To see your work and be able to check if you have messed up on anything
Step-by-step explanation:
Here
BC=P=10
AC=B=b
AB=H=15
then using fourmula
h^2= p^2+ b^2
b^2= 15×15 - 10× 10
b^2= 125
b=11.2
Answer:
it would be everything multiplied by 10
so
500
280
200
Step-by-step explanation:
Perimeter of rectangle = length + length + width + width
To find the combinations, think of two numbers that each multiplied by 2 and added up to give 12 or 14
Rectangle with perimeter 12
Say we take length = 2 and width = 3
Multiply the length by 2 = 2 × 2 = 4
Multiply the width by 3 = 2 × 3 = 6
Then add the answers = 4 + 6 = 10
This doesn't give us perimeter of 12 so we can't have the combination of length = 2 and width = 3
Take length = 4 and width = 2
Perimeter = 4+4+2+2 = 12
This is the first combination we can have
Take length = 5 and width = 1
Perimeter = 5+5+1+1 = 12
This is the second combination we can have
The question doesn't specify whether or not we are limited to use only integers, but if it is, we can only have two combinations of length and width that give perimeter of 12
length = 4 and width = 2
length = 5 and width = 1
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Rectangle with perimeter of 14
Length = 4 and width = 3
Perimeter = 4+4+3+3 = 14
Length = 5 and width = 2
Perimeter = 5+5+2+2 = 14
Length = 6 and width = 1
Perimeter = 6+6+1+1 = 14
We can have 3 different combinations of length and width
