Answer:
6(8-5)
Step-by-step explanation:
Answer:
174
Step-by-step explanation:
Step-by-step explanation:
The dimensions of the cuboid are 5 cm, 7 cm and 9 cm.
The formula for the surface area of the cuboid is as follows :
![A=2(lb+bh+hl)](https://tex.z-dn.net/?f=A%3D2%28lb%2Bbh%2Bhl%29)
We have,
b = 9 cm, l = 7 cm and h = 5 cm
So,
![A=2(9\times 7+9\times 5+7\times 5)\\\\A=2\times 143\\\\A=286\ cm^2](https://tex.z-dn.net/?f=A%3D2%289%5Ctimes%207%2B9%5Ctimes%205%2B7%5Ctimes%205%29%5C%5C%5C%5CA%3D2%5Ctimes%20143%5C%5C%5C%5CA%3D286%5C%20cm%5E2)
Jamie forgets to multiply 143 by 2 as the formula for surface area contains 2 as well.
Answer:
Step-by-step explanation:
<h3>To prove quadrilateral is a square:</h3>
a) Slope of CB
C(-3,-1) ; B = (0,3)
![\sf \boxed{Slope = \dfrac{y_2-y_1}{x_2-x_1}}](https://tex.z-dn.net/?f=%5Csf%20%5Cboxed%7BSlope%20%3D%20%5Cdfrac%7By_2-y_1%7D%7Bx_2-x_1%7D%7D)
![\sf = \dfrac{3-[-1]}{0-[-3]}\\\\ =\dfrac{3+1}{0+3}\\\\ = \dfrac{4}{3}](https://tex.z-dn.net/?f=%5Csf%20%3D%20%5Cdfrac%7B3-%5B-1%5D%7D%7B0-%5B-3%5D%7D%5C%5C%5C%5C%20%3D%5Cdfrac%7B3%2B1%7D%7B0%2B3%7D%5C%5C%5C%5C%20%3D%20%5Cdfrac%7B4%7D%7B3%7D)
![\sf \text{\bf slope of CB = $\dfrac{4}{3}$}](https://tex.z-dn.net/?f=%5Csf%20%5Ctext%7B%5Cbf%20slope%20of%20CB%20%3D%20%24%5Cdfrac%7B4%7D%7B3%7D%24%7D)
b) D(1,-4) ; A(4,0)
![Slope \ of \ DA = \dfrac{0-[-4]}{4-1}\\](https://tex.z-dn.net/?f=Slope%20%5C%20of%20%5C%20DA%20%3D%20%5Cdfrac%7B0-%5B-4%5D%7D%7B4-1%7D%5C%5C)
![= \dfrac{0+4}{3}\\\](https://tex.z-dn.net/?f=%3D%20%5Cdfrac%7B0%2B4%7D%7B3%7D%5C%5C%5C)
![\sf \text{\bf Slope of DA = $\dfrac{4}{3}$}](https://tex.z-dn.net/?f=%5Csf%20%5Ctext%7B%5Cbf%20Slope%20of%20DA%20%3D%20%24%5Cdfrac%7B4%7D%7B3%7D%24%7D)
Slope of CB = slope of DA
c) C(-3,-1) ; D(1 , -4)
![\sf Slope \ of \ CD =\dfrac{-4-[-1]}{1-[-3]}](https://tex.z-dn.net/?f=%5Csf%20Slope%20%5C%20of%20%5C%20CD%20%3D%5Cdfrac%7B-4-%5B-1%5D%7D%7B1-%5B-3%5D%7D)
![\sf = \dfrac{-4+1}{1+3}\\\\ = \dfrac{-3}{4}\\](https://tex.z-dn.net/?f=%5Csf%20%3D%20%5Cdfrac%7B-4%2B1%7D%7B1%2B3%7D%5C%5C%5C%5C%20%20%20%20%3D%20%5Cdfrac%7B-3%7D%7B4%7D%5C%5C)
![\sf Slope \ of \ CD * Slope of CB = \dfrac{-3}{4}*\dfrac{4}{3}=-1](https://tex.z-dn.net/?f=%5Csf%20Slope%20%5C%20of%20%5C%20CD%20%2A%20Slope%20of%20CB%20%3D%20%20%5Cdfrac%7B-3%7D%7B4%7D%2A%5Cdfrac%7B4%7D%7B3%7D%3D-1)
So, CD is perpendicular to CB
d) B(0,3) ; D(1,-4)
![Slope \ of \ BD = \dfrac{-4-3}{1-0}\\\\=\dfrac{-7}{1}\\\\=-7](https://tex.z-dn.net/?f=Slope%20%5C%20of%20%5C%20BD%20%3D%20%5Cdfrac%7B-4-3%7D%7B1-0%7D%5C%5C%5C%5C%3D%5Cdfrac%7B-7%7D%7B1%7D%5C%5C%5C%5C%3D-7)
e) C(-3,-1) ; A(4,0)
![\sf Slope \ of \ CA = \dfrac{0-[-1]}{4-[-3]}\\](https://tex.z-dn.net/?f=%5Csf%20Slope%20%5C%20of%20%5C%20CA%20%3D%20%5Cdfrac%7B0-%5B-1%5D%7D%7B4-%5B-3%5D%7D%5C%5C)
![=\dfrac{0+1}{4+3}\\\\=\dfrac{1}{7}](https://tex.z-dn.net/?f=%3D%5Cdfrac%7B0%2B1%7D%7B4%2B3%7D%5C%5C%5C%5C%3D%5Cdfrac%7B1%7D%7B7%7D)
![\text{Slope of CA *Slope of BD = $\dfrac{1}{7}$*(-7)}=-1](https://tex.z-dn.net/?f=%5Ctext%7BSlope%20of%20CA%20%2ASlope%20of%20BD%20%3D%20%24%5Cdfrac%7B1%7D%7B7%7D%24%2A%28-7%29%7D%3D-1)
So, CA is perpendicular to BD
![\sf \text{\bf Slope of DA = \dfrac{4}{3}}](https://tex.z-dn.net/?f=%5Csf%20%5Ctext%7B%5Cbf%20Slope%20of%20DA%20%3D%20%5Cdfrac%7B4%7D%7B3%7D%7D)
Depends on how long she works and when she works.