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Mathematics

1 answer:

belka [17]2 years ago

4 0

We can solve this problem in three steps.

1) find the area of the square

$a=lw\\\\a=(7.5)(6)\\\\a=45ft^{2}$

2) find the area of the triangle

$a=\frac{1}{2}bh\\\\a=\frac{1}{2} (6)(4)\\\\a=12ft^{2}$

3) add the areas together

$a=45+12$

a = 57 ft²

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Q1.Simplify:

solniwko [45]

$\textbf{(i)}\\\\\{1^3 +2^3 \} \times \left(\dfrac 13 \right)^2\\\\=(1+8) \left(\dfrac 19 \right)\\\\=9\left(\dfrac 19 \right)\\\\=1\\\\$

$\textbf{(ii)}\\\\\{5^{-1}\times 4^{-1}\}^2\\\\=\left(5^{-1} \right)^2 \times \left(4^{-1} \right)^2~~~~~~~~~~~;[(ab)^m = a^mb^m]\\\\=5^{-2}\times 4^{-2}~~~~~~~~~~~~~~~~~~~~;[(a^m)^n = a^{mn}]\\\\=\dfrac 1{5^2} \times \dfrac 1{4^2}~~~~~~~~~~~~~~~~~~~~~~;\left[a^{-m} = \dfrac 1{a^m},~ a\neq 0 \right]\\\\=\dfrac{1}{400}\\\\=0.0025\\\\$

$\textbf{(iii)}\\\\\left\{ \left(\dfrac 13 \right)^{-2} \times \left( \dfrac 12 \right)^{-2}\right\}\div \left(\dfrac 14 \right)^{-3}\\\\\\=\left( \dfrac 13 \times \dfrac 12 \right)^{-2} \div\left(4^{-1}\right)^{-3}\\\\\\=\left(\dfrac 16 \right)^{-2} \div 4^3\\\\\\=\left(6^{-1} \right)^{-2}\div 4^3\\\\\\=6^2 \div 4^3\\\\\\=36\div 64\\\\\\=\dfrac{9}{16}\\\\\\=0.5625$