Why is it helpful to write a number as a product of primes?

Evaluate: x(y – z) for x = 3, y = 9, and z = 2 O 33 O 45 O-21 O 21

ad-work [718]

Answer:

21

Step-by-step explanation:

Hi there!

We are given the expression x(y-z), and we want to evaluate it if x=3, y=9, and z=2

As we're given the values of the variables, we can substitute those values into the expression to find out what the value of the expression is.

So substitute 3 as x, 9 as y, and 2 as z in the expression above

x(y-z)

3(9-2)

Now we must follow the order of operations; start by subtracting 2 from 9 (do the stuff inside the parentheses first)

3(7)

Now multiply 3 and 7 together

21

Hope this helps!

3 0

3 years ago

An art class is making a mural for the school that has a triangle drawn in the middle. The length of the bottom of the triangle

Arte-miy333 [17]

Answer:

7

Step-by-step explanation:

4 0

3 years ago

Question 1,2, and 3 how do i factor those? Can you show the work and explain how?

DiKsa [7]

1: $3n^{2}+9n+6$

notice that each part is divisible by 3

$3n^{2}$ ÷ 3 = $n^{2}$

9n ÷ 3 = 3n

6 ÷ 3 = 2

so it becomes $3(n^{2} +3n+2)$

3n can be rewritten as 2n+n

-you want to rewrite it into two numbers that multiply to the number that's alone (in this case 2)

which would get you

$3(n^{2} +2n+n+2)$

Now that it's rewritten, you can factor out n + 2 from the equation.

the answer is

3(n+2)(n+1)

And you can check that by multiplying (n+2)(n+1) which is $n^{2} +2n+n+2$ and then each of those by 3, which is $3n^{2} +6n+3n+6$ or $3n^{2}+9n+6$ , our origional equation

2: $28+x^{2} -11x$

So I rewrote this as $x^{2} -11x+28$ (it's the same thing, just reordered using the commutative property)

now -11x can be rewritten as -4x-7x

(remember, the two numbers should multiply to equal 28, which is our constant.)

$x^{2} -4x-7x+28$

now we can factor out x from the first expression and -7 from the second

$x(x-4)-7(x-4)$

and lastly you factor out x-4,

which would give you

(x-4)(x-7)

Make sure to check your work and make sure it multiplies to $x^{2} -11x+28$

3: $9x^{2} -12x+4$

The first thing I notice when looking at this problem is that both 9 and 4 are perfect squares. Not only that, but they are the squares of 2 and 3, which are products of -12

So if you rewrite 9 as $3^{2}$ and 4 as $2^{2}$ , the equation becomes