Subtract the portion used from the whole.
= 3 m - 5/6m
must have common denominators
= (3*6)/(1*6) - 5/6
= 18/6 - 5/6
= 13/6 m
= 2 1/6 m
ANSWER: 2 1/6 meters left
Hope this helps! :)
You can make 3 tens blocks and then 3 tens blocks but only two should be filled than for the third one fill 7 only
Answer: it increased by $403. He ended with $6603
Step-by-step explanation: since he started off with $6200, you would move the decimal over two places (.065) and then multiply 6200 and .065 to get $403.
Answer:
70.7
Step-by-step explanation:
I put it all together and got 70.7106781187 i cut it
to 70.7
Let the lengths of the east and west sides be x and the lengths of the north and south sides be y. the dimensions you want are therefore x and y.
The cost of the east and west fencing is $4*2*x; the cost of the north and south fencing is $2*2*y. We have to put in that "2" because there are 2 sides that run from east to west and 2 sides that run from north to south.
The total cost of all this fencing is $4(2)(x) + $2(2)(y) = $128. Let's reduce this by dividing all three terms by 4: 2x + y = 32.
Now we are to maximize the area of the vegetable patch, subject to the constraint that 2x + y = 32. The formula for area is A = L * W. Solving 2x + y = 32 for y, we get y = -2x + 32.
We can now eliminate y. The area of the patch is (x)(-2x+32) = A. We want to maximize A.
If you're in algebra, find the x-coordinate of the vertex of this quadratic equation. Remember the formula x = -b/(2a)? Once you have calculated this x, subst. your value into the formula for y: y= -2x + 32.
Now multiply together your x and y values to obtain the max area of the patch.
If you're in calculus, differentiate A = x(-2x+32) with respect to x and set the derivative equal to zero. This approach should give you the same x value as before; the corresponding y value will be the same; y=-2x+32.
Multiply x and y together. That'll give you the maximum possible area of the garden patch.
