Volume of a cube = l^3, where l is the side length (note that the length, width and height in a cube are equal)
Volume of cube = 6^3 = 216 m^3
AB = 6 cm, AC = 12 cm, CD = ?
In triangle ABC, ∠CBA = 90°, therefore in triangle BCD ∠CBD = 90° also.
Since ∠BDC = 55°, ∠CBD = 90°, and there are 180 degrees in a triangle, we know ∠DCB = 180 - 55 - 90 = 35°
In order to find ∠BCA, use the law of sines:
sin(∠BCA)/BA = sin(∠CBA)/CA
sin(∠BCA)/6 cm = sin(90)/12 cm
sin(∠BCA) = 6*(1)/12 = 0.5
∠BCA = arcsin(0.5) = 30° or 150°
We know the sum of all angles in a triangle must be 180°, so we choose the value 30° for ∠BCA
Now add ∠BCA (30°) to ∠DCB = 35° to find ∠DCA.
∠DCA = 30 + 35 = 65°
Since triangle DCA has 180°, we know ∠CAD = 180 - ∠DCA - ∠ADC = 180 - 65 - 55 = 60°
In triangle DCA we now have all three angles and one side, so we can use the law of sines to find the length of DC.
12cm/sin(∠ADC) = DC/sin(∠DCA)
12cm/sin(55°) = DC/sin(60°)
DC = 12cm*sin(60°)/sin(55°)
DC = 12.686 cm
Answer:
first do 9x40 and get your answer then do 18x3 and get that answer and then add to do answers together
![x= \sqrt[8]{y} \\ y = z^{16} ](https://tex.z-dn.net/?f=x%3D%20%5Csqrt%5B8%5D%7By%7D%0A%5C%5C%20y%20%3D%20z%5E%7B16%7D%0A%0A%20)
Plug "y" into the first equation.
![x = \sqrt[8]{z^{16}} = (z^{16})^{1/8} = z^{16/8} = z^2](https://tex.z-dn.net/?f=x%20%3D%20%20%5Csqrt%5B8%5D%7Bz%5E%7B16%7D%7D%20%3D%20%28z%5E%7B16%7D%29%5E%7B1%2F8%7D%20%3D%20z%5E%7B16%2F8%7D%20%3D%20z%5E2)
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