<span>So, L*W=A Because it is 4 cm longer, L=W+4 Because the area is 96, LW=96 Substitute to get W(W+4)=96 Multiply it out. W^2+4W-96=0 By solving the quadratic, W+12(W-8)=0 so either W+12 or W-8 is zero. The width must be positive, so the width is 8. Therefore the length is 12.
Hope this helps.</span>
Answer:
9) 844cm^2 10) 15,000in^3 11) 240cm^3 12) 480in^3
Step-by-step explanation:
9) A= 200 A=160 A=192 A=192 A=100
200+160= 360+192= 552+192= 744+100=844cm^2
for #9 i divided the figure into 5 parts and got the area of each one and added all of the totally areas to get the final answer
10) A= 20*30= 600*25= 15,000in^3
11) A= 2*5= 10*4= 40cm^2 (cheese) A= 40*6= 240cm^3
12) 12*10= 120*4= 480in^3
Answer:
3) 5/2
Step-by-step explanation:
2y = 5x + 4
y = 5/2x + 2
slope = 5/2
(y = mx + b, m being slope, b being y intercept)
The limit of a function f at a point a exists if:
![\lim_{x\to a^-}f(x)=\lim_{x\to a^+}f(x)](https://tex.z-dn.net/?f=%5Clim_%7Bx%5Cto%20a%5E-%7Df%28x%29%3D%5Clim_%7Bx%5Cto%20a%5E%2B%7Df%28x%29)
![\begin{gathered} \text{ From the given image, the value of the left limit of f at x=1 } \\ \lim_{x\to a^-}f(x)\lt1 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20%5Ctext%7B%20From%20the%20given%20image%2C%20the%20value%20of%20the%20left%20limit%20of%20f%20at%20x%3D1%20%7D%20%5C%5C%20%5Clim_%7Bx%5Cto%20a%5E-%7Df%28x%29%5Clt1%20%5Cend%7Bgathered%7D)
Also,
![\begin{gathered} \text{ From the given image, the value of the right limit of f at x=1 } \\ \lim_{x\to a^+}f(x)\gt1 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20%5Ctext%7B%20From%20the%20given%20image%2C%20the%20value%20of%20the%20right%20limit%20of%20f%20at%20x%3D1%20%7D%20%5C%5C%20%5Clim_%7Bx%5Cto%20a%5E%2B%7Df%28x%29%5Cgt1%20%5Cend%7Bgathered%7D)
Therefore,
![\lim_{x\to1^-}f(x)\lim_{x\to1^+}f(x)](https://tex.z-dn.net/?f=%5Clim_%7Bx%5Cto1%5E-%7Df%28x%29%5Clim_%7Bx%5Cto1%5E%2B%7Df%28x%29)
Hence, the limit does not exist