10x^3 y^-5 z^-2 if x=3 y=2 and z=5 express your answer as a reduced fraction

Mathematics

1 answer:

sineoko [7]2 years ago

7 0

Answer:

$\large\boxed{\dfrac{27}{80}}$

Step-by-step explanation:

Put x = 3, y = 2 and z = 5 to the given expression $10x^3y^{-5}z^{-2}$ :

$10(3^3)(2^{-5})(5^{-2})\qquad\text{use}\ a^{-n}=\dfrac{1}{a^n}\\\\=\dfrac{10(27)}{(2^5)(5^2)}=\dfrac{270}{(32)(25)}=\dfrac{270}{800}=\dfrac{27}{80}$

You might be interested in

How many solutions does this system of equations have?

azamat

Answer:

its one solution

Step-by-step explanation:

3 0

2 years ago

Find all values of c such that c/(c - 5) = 4/(c - 4). If you find more than one solution, then list the solutions you find separ

Deffense [45]

Answer:

$\large\boxed{\bold{NO\ REAL\ SOLUTIONS}}\\\boxed{x=4-2i,\ x=4+2i}$

Step-by-step explanation:

$Domain:\\c-4\neq0\ \wedge\ c-5\neq0\Rightarrow c\neq4\ \wedge\ c\neq5\\\\\dfrac{c}{c-5}=\dfrac{4}{c-4}\qquad\text{cross multiply}\\\\c(c-4)=4(c-5)\qquad\text{use the distributive property}\\\\c^2-4c=4c-20\qquad\text{subtract}\ 4c\ \text{from both sides}\\\\c^2-8c=-20\qquad\text{add 20 to both sides}\\\\c^2-8c+20=0\qquad\text{use the quadratic formula}$

$\text{for}\ ax^2+bx+c=0\\\\\text{if}\ b^2-4ac0,\ \text{then the equation has two solutions}\ x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

$x^2-8x+20=0\\\\a=1,\ b=-8,\ c=20\\\\b^2-4ac=(-8)^2-4(1)(20)=64-80=-16$

$\text{In the set of complex numbers:}\\\\i=\sqrt{-1}\\\\\text{therefore}\ \sqrt{b^2-4ac}=\sqrt{-16}=\sqrt{(16)(-1)}=\sqrt{16}\cdot\sqrt{-1}=4i\\\\x=\dfrac{-(-8)\pm4i}{2(1)}=\dfrac{8\pm4i}{2}=\dfrac{8}{2}\pm\dfrac{4i}{2}=4\pm2i$