All categories

3 years ago

Factorise 12u-uv please help lol

Mathematics

1 answer:

max2010maxim [7]3 years ago

6 0

Answer:

u(12-v)

Step-by-step explanation:

12u and -uv both have a factor of u in common, so we can pull it out to give us the factored expression u(12-v).

Note on factoring:

Remember, it's just using the distributive property in reverse! We can get our original expression back by distributing the u to the 12 and the -v.

You might be interested in

How do I set this up to solve as a linear equation?

NISA [10]

I don't see any numbers to set up in an equation or letters

4 0

3 years ago

Which expression is equivalent to 2j 5j?

Oliga [24]

2j+5j=7j Just add the two together.

6 0

3 years ago

Last Math problem, I need super extra help solving this math problem. 3y-4-6-3x-7y+6x+1

pantera1 [17]

Answer:

3x-4y-9

Step-by-step explanation:

Combine like terms

5 0

2 years ago

How do you solve m=10x-x for x

cricket20 [7]

m=10x-x for x
first subtract x from 10x

m=9x
divide both sides by 9

m/9=x

x=m/9

3 0

3 years ago

Given that 'n' is a natural number. Prove that the equation below is true using mathematical induction.

LenaWriter [7]

<h3>To ProvE :- </h3>

1 + 3 + 5 + ..... + (2n - 1) = n²

Method :-

If P(n) is a statement such that ,

P(n) is true for n = 1
P(n) is true for n = k + 1 , when it's true for n = k ( k is a natural number ) , then the statement is true for all natural numbers .

$\sf\to \textsf{ Let P(n) : 1 + 3 + 5 + $\dots$ +(2n-1) = n$^{\sf 2}$ }$

Step 1 : Put n = 1 :-

$\sf\longrightarrow LHS = \boxed{\sf 1 } \\$

$\sf\longrightarrow RHS = n^2 = 1^2 = \boxed{\sf 1 }$

Step 2 : Assume that P(n) is true for n = k :-

$\sf\longrightarrow 1 + 3 + 5 + \dots + (2k - 1 ) = k^2$

Add (2k +1) to both sides .

$\sf\longrightarrow 1 + 3+5+\dots+(2k-1)+(2k+1)=k^2+(2k+1)$

RHS is in the form of ( a + b)² = a²+b²+2ab .

$\sf\longrightarrow 1 + 3+5+\dots+(2k-1)+(2k+1)= (k +1)^2$

Adding and subtracting 1 to LHS .

$\sf\longrightarrow 1 + 3+5+\dots+(2k-1)+(2k+1) + 1 -1 = (k +1)^2 \\$

$\sf\longrightarrow 1 + 3+5+\dots+(2k-1)+(2k+2) - 1 = (k +1)^2$

Take out 2 as common .

$\sf\longrightarrow 1 + 3+5+\dots+(2k-1)+\{2(k+1)-1\}= (k +1)^2$

P(n) is true for n = k + 1 .

Hence by the principal of Mathematical Induction we can say that P(n) is true for all natural numbers 'n' .

**Edits are welcomed**

8 0

2 years ago

Read 2 more answers