keeping in mind that anything raised at the 0 power, is 1, with the sole exception of 0 itself.
![\bf ~~~~~~~~~~~~\textit{negative exponents} \\\\ a^{-n} \implies \cfrac{1}{a^n} \qquad \qquad \cfrac{1}{a^n}\implies a^{-n} \qquad \qquad a^n\implies \cfrac{1}{a^{-n}} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \cfrac{(r^{-7}b^{-8})^0}{t^{-4}w}\implies \cfrac{1}{t^{-4}w}\implies \cfrac{1}{t^{-4}}\cdot \cfrac{1}{w}\implies t^4\cdot \cfrac{1}{w}\implies \cfrac{t^4}{w}](https://tex.z-dn.net/?f=%20%5Cbf%20~~~~~~~~~~~~%5Ctextit%7Bnegative%20exponents%7D%0A%5C%5C%5C%5C%0Aa%5E%7B-n%7D%20%5Cimplies%20%5Ccfrac%7B1%7D%7Ba%5En%7D%0A%5Cqquad%20%5Cqquad%0A%5Ccfrac%7B1%7D%7Ba%5En%7D%5Cimplies%20a%5E%7B-n%7D%0A%5Cqquad%20%5Cqquad%20a%5En%5Cimplies%20%5Ccfrac%7B1%7D%7Ba%5E%7B-n%7D%7D%0A%5C%5C%5C%5C%5B-0.35em%5D%0A%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%0A%5Ccfrac%7B%28r%5E%7B-7%7Db%5E%7B-8%7D%29%5E0%7D%7Bt%5E%7B-4%7Dw%7D%5Cimplies%20%5Ccfrac%7B1%7D%7Bt%5E%7B-4%7Dw%7D%5Cimplies%20%5Ccfrac%7B1%7D%7Bt%5E%7B-4%7D%7D%5Ccdot%20%5Ccfrac%7B1%7D%7Bw%7D%5Cimplies%20t%5E4%5Ccdot%20%5Ccfrac%7B1%7D%7Bw%7D%5Cimplies%20%5Ccfrac%7Bt%5E4%7D%7Bw%7D%20)
Answer:
area = 102 ft²
Step-by-step explanation:
(8x6) + (9x6) = 48 + 54 = 102 ft²
Answer:
50°
Step-by-step explanation:
Let measure of one equal side be x.
80° +x +x = 180°[ sum of interior angle of triangle is 180°]
= 2x = 180° -80°
= x = 100°/2
=x = 50°
It would last 13.5 weeks because you would make a proportion:
12/16 = x/18 and then cross multiply. You would get 216=16x, then you have to divide both sides by 16 to get x = 13.5
Answer:
Step-by-step explanation:
From the given information, the symmetric equations for the line pass through(4, -5, 2) i.e (
) and are parallel to 
The parallel vector to the line i + zj+k = ai + bj + ck
Hence, the equation for the line is :

x = 4 + t
y = -5 + 2t
z = 2 + t
Thus, x, y, z = ( 4+t, -5+2t, 2+t )
The symmetric equation can now be as follows:



∴
