Mathematics, 19.02.2021 18:50 cay88

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Answer from: Quest

62926281916 inches of rain: )

Answer from: Quest

$4\left(625^4\right)\left(y^6\right)$

step-by-step explanation:

given expression is $4\sqrt{625^8y^{12}}$ .

now we need to simplify the given expression $\sqrt{20y^8}$ .

we know that square root is equivalent to the power 1/2. so that means we can take half of the exponents to get the square root of given expression.

$4\sqrt{625^8y^{12}}$

$=4\left(625^8y^{12}\right)^{\frac{1}{2}}$

$=4\left(625^8\right)^{\frac{1}{2}}\left(y^{12}\right)^{\frac{1}{2}}$

$=4\left(625^4\right)\left(y^6\right)$

expanding that will give a very very large number so let's keep answer in exponent form.

Answer from: Quest

i think it's eleven too, don't woosh me here, but according to the "experts" it's 2.

the proof starts from the peano postulates, which define the natural

numbers n. n is the smallest set satisfying these postulates:

p1. 1 is in n.

p2. if x is in n, then its "successor" x' is in n.

p3. there is no x such that x' = 1.

p4. if x isn't 1, then there is a y in n such that y' = x.

p5. if s is a subset of n, 1 is in s, and the implication

(x in s => x' in s) holds, then s = n.

then you have to define addition recursively:

def: let a and b be in n. if b = 1, then define a + b = a'

(using p1 and p2). if b isn't 1, then let c' = b, with c in n

(using p4), and define a + b = (a + c)'.

then you have to define 2:

def: 2 = 1'

2 is in n by p1, p2, and the definition of 2.

theorem: 1 + 1 = 2

proof: use the first part of the definition of + with a = b = 1.

then 1 + 1 = 1' = 2 q.e.d.

note: there is an alternate formulation of the peano postulates which

replaces 1 with 0 in p1, p3, p4, and p5. then you have to change the

definition of addition to this:

def: let a and b be in n. if b = 0, then define a + b = a.

if b isn't 0, then let c' = b, with c in n, and define

a + b = (a + c)'.

you also have to define 1 = 0', and 2 = 1'. then the proof of the

theorem above is a little different:

proof: use the second part of the definition of + first:

1 + 1 = (1 + 0)'

now use the first part of the definition of + on the sum in

parentheses: 1 + 1 = (1)' = 1' = 2 q.e.d.

Answer from: Quest

hence, the required result is $b-a=\left[\begin{array}{ccc}1& 1& -2\\1& 1& -2\\1& 1& -2\end{array}\right]$

step-by-step explanation:

suppose we have two matrices a and b

we will assume the two matrices

$a=\left[\begin{array}{ccc}1& 2& 3\\4& 5& 6\\7& 8& 9\end{array}\right]$

$b=\left[\begin{array}{ccc}2& 3& 1\\5& 6& 4\\8& 9& 7\end{array}\right]$

we have to compute b-a which means subtracting corresponding elements of a from b.

$b=\left[\begin{array}{ccc}2& 3& 1\\5& 6& 4\\8& 9& 7\end{array}\right]$ - $a=\left[\begin{array}{ccc}1& 2& 3\\4& 5& 6\\7& 8& 9\end{array}\right]$

$b-a=\left[\begin{array}{ccc}1& 1& -2\\1& 1& -2\\1& 1& -2\end{array}\right]$

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