Answer:
perimeter =22m
area=28square meter
perimeter of rectangle=4l
22m=4l
22÷4=l
5.5=l
l=5.5m
area of rectangle =l×b
28=5.5×b
28÷5.5=b
5.09=b
b=5.09m
length =5.5m and breadth =5.09m answer !!!!
so hmmm seemingly the graphs meet at -2 and +2 and 0, let's check

so f(x) = g(x) at those points, so let's take the integral of the top - bottom functions for both intervals, namely f(x) - g(x) from -2 to 0 and g(x) - f(x) from 0 to +2.
![\stackrel{f(x)}{2x^3-x^2-5x}~~ - ~~[\stackrel{g(x)}{-x^2+3x}]\implies 2x^3-x^2-5x+x^2-3x \\\\\\ 2x^3-8x\implies 2(x^3-4x)\implies \displaystyle 2\int\limits_{-2}^{0} (x^3-4x)dx \implies 2\left[ \cfrac{x^4}{4}-2x^2 \right]_{-2}^{0}\implies \boxed{8} \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Cstackrel%7Bf%28x%29%7D%7B2x%5E3-x%5E2-5x%7D~~%20-%20~~%5B%5Cstackrel%7Bg%28x%29%7D%7B-x%5E2%2B3x%7D%5D%5Cimplies%202x%5E3-x%5E2-5x%2Bx%5E2-3x%20%5C%5C%5C%5C%5C%5C%202x%5E3-8x%5Cimplies%202%28x%5E3-4x%29%5Cimplies%20%5Cdisplaystyle%202%5Cint%5Climits_%7B-2%7D%5E%7B0%7D%20%28x%5E3-4x%29dx%20%5Cimplies%202%5Cleft%5B%20%5Ccfrac%7Bx%5E4%7D%7B4%7D-2x%5E2%20%5Cright%5D_%7B-2%7D%5E%7B0%7D%5Cimplies%20%5Cboxed%7B8%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
![\stackrel{g(x)}{-x^2+3x}~~ - ~~[\stackrel{f(x)}{2x^3-x^2-5x}]\implies -x^2+3x-2x^3+x^2+5x \\\\\\ -2x^3+8x\implies 2(-x^3+4x) \\\\\\ \displaystyle 2\int\limits_{0}^{2} (-x^3+4x)dx \implies 2\left[ -\cfrac{x^4}{4}+2x^2 \right]_{0}^{2}\implies \boxed{8} ~\hfill \boxed{\stackrel{\textit{total area}}{8~~ + ~~8~~ = ~~16}}](https://tex.z-dn.net/?f=%5Cstackrel%7Bg%28x%29%7D%7B-x%5E2%2B3x%7D~~%20-%20~~%5B%5Cstackrel%7Bf%28x%29%7D%7B2x%5E3-x%5E2-5x%7D%5D%5Cimplies%20-x%5E2%2B3x-2x%5E3%2Bx%5E2%2B5x%20%5C%5C%5C%5C%5C%5C%20-2x%5E3%2B8x%5Cimplies%202%28-x%5E3%2B4x%29%20%5C%5C%5C%5C%5C%5C%20%5Cdisplaystyle%202%5Cint%5Climits_%7B0%7D%5E%7B2%7D%20%28-x%5E3%2B4x%29dx%20%5Cimplies%202%5Cleft%5B%20-%5Ccfrac%7Bx%5E4%7D%7B4%7D%2B2x%5E2%20%5Cright%5D_%7B0%7D%5E%7B2%7D%5Cimplies%20%5Cboxed%7B8%7D%20~%5Chfill%20%5Cboxed%7B%5Cstackrel%7B%5Ctextit%7Btotal%20area%7D%7D%7B8~~%20%2B%20~~8~~%20%3D%20~~16%7D%7D)
Answer:
161 cm
Step-by-step explanation:
Since 115 cm is 5 boxes, we take 115 and divide it by 5. This will get us the cm of 1 box. 115/5=23
Now we take 23 and multiply it by 7 to get the final height. 23x7=161
Answer:
cement by 5 Kg
Step-by-step explanation:
add the parts of the ratio, 5 + 4 + 1 = 10 parts
Divide the quantity of concrete by 10 to find the value of one part of the ratio.
110Kg ÷ 10 = 11Kg ← value of 1 part of ratio , thus
5 parts = 5 × 11Kg = 55Kg ← cement required
4 parts = 4 × 11Kg = 44Kg ← sand required
1 part = 11Kg ← gravel required
He requires 55Kg of concrete but only has 50Kg.
He requires 44Kg of sand and has 55Kg
He requires 11Kg of gravel and has 15Kg
Thus he is 5Kg short of cement.
Answer:
56
Step-by-step explanation: