W=(-2/7)xx=(-9/4),y=3/4 ,and z=(-9/2) evaluate wx-yz

Can someone please help me with this

postnew [5]

Put the given values of p and q in the factored form equation.

... f(x) = (x -(-1))(x -(-2)) . . . . p and q values put in

... f(x) = (x +1)(x +2) . . . . . . .simplified

Multiplying the factors, we have

... f(x) = x(x +2) + 1(x +2) = x² +2x +1x +2

... f(x) = x² +3x +2

We want to factor x³ -x² -6x. We notice first of all that x is a factor of all terms. Thus we have

... = x(x² -x -6)

Now, we're looking for factors of -6 that add up to -1. Those are -3 and 2. Thus the factorization is ...

... = x(x -3)(x +2)

We want a description of the structure and an equivalent expression for

... 64x⁹ -216

We note that 64, 216, and x⁹ are all cubes, so this expression is ...

... the difference of cubes.

It can be rewritten to

... = 8((2x³)³ -3³)

and so can be factored as

... = 8(2x³ -3)(4x⁶ +6x³ +9)

7 0

3 years ago

Problems 2.21, 2.22, 2.23

RoseWind [281]

Statements can be proved by contrapositive, contradiction or by induction.

<em>2.21 and 2.23 are proved by contrapositive</em>
<em>2.22 is proved by induction</em>

<u />

<u>2.21: If </u> $n^3$ <u> is even, then n is even (By contrapositive)</u>

The contrapositive of the above statement is that:

Represent the value of n as:

$n = 2k + 1$ , where $k \ge 0$

Take the cube of both sides

$n^3 = (2k + 1)^3$

Expand

$n^3 = 8k^3 + 6k^2 + 6k + 1$

Group

$n^3 =[ 8k^3 + 6k^2 + 6k] + 1$

Factor out 2

$n^3 =2[4k^3 + 3k^2 + 3k] + 1$

Assume w is an integer; where:

$w =4k^3 + 3k^2 + 3k$

So, we have:

$n^3 =2w + 1$

The constant term (i.e. 1) means that $n^3$ is odd.

Hence, the statement has been proved by contrapositive.

<u />

We have: $3n + 4$ <u />

<u />

Assume that: $n = 2k + 2$ for $k \ge 0$

So, we have:

$3n + 4 = 3(2k + 2) + 4$

Open bracket

$3n + 4 = 6k + 6 + 4$

$3n + 4 = 6k + 10$

Factorize

$3n + 4 = 2(3k + 5)$

The factor of 2 means that $3n + 4$ is even.

<em>Hence, </em> $3n + 4$ <em> is even, if and only if n is even </em>

<em />

<u />

To do this, we prove by contrapositive.

The contrapositive of the above statement is:

If $s = -1$ and $t=-1$ , then $s + t + st = -1$

We have:

$s + t + st = -1$

Substitute the values of s and t in: $s + t + st = -1$

$-1 -1 -1 \times -1 = -1$

$-1 -1 + 1 = -1$

$-1 = -1$

Hence, by contrapositive:

If $s = -1$ and $t=-1$ , then $s + t + st = -1$

Read more about proofs at:

brainly.com/question/19643658

7 0

3 years ago

How to work this out ?

Free_Kalibri [48]

Answer:

The solution is (-3, -2).

Step-by-step explanation:

Please note that if you simply add these 2 equations together, 9x - 9x = 0, and you are then left with -5y = 10. Thus, y = -2. Subbing -2 for y in the first equation, we get 9x - 8(-2) = -11, or 9x + 16 = -11.

Subtracting 16 from both sides, we get 9x = -27, so that x = -3.

The solution is (-3, -2).

6 0

2 years ago

Aidan has four cards with +, –, ×, ÷ printed on them.

fiasKO [112]

Answer:

5 x 4 + 2 - 3 / 1

Step-by-step explanation:

Took me a lot of tries but got it! :) Hope this helps

7 0

3 years ago

I DONT KNOW WHAT IS 89090x908

never [62]

Answer:

80 893 720

Step-by-step explanation:

hope dis helps!!!!

6 0

2 years ago