Answer:
![15,000\:\mathrm{mm^3}](https://tex.z-dn.net/?f=15%2C000%5C%3A%5Cmathrm%7Bmm%5E3%7D)
Step-by-step explanation:
The composite figure consists of a square prism and a trapezoidal prism. By adding the volume of each, we obtain the volume of the composite figure.
The volume of the square prism is given by
, where
is the base length and
is the height. Substituting given values, we have: ![V=14^2\cdot 30=196\cdot 30=5,880\:\mathrm{mm^3}](https://tex.z-dn.net/?f=V%3D14%5E2%5Ccdot%2030%3D196%5Ccdot%2030%3D5%2C880%5C%3A%5Cmathrm%7Bmm%5E3%7D)
The volume of a trapezoidal prism is given by
, where
and
are bases of the trapezoid,
is the length of the height of the trapezoid and
is the height. This may look very confusing, but to break it down, we're finding the area of the trapezoid (base) and multiplying it by the height. The area of a trapezoid is given by the average of the bases (
) multiplied by the trapezoid's height (
).
Substituting given values, we get:
![V=\frac{14+24}{2}\cdot (30-14)\cdot 30,\\V=19\cdot 16\cdot 30=9,120\:\mathrm{mm^3}}](https://tex.z-dn.net/?f=V%3D%5Cfrac%7B14%2B24%7D%7B2%7D%5Ccdot%20%2830-14%29%5Ccdot%2030%2C%5C%5CV%3D19%5Ccdot%2016%5Ccdot%2030%3D9%2C120%5C%3A%5Cmathrm%7Bmm%5E3%7D%7D)
Therefore, the total volume of the composite figure is
(ah, perfect)
Alternatively, we can break the figure into a larger square prism and a triangular prism to verify the same answer:
![V=30^2\cdot 14+\frac{1}{2}\cdot10\cdot 16\cdot 30=\boxed{15,000\:\mathrm{mm^3}}\checkmark](https://tex.z-dn.net/?f=V%3D30%5E2%5Ccdot%2014%2B%5Cfrac%7B1%7D%7B2%7D%5Ccdot10%5Ccdot%2016%5Ccdot%2030%3D%5Cboxed%7B15%2C000%5C%3A%5Cmathrm%7Bmm%5E3%7D%7D%5Ccheckmark)
Answer:
The unit rate is 2 cups of flour per cup of sugar.
Step-by-step explanation:
-10y^2 + (-3y^2) - 4y^2 - (-6y^2) =
-10y^2 - 3y^2 - 4y^2 + 6y^2 =
-17y^2 + 6y^2 =
- 11y^2
The equation of parabola is
![x^2=\dfrac{3}{4}y](https://tex.z-dn.net/?f=x%5E2%3D%5Cdfrac%7B3%7D%7B4%7Dy)
.
The canonical equation of parabola is
![y^2=2p x](https://tex.z-dn.net/?f=y%5E2%3D2p%20x)
and this parabola has branches that go in positive direction of x-axis.
Since in your equation x is changed to y and y to x, then the <span>branches of the parabola go in positive y-axis direction, the vertex is placed at the origin.
</span>
<span />
It is given that z varies directly with x and inversely with the square of y so it follows:
![z=k\frac{x}{y^2}](https://tex.z-dn.net/?f=z%3Dk%5Cfrac%7Bx%7D%7By%5E2%7D)
It is also given that z=18 when x=6 and y=2 so it follows:
![\begin{gathered} 18=k\frac{6}{2^2} \\ k=\frac{18\times4}{6} \\ k=12 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%2018%3Dk%5Cfrac%7B6%7D%7B2%5E2%7D%20%5C%5C%20k%3D%5Cfrac%7B18%5Ctimes4%7D%7B6%7D%20%5C%5C%20k%3D12%20%5Cend%7Bgathered%7D)
So the equation of variation becomes:
![z=12\frac{x}{y^2}](https://tex.z-dn.net/?f=z%3D12%5Cfrac%7Bx%7D%7By%5E2%7D)
Therefore the value of z when x=7 and y=7 is given by:
![\begin{gathered} z=\frac{12\times7}{7^2} \\ z=\frac{12}{7} \\ z\approx1.7143 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20z%3D%5Cfrac%7B12%5Ctimes7%7D%7B7%5E2%7D%20%5C%5C%20z%3D%5Cfrac%7B12%7D%7B7%7D%20%5C%5C%20z%5Capprox1.7143%20%5Cend%7Bgathered%7D)
Hence the value of z is 12/7 or 1.7143.