Answer:
![\left[\begin{array}{c}-\frac{8}{\sqrt{117} } \\\frac{7}{\sqrt{117} }\\\frac{2}{\sqrt{117} }\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D-%5Cfrac%7B8%7D%7B%5Csqrt%7B117%7D%20%7D%20%5C%5C%5Cfrac%7B7%7D%7B%5Csqrt%7B117%7D%20%7D%5C%5C%5Cfrac%7B2%7D%7B%5Csqrt%7B117%7D%20%7D%5Cend%7Barray%7D%5Cright%5D)
Step-by-step explanation:
We are required to find a unit vector in the direction of:
![\left[\begin{array}{c}-8\\7\\2\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D-8%5C%5C7%5C%5C2%5Cend%7Barray%7D%5Cright%5D)
Unit Vector, ![\hat{a}=\dfrac{\overrightarrow{a}}{|\overrightarrow{a}|}](https://tex.z-dn.net/?f=%5Chat%7Ba%7D%3D%5Cdfrac%7B%5Coverrightarrow%7Ba%7D%7D%7B%7C%5Coverrightarrow%7Ba%7D%7C%7D)
The Modulus of
=![\sqrt{(-8)^2+7^2+(-2)^2}=\sqrt{117}](https://tex.z-dn.net/?f=%5Csqrt%7B%28-8%29%5E2%2B7%5E2%2B%28-2%29%5E2%7D%3D%5Csqrt%7B117%7D)
Therefore, the unit vector of the matrix is given as:
![\left[\begin{array}{c}-\frac{8}{\sqrt{117} } \\\frac{7}{\sqrt{117} }\\\frac{2}{\sqrt{117} }\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D-%5Cfrac%7B8%7D%7B%5Csqrt%7B117%7D%20%7D%20%5C%5C%5Cfrac%7B7%7D%7B%5Csqrt%7B117%7D%20%7D%5C%5C%5Cfrac%7B2%7D%7B%5Csqrt%7B117%7D%20%7D%5Cend%7Barray%7D%5Cright%5D)
Answer:
Im sosorry my internet is so poor sorry.
Answer:
Step-by-step explanation:
-3(3/6) + 4(3/5)
-3(1/2) + 4(3/5)
-3/2 + 12/5
-15/10 + 24/10
9/10
We are given
Jim's backyard:
Length is
![L=15\frac{5}{6} =\frac{95}{6} yards](https://tex.z-dn.net/?f=L%3D15%5Cfrac%7B5%7D%7B6%7D%20%3D%5Cfrac%7B95%7D%7B6%7D%20yards)
width is
![W=10\frac{2}{5} =\frac{52}{5} yards](https://tex.z-dn.net/?f=W%3D10%5Cfrac%7B2%7D%7B5%7D%20%3D%5Cfrac%7B52%7D%7B5%7D%20yards)
Since, this is rectangle
so, we can find area of rectangle
![A=L*W](https://tex.z-dn.net/?f=A%3DL%2AW)
![A=(\frac{95}{6})*(\frac{52}{5})](https://tex.z-dn.net/?f=A%3D%28%5Cfrac%7B95%7D%7B6%7D%29%2A%28%5Cfrac%7B52%7D%7B5%7D%29)
![A=\frac{494}{3} yard^2](https://tex.z-dn.net/?f=A%3D%5Cfrac%7B494%7D%7B3%7D%20yard%5E2)
Area of one sod:
length is
![l=1\frac{1}{3} =\frac{4}{3} yards](https://tex.z-dn.net/?f=l%3D1%5Cfrac%7B1%7D%7B3%7D%20%3D%5Cfrac%7B4%7D%7B3%7D%20yards)
width is
![w=1\frac{1}{3} =\frac{4}{3} yards](https://tex.z-dn.net/?f=w%3D1%5Cfrac%7B1%7D%7B3%7D%20%3D%5Cfrac%7B4%7D%7B3%7D%20yards)
Since, it is rectangle in shape
so,
![Area=l*w](https://tex.z-dn.net/?f=Area%3Dl%2Aw)
![A=\frac{4}{3}*\frac{4}{3}](https://tex.z-dn.net/?f=A%3D%5Cfrac%7B4%7D%7B3%7D%2A%5Cfrac%7B4%7D%7B3%7D)
![A=\frac{16}{9}yard^2](https://tex.z-dn.net/?f=A%3D%5Cfrac%7B16%7D%7B9%7Dyard%5E2)
Number of pieces of sod:
we can use formula
Number of pieces of sod = (area of Jim's backyard)/(area of one sod)
![N=\frac{\frac{494}{3} }{\frac{16}{9} }](https://tex.z-dn.net/?f=N%3D%5Cfrac%7B%5Cfrac%7B494%7D%7B3%7D%20%7D%7B%5Cfrac%7B16%7D%7B9%7D%20%7D)
now, we can simplify it
pieces need ..............Answer