Question

1) What is the Multiplicative inverse of $10^{-100}$ ?

2) The sum of the two numbers is 35 and their difference is 7. Therefore, the sum of their cubes is?

Answer both the questions.

Answer 1

$\bf\huge\color{red}{SOLUTIONS;}$

1☞The multiplicative inverse of 10^-100 = 10^100

2☞The first thing to do in algebraic equations is to assign variables to what you don't know. In this case, we don't know either number so we'll call them x and y.

The problem gives us two key bits of info. One, these numbers have a difference of 7; so when you subtract them, you get 7:
x−y=7

Also, they have a sum of 35; so when you add them, you get 35:
x+y=35

We now have a system of two equations with two unknowns:
x−y=7
x+y=35

If we add them together, we see we can cancel the ys:
Xx−y=7
+x+y=35−−−−−−−−−−
X2x+0y=42
→2x=42

Now divide by 2 and we have x=21. From the equation x+y=35, we can see that y=35−x. Using this and the fact that x=21, we can solve for y:
y=35−x
→y=35−21=14

So the two numbers are 21 and 14, which do indeed add to 35 and have a difference of 7.

Now , you could find the cubes :- 21^3 and 14^3 = Your Answer.

Answer 2

heya dear
_______________☺☺
❤❤❤

(2)
sum of two no. is=35

let the no. be =x+y=35..........1

their difference is x-y=7.........2

let do addition of both equations,
x+y=35
x-y=7
2x= 42
x=21

y=35-21
=14

now sum of their cube s
12005
(1)

l
${ \frac{1}{10} }^{100}$