Answer:
x=16 and y=9
Step-by-step explanation:
2y+30=3y+21[Diagonals of parallelogram are equal]
30-21=3y-2y
y=9
3y=2x-5[Diagonals of parallelogram are equal]
3×9=2x-5
27=2x-5
2x=27+5
x=32/2=16
B^2 = a^2 - c^2
b^2 = 9^2 - 5^2
b^2 = 56
b = 7.5
Answer:
(a) The unit of 70.5 is lbm/ft^3 and the unit of 8.27×10^-7 is in^2/lbf
(b) density = 0.1206g/cm^3
(c) rho = 0.1206exp(8.27×10^-7P)
Step-by-step explanation:
(a) The unit of 70.5 is the same as the unit of rho which is lbm/ft^3. The unit of 8.27×10^-7 is the inverse of the unit of P (lbf/in^2) because exp is found of a constant. Therefore, the unit of 8.27×10^-7 is in^2/lbf
(b) P = 9×10^6N/m^2
rho = 70.5exp(8.27×10^-7× 9×10^6) = 70.5exp7.443 = 70.5×1.71 = 120.6kg/m^3
rho = 120.6kg/m^3 × 1000g/1kg × 1m^3/10^6cm^3 = 0.1206g/cm^3
(c) Formula for rho (g/cm^3) as a function of P (N/m^2) is
rho = 0.1206exp(8.27×10^-7P) (the unit of 0.1206 is g/cm^3)
From the right hand side, we will need to find a way to rewriting 3x²y in terms of cube roots.
We know that 27 is 3³, so if we were to rewrite it in terms of cube roots, we will need to multiply everything by itself two more twice. (ie we can rewrite it as ∛(3x²y)³)
Hence, we can say that it's:
![\sqrt[3]{162x^{c}y^{5}} = \sqrt[3]{(3x^{2}y)^{3}} * \sqrt[3]{6y^{d}}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B162x%5E%7Bc%7Dy%5E%7B5%7D%7D%20%3D%20%5Csqrt%5B3%5D%7B%283x%5E%7B2%7Dy%29%5E%7B3%7D%7D%20%2A%20%5Csqrt%5B3%5D%7B6y%5E%7Bd%7D%7D)
![= \sqrt[3]{162x^{6}y^{3+d}}](https://tex.z-dn.net/?f=%3D%20%5Csqrt%5B3%5D%7B162x%5E%7B6%7Dy%5E%7B3%2Bd%7D%7D)
Hence, c = 6 and d = 2