Help me please! will give brainliest!

Mathematics

1 answer:

GenaCL600 [577]3 years ago

7 0

Answer:right

Step-by-step explanation:

You might be interested in

Quiz Questions

gavmur [86]

1. 6:3
2. 6:2
3. 3:18
4:5:6
5:13
6:x
7:x
8: check
9:check
10: x

6 0

2 years ago

Joe is building a post for his mailbox. To find the correct dimensions, he needs to expand this expression: (x-3) (x-9) (x-4) Se

kolbaska11 [484]

Answer:

$(x-3)(x-9)(x-4)=x^3-16x^2+75x-108$

Step-by-step explanation:

Given:

The expression to expand is given as:

$(x-3)(x-9)(x-4)$

Let us expand the first two binomials of the given expression using FOIL method.

The FOIL method states that:

$(a + b)(c + d) = ac + ad + bc + bd$

$(x-3)(x-9)=(x\times x)+(x\times -9)+(-3\times x)+(-3\times -9)\\\\(x-3)(x-9)=x^2-9x-3x+27=x^2-12x+27$

Now, let us multiply the result with the remaining binomial. Multiplying each term of the trinomial with each term of the binomial, we get:

$(x^2-12x+27)(x-4)\\\\= (x^2\times x)+ (x^2\times -4)+(-12x\times x)+(-12x\times -4)+(27\times x)+(27\times -4)\\\\=x^3-4x^2-12x^2+48x+27x-108\\\\=x^3-16x^2+75x-108...........(\because-12x^2-4x^2=-16x^2,48x+27x=75x)$

Therefore, the equivalent expression after expanding is given as:

$(x-3)(x-9)(x-4)=x^3-16x^2+75x-108$

3 0

2 years ago

Simplify.<br><br> 1/2b - 2/5b<br><br> please i am begging u

DerKrebs [107]

First, we need to find the LCD (Least Common Denominator) so we can subtract them much easier.

An easy way to find the LCD is to just multiply the denominators together: So 2 • 5 = 10

We have to do the same to the numerators too though.

1/2 (5) = 5/10

2/5 (2) = 4/10

Now, the remade expression is:

5/10 - 4/10 which will equal 1/10.

Hope it helped! If it did, please mark as Brainliest! :)

8 0

3 years ago

Read 2 more answers

Which equation shows the commutative property of addition?

anygoal [31]

Answer:

4+3=3+4

Step-by-step explanation:

The commutative property of addition says that, the order in which you add two real numbers does not matter.

Let $a,b\in \mathbb R$ , then