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kvv77 [185]
3 years ago
10

F(x) = x2 − 2x How would I factor this?

Mathematics
2 answers:
cupoosta [38]3 years ago
3 0

The only way to factor an equation such as this is by pulling out the greatest common factor. When we look at what is the largest term in both of these, we discover both have an x, but nothing else. So, we need to extract an x out of each piece.


f(x) = x^2 - 2x

f(x) = x(x - 2)


This will give you the fully factored form for this equation.

loris [4]3 years ago
3 0

f(x) = x^2 - 2x ← Factor out a common factor in each, which is a <em>x</em>

= x(x - 2)

And there you go!

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Please someone help me! I know that the answer is close to 17, so if u know the answer please show me how to do it!
gogolik [260]

Answer:

Ask ur mom she's nice

Step-by-step explanation:


8 0
3 years ago
a set v is given, together with definitions of addition and scalar multiplication. determine which properties of a vector space
agasfer [191]

The properties of a vector space are satisfied Properties 1,2, 5(a) and 5(c) are satisfied, the relaxation of the homes aren't legitimate are ifv = x ^ 2 1× v=1^ ×x ^ 2 = 1 #V

Property three does now no longer follow: Suppose that Property three is legitimate, shall we namev = a * x ^ 2 +bx +cthe neuter of V. Since v is the neuter, then O have to be constant with the aid of using the neuted, consequently 0 = O + v = (O  x ^ 2 + Ox + O) + (a × x ^ 2 + bx + c) = c × x ^ 2 + b ^ 2 + a

= 0 If O is the neuter, then it ought to restore x², but 0+ x² = (0x²+0x+zero) + (x²+0x+zero) = 1.This is a contradiction due to the fact x² isn't 1. We finish that V doesnt have a neuter vector. This additionally method that belongings four would not observe either. A set with out 0 cant

have additive inverse

Let r= v ×2x ^ 2 + v × 1x +v0 , w= w ×2x ^ 2 + w × 1x +w0 . We have that\\v+w= (vO + wO) ^  x^ 2 +(vl^ × wl)^  x+ ( v 2^ × w2)• w+v= (wO + vO) ^x^ 2 +(wl^ × vl)x+ ( w 2^ ×v2)

Since the sum of actual numbers is commutative, we finish that v + w = w + v Therefore, belongings 5(a) is valid.

Property 5(b) isn't valid: we are able to introduce

a counter example. we could use z = 1 thenv = x ^ 2 w = x ^ 2 + 1\\(v + w) + z = (x ^ 2 + 2) + 1 = 3x ^ 2 + 1

v + (w + z) = x ^ 2 + (2x ^ 2 + 1) = x ^ 2 + 3

Since 3x ^ 2 +1 ne x^ 2 +3. then the associativity rule doesnt hold.

(1+2)^ * (x^ 2 +x)=3^ * (x ^ 2 + x) = 3x + 3\\1^ × (x^ 2 +x)+2^ × (x ^ 2 + x) = (x + 1) + (2x + 2) = 3x ^ 2 + x ( ne 3x + 3 )\\(1^ ×2)^ ×(x^ 2 +x)=2^ × (x ^ 2 + x) = 2x + 2\\1^ × (2^ × (x ^ 2 + x) )=1^ × (2x+2)=2x^ 2 +2x( ne2x+2)

Property f doesnt observe because of the switch of variables. for instance, if v = x ^ 2 1 × v=1^ × x ^ 2 = 1 #V

Properties 1,2, 5(a) and 5(c) are satisfied, the relaxation of the homes arent legitimate.

Step-with the aid of using-step explanation:

Note that each sum and scalar multiplication entails in replacing the order from that most important coefficient with the impartial time period earlier than doing the same old sum/scalar multiplication.

Property three does now no longer follow: Suppose that Property three is legitimate, shall we name v = a × x ^ 2 +bx +c the neuter of V. Since v is the neuter, then O have to be constant with the aid of using the neuted, consequently0 = O + v = (O × x ^ 2 + Ox + O) + (a × x ^ 2 + bx + c) = c × x ^ 2 + b ^ 2 + a

= zero If O is the neuter, then it ought to restore x², but zero + x² = (0x²+0x+zero) + (x²+0x+zero) = 1.This is a contradiction due to the fact x² isn't 1. We finish that V doesnt have a neuter vector. This additionally method that belongings four would not observe either. A set with out 0 cant have additive inverse

Let r= v × 2x ^ 2 + v × 1x +v0 , w= w2x ^ 2 + w × 1x +w0 . \\We have thatv+w= (vO + wO) ^ x^ 2 +(vl^ wl)^x+ ( v 2^ w2)w+v= (wO + vO) ^ x^ 2 +(wl^ vl)x+ ( w 2^v2)

Since the sum of actual numbers is commutative, we finish that v + w = w + v Therefore, belongings 5(a) is valid.

Property 5(b) isn't valid: we are able to introduce

a counter example. we could usez = 1 then v = x ^ 2 w = x ^ 2 + 1(v + w) + z = (x ^ 2 + 2) + 1 = 3x ^ 2 + 1v + (w + z) = x ^ 2 + (2x ^ 2 + 1) = x ^ 2 + 3\\Since 3x ^ 2 +1 ne x^ 2 +3.then the associativity rule doesnt hold.

Note that each expressions are same because of the distributive rule of actual numbers. Also, you could be aware that his assets holds due to the fact in each instances we 'switch variables twice.

· (1+2)^ * (x^ 2 +x)=3^ * (x ^ 2 + x) = 3x + 31^ * (x^ 2 +x)+2^ * (x ^ 2 + x) = (x + 1) + (2x + 2) = 3x ^ 2 + x ( ne 3x + 3 )(1^ * 2)^ * (x^ 2 +x)=2^ * (x ^ 2 + x) = 2x + 21^ * (2^ * (x ^ 2 + x) )=1^ ×* (2x+2)=2x^ 2 +2x( ne2x+2)

Read more about polynomials :

brainly.com/question/2833285

#SPJ4

8 0
10 months ago
You usually buy a 16.5-ounce bottle of shampoo. There is a new bottle that says it gives you 30% more free.
Katarina [22]

Answer:

\frac{x-16.5}{16.5}=0.30

Hopefully, this is your desired setup. (I noticed the formula you have to fill in there.)

Have a good day.

Step-by-step explanation:

Hi.

You could use the percent change formula.

Since we know it is a percent increase then we will do new-old instead of old-new:

\frac{\text{new}-\text{old}}{\old}

x is the new amount of shampoo.

16.5 is the original amount (old) of shampoo.

The percent increase is 30%=0.30 .

So we have the following equation:

\frac{x-16.5}{16.5}=0.30

We could have found the equation like this:

16.5+16.5(0.30)=x

Subtract 16.5 on both sides:

16.5(0.30)=x-16.5

Divide both sides by 16.5:

0.30=\frac{x-16.5}{16.5}

By us of symmetric property of equality:

\frac{x-16.5}{16.5}=0.30

5 0
3 years ago
Read 2 more answers
What is the absolute value of -135​
sammy [17]

135

with math any absolute number cant be negative so you just make it positive

7 0
3 years ago
Read 2 more answers
Please help me 4t−t+2
podryga [215]

Answer:

The answer is 3t + 2

Step-by-step explanation:

Subtract t from 4t.

hoped this helped!

brainly, please?

3 0
2 years ago
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