Draw the tape diagram: a long rectangle and divide it into 4 squares in a row.The whole rectangle represents 24 (as a whole number), you're dividing it into 4 parts because you want to know what is 1/4 of it.Now, to find the value of each square: 24:4 = 6.Put the number 6 in each square, and you have found your result.1/4 of 24 is 6.
Gross=amount earned not minusing tax
net=amount earned minus tax
gross=558
net=423.65
deduction=gross-net
deduction=558-423.65=134.35
deduction=$134.35
=4m - 17+ 3m - 11
combine like terms
=(4m + 3m) + (-17 - 11)
=7m - 28
This can be factored/simplified by dividing everything by 7
=7(m - 4)
Hope this helps! :)
The function is L = 10m + 50
Here, we want to find out which of the functions is required to determine the number of lunches L prepared after m minutes
In the question, we already had 50 lunches prepared
We also know that he prepares 10 lunches in one minute
So after A-lunch begins, the number of lunches prepared will be 10 * m = 10m
Adding this to the 50 on ground, then we have the total L lunches
Mathematically, that would be;
L = 10m + 50
Answer: ![3x^2y\sqrt[3]{y}\\\\](https://tex.z-dn.net/?f=3x%5E2y%5Csqrt%5B3%5D%7By%7D%5C%5C%5C%5C)
Work Shown:
![\sqrt[3]{27x^{6}y^{4}}\\\\\sqrt[3]{3^3x^{3+3}y^{3+1}}\\\\\sqrt[3]{3^3x^{3}*x^{3}*y^{3}*y^{1}}\\\\\sqrt[3]{3^3x^{2*3}*y^{3}*y}\\\\\sqrt[3]{\left(3x^2y\right)^3*y}\\\\\sqrt[3]{\left(3x^2y\right)^3}*\sqrt[3]{y}\\\\3x^2y\sqrt[3]{y}\\\\](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B27x%5E%7B6%7Dy%5E%7B4%7D%7D%5C%5C%5C%5C%5Csqrt%5B3%5D%7B3%5E3x%5E%7B3%2B3%7Dy%5E%7B3%2B1%7D%7D%5C%5C%5C%5C%5Csqrt%5B3%5D%7B3%5E3x%5E%7B3%7D%2Ax%5E%7B3%7D%2Ay%5E%7B3%7D%2Ay%5E%7B1%7D%7D%5C%5C%5C%5C%5Csqrt%5B3%5D%7B3%5E3x%5E%7B2%2A3%7D%2Ay%5E%7B3%7D%2Ay%7D%5C%5C%5C%5C%5Csqrt%5B3%5D%7B%5Cleft%283x%5E2y%5Cright%29%5E3%2Ay%7D%5C%5C%5C%5C%5Csqrt%5B3%5D%7B%5Cleft%283x%5E2y%5Cright%29%5E3%7D%2A%5Csqrt%5B3%5D%7By%7D%5C%5C%5C%5C3x%5E2y%5Csqrt%5B3%5D%7By%7D%5C%5C%5C%5C)
Explanation:
As the steps above show, the goal is to factor the expression under the root in terms of pulling out cubed terms. That way when we apply the cube root to them, the exponents cancel. We cannot factor the y term completely, so we have a bit of leftovers.
