earthworms he can have if his square of compost has a side length that is 3 times longer .
<u>Step-by-step explanation:</u>
Here we have , Hiram raises earthworms. in a square of compost 4ft by 4ft he can have 1000 earthworms. We need to find how many earthworms can he have if his square of compost has a side length that is 3 times longer .
Area of square compost =
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In this , square compost with can have 1000 earthworms ! if side is 3 times longer i.e. New side = , So , new area will be :
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Now , can have 1000 earthworms ,So will have :
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Therefore, earthworms he can have if his square of compost has a side length that is 3 times longer .
B.)M is the correct answer I thons , because mostly we use capital letters for naming points.
Answer:
Length = 5
Step-by-step explanation:
width = 2
length =5
because 5 x 2 is 10. area is length times width
Answer:
City A
Step-by-step explanation:
200 + 45 = 245
100 + 20 = 120
120 + 110 = 230
50 + 55 = 110
City A has the greatest tendency.
6x+15y=-45
<span>6x+5y=-35
subtract second equation from the first
</span>6x+15y=-45
<span>6x+5y=-35
</span><u>- - +
</u> 10y = -10
⇒ y = (-10/10) = 1
put value of y in one of the equation,
6x+5y = -35
⇒ 6x + 5(-1) = -35
⇒ 6x - 5 =-35
⇒ 6x = -35 + 5 = 30
⇒ x = 30/6
⇒ x = 5
So x=5 and y=-1.