The required model is attached in the image and the division of 3/4 by 3 /12 is 3.
Given that,
To draw the model to represent 3/4 divided by 3/12.
In mathematics, it deals with numbers of operations according to the statements.
Model hs been attached below for the given problem of division,
= 3 * 4 / 3 * 12
For a rectangle space, drawing 4 column and 12 raws,
Now for 3 / 4 3 of the 4 column to be filled by red, for 3 / 12 every 3 cells in the row have been filled by red.
Calculating number of cells in 3 / 4 = 36
calculating number of cells in 3 / 12 = 12
Now ratio of the cells = 36 / 12 = 3
Thus, the required model is attached in the image and the division of 3/4 by 3 /12 is 3.
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The answer you are looking for is 1/3.
First, you must look at the denominators of each fraction, to see if you can simply divide straight across. In this case, you cannot, which means you must multiply by the reciprocal or "flipped" number. Generally when you multiply by the reciprocal, you flip the number with the smaller denominator to multiply.
So, you have 4/3 * 3/12. You then multiply straight across (4 * 3)/(3 * 12), to get an answer of 12/36, which simplified (divided by 3/3) gives you a final answer of 1/3.
I hope this helps!
Answer:
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Step-by-step explanation:
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To find the instantaneous rate of change of the function f(x,y) = x^2 + ln(y) at (3,1) to (1,2), we can use the partial derivatives with respect to x and y:
fx(x,y) = 2x
fy(x,y) = 1/y
Then, we can use the gradient vector to find the direction of maximum increase:
∇f(x,y) = <fx(x,y), fy(x,y)> = <2x, 1/y>
At point (3,1), the gradient vector is:
∇f(3,1) = <6, 1>
At point (1,2), the gradient vector is:
∇f(1,2) = <2, 1/2>
To find the instantaneous rate of change from (3,1) to (1,2), we can use the formula for directional derivative:
Dv(f) = ∇f(x,y) · v
where v is the unit vector in the direction from (3,1) to (1,2). The direction vector v is given by:
v = <1, 2> - <3, 1> = <-2, 1>
To make v a unit vector, we need to normalize it by dividing it by its length:
|v| = sqrt((-2)^2 + 1^2) = sqrt(5)
u = v/|v| = <-2/sqrt(5), 1/sqrt(5)>
Then, the instantaneous rate of change from (3,1) to (1,2) is:
Dv(f) = ∇f(3,1) · u = <6, 1> · <-2/sqrt(5), 1/sqrt(5)> = (-12/sqrt(5)) + (1/sqrt(5)) = -11/sqrt(5)
Therefore, the instantaneous rate of change of the function f(x,y) = x^2 + ln(y) from (3,1) to (1,2) is -11/sqrt(5).
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