If u have 13 tens . How can you write it in different ways

Mathematics

2 answers:

MrMuchimi2 years ago

7 0

You have 130 or you have 1 one hundred and 3 tens

denis-greek [22]2 years ago

6 0

You can write:

10 tens and 30 ones

You might be interested in

If x+1/x= 3, then prove that m^5+1/m^5= 123

alina1380 [7]

9514 1404 393

Explanation:

We can start with the relations ...

$\displaystyle\left(x+\frac{1}{x}\right)^3=\left(x^3+\frac{1}{x^3}\right)+3\left(x+\frac{1}{x}\right)\\\\\left(x+\frac{1}{x}\right)^5=\left(x^5+\frac{1}{x^5}\right)+5\left(x^3+\frac{1}{x^3}\right)+10\left(x+\frac{1}{x}\right)\\\\\textsf{From these, we can derive ...}\\\\x^5+\frac{1}{x^5}=\left(x+\frac{1}{x}\right)^5-5\left(\left(x+\frac{1}{x}\right)^3-3\left(x+\frac{1}{x}\right)\right)-10\left(x+\frac{1}{x}\right)$

$\displaystyle x^5+\frac{1}{x^5}=\left(x+\frac{1}{x}\right)^5-5\left(x+\frac{1}{x}\right)^3+5\left(x+\frac{1}{x}\right)\right)\\\\x^5+\frac{1}{x^5}=3^5 -5(3^3)+5(3)\\\\=((3^2-5)3^2+5)\cdot3=(4\cdot9+5)\cdot3=(41)(3)\\\\=\boxed{123}$

7 0

2 years ago

Question 6

forsale [732]

Answer:

12.35 should be the 10th term sorry if im wrong tho

Step-by-step explanation:

8 0

2 years ago

F(x)=x^3-2x+6 g(x)=2x^3+3x^2-4x+2 Find (f-g)(x).

____ [38]

(f-g)(x) = f(x) - g(x)
= (x^3 -2x+6) - (2x^3+3x^2-4x+2)
= x^3 -2x +6 -2x^3 -3x^2 +4x -2 . . . . distribute the negative sign
= (1-2)x^3 -3x^2 +(-2+4)x +(6-2) . . . . . combine like terms

(f-g)(x) = -x^3 -3x^2 +2x +4

7 0

2 years ago

Erggkhbqea <br><br>cbse class 7 maths chapter 8 exercise 8.3 solutions in

madam [21]

Answer:

am I supposed to join your meet with that C0DE

5 0

1 year ago

How do you solve? Help?

Vinvika [58]

Plug in a -20 wherever you see a b
(-20)^2 +4*-20-89/13
400 + -80-89/13
400 - 169/13
400-13
=387

Hope this helps

4 0

2 years ago