<u>green</u>: 2 pairs of parallel sides that are the same length--4 sides and 4 right angles. Adjacent sides are different length.
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The only feature in common is "4 sides."
Answer:
Hi, I don't see an attachment with your question. But without any attachments to refer to. You use Pyrhagoras Theorem to solve this question.
Step-by-step explanation:
A^2 (Short side) + B^2 (Short side) = C^2 (Long side)
Stay safe and Merry Christmas! :)
Answer:
Banana = 6, the rest is in the picture
Step-by-step explanation:
One simple way to approach this is to subtract the number that DOESN’T have the variable with the solution by the sum.
Lets use problem 1 as the example:
1.) N+8=13
In this case, we subtract 13 by 8 and get 5, therefore N= 5.
2.) B+17=62
62-17= 45
B= 45
3.) X+5/6= 7/16
7/16-5/6= -19/48
X= -19/48
4.) y+1/7=5/8
5/8-1/7= 27/56
Y= 27/56
5.) 14+y=69
69-14=55
Y=55
6.) c+3.6=4.9
4.9-3.6=1.3
C= 1.3
Now for the word problems:
Lets use 7 as the example-
7.) a number increased by 19 is equal to 221.
Since we don’t know the number being increased by 19, that will be our variable. It will be re-written as so:
N + 19 = 221
Where N is the unknown number.
Using the same method of subtracting by the sum and the number that’s not the variable, we’ll get the solution to N
221-19= 202
N= 202
8.) N + 1 2/3 = 8
8 - 1 2/3 = 19/3= 6 1/3
N= 6 1/3
I really hope this helps
