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  1. Asked: December 20, 2020In:

    a square tile measures 20cm by 20cm. How many of such tiles will cover a floor measuring 5m by 4m?

    Odinaka_O
    Odinaka_O Tyro

    floor area= $5\times 4=20m^{2}$ square tile area=$0.2\times 0.2=0.04m^{2}$ number of tiles that will cover the floor=$\dfrac{20}{0.04}=500$ tiles

    floor area= 5\times 4=20m^{2}
    square tile area=0.2\times 0.2=0.04m^{2}
    number of tiles that will cover the floor=\dfrac{20}{0.04}=500 tiles

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  2. Asked: December 20, 2020In:

    The angle of a sector of a circle of radius 4.2cm is 150^{0}. if \Pi =\dfrac{22}{7}, what is the perimeter of the sector?

    Odinaka_O
    Odinaka_O Tyro

    Perimeter of sector=$2r + Length of arc$ Perimeter of sector=$2r + \dfrac{\theta }{360}\times 2\Pi r$ $\theta =150^{0},r=4.2cm, \Pi =\dfrac{22}{7}$ $\Rightarrow P=2\times 4.2+\dfrac{150}{360}\times 2\times \dfrac{22}{7} \times 4.2$ $\Rightarrow P=8.4 + 0.417\times 2 \times 3.142\times 4.2$ $\RightarRead more

    Perimeter of sector=2r + Length of arc
    Perimeter of sector=2r + \dfrac{\theta }{360}\times 2\Pi r
    \theta =150^{0},r=4.2cm, \Pi =\dfrac{22}{7}
    \Rightarrow P=2\times 4.2+\dfrac{150}{360}\times 2\times \dfrac{22}{7} \times 4.2
    \Rightarrow P=8.4 + 0.417\times 2 \times 3.142\times 4.2
    \Rightarrow P=8.4 + 11.0058
    \therefore P=19.4058cm

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  3. Asked: December 21, 2020In:

    Solve \begin{vmatrix} 8 & 7\\ 16 & 11 \end{vmatrix}

    Odinaka_O
    Odinaka_O Tyro

    $\begin{vmatrix} 8 & 7\\ 16 & 11 \end{vmatrix}$ $\Rightarrow (8\times 11)-(7\times 16)$ $\Rightarrow 88-112=-24$

    \begin{vmatrix} 8 & 7\\ 16 & 11 \end{vmatrix}
    \Rightarrow (8\times 11)-(7\times 16)
    \Rightarrow 88-112=-24

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  4. Asked: December 21, 2020In:

    Solve \begin{vmatrix} 4 & 1\\ 8 & 1 \end{vmatrix}

    Odinaka_O
    Odinaka_O Tyro

    $\begin{vmatrix} 4 & 1\\ 8 & 1 \end{vmatrix}$ $\Rightarrow (4\times 1)-(8\times 1)$ $\Rightarrow 4-8=-4$

    \begin{vmatrix} 4 & 1\\ 8 & 1 \end{vmatrix}
    \Rightarrow (4\times 1)-(8\times 1)
    \Rightarrow 4-8=-4

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  5. Asked: December 21, 2020In:

    Solve \begin{vmatrix} 3 & 2\\ -3 & 6 \end{vmatrix}

    Odinaka_O
    Odinaka_O Tyro

    $\begin{vmatrix} 3 & 2\\ -3 & 6 \end{vmatrix}$ $\Rightarrow (6\times 3)-(-3\times 2)$ $\Rightarrow 18-(-6)=24$

    \begin{vmatrix} 3 & 2\\ -3 & 6 \end{vmatrix}
    \Rightarrow (6\times 3)-(-3\times 2)
    \Rightarrow 18-(-6)=24

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  6. Asked: December 21, 2020In:

    Solve \begin{vmatrix} 2 & 7\\ -1 & 3 \end{vmatrix}

    Odinaka_O
    Odinaka_O Tyro

    $\begin{vmatrix} 2 & 7\\ -1 & 3 \end{vmatrix}$ $\Rightarrow (2\times 3)-(-1\times 7)$ $\Rightarrow 6-(-7)=13$

    \begin{vmatrix} 2 & 7\\ -1 & 3 \end{vmatrix}
    \Rightarrow (2\times 3)-(-1\times 7)
    \Rightarrow 6-(-7)=13

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  7. Asked: December 21, 2020In:

    Solve \begin{vmatrix} 6 & -3\\ -3 & 7 \end{vmatrix}

    Odinaka_O
    Odinaka_O Tyro

    $\begin{vmatrix} 6 & -3\\ -3 & 7 \end{vmatrix}$ $\Rightarrow (6\times 7)-(-3\times -3)$ $\Rightarrow 42-9=33$

    \begin{vmatrix} 6 & -3\\ -3 & 7 \end{vmatrix}
    \Rightarrow (6\times 7)-(-3\times -3)
    \Rightarrow 42-9=33

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  8. Asked: December 21, 2020In:

    Solve \begin{vmatrix} 0 & -1\\ -3 & 1 \end{vmatrix}

    Odinaka_O
    Odinaka_O Tyro

    $\begin{vmatrix} 0 & -1\\ -3 & 1 \end{vmatrix}$ $\Rightarrow (0\times 1)-(-3\times -1)$ $\Rightarrow 0-3=-3$

    \begin{vmatrix} 0 & -1\\ -3 & 1 \end{vmatrix}
    \Rightarrow (0\times 1)-(-3\times -1)
    \Rightarrow 0-3=-3

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  9. Asked: December 21, 2020In:

    Solve \begin{vmatrix} 9 & -3\\ 4 & 2 \end{vmatrix}

    Odinaka_O
    Odinaka_O Tyro

    $\begin{vmatrix} 9 & -3\\ 4 & 2 \end{vmatrix}$ $\Rightarrow (9\times 2)-(-3\times 4)$ $\Rightarrow 18-(-12)=30$

    \begin{vmatrix} 9 & -3\\ 4 & 2 \end{vmatrix}
    \Rightarrow (9\times 2)-(-3\times 4)
    \Rightarrow 18-(-12)=30

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