Ask question

Ask question

All categories

3 years ago

11

Solve the following equation (7X3+4X2+3X+2) + (2X2+X+5)=

1 answer:

rjkz [21]3 years ago

5 0

The way it's written it's
Simplify the following:
7 X^3 + 4 X^2 + 2 X^2 + 3 X + X + 2 + 5
Grouping like terms, 7 X^3 + 4 X^2 + 2 X^2 + 3 X + X + 2 + 5 = 7 X^3 + (4 X^2 + 2 X^2) + (3 X + X) + (2 + 5):7 X^3 + (4 X^2 + 2 X^2) + (3 X + X) + (2 + 5)
4 X^2 + 2 X^2 = 6 X^2:
7 X^3 + 6 X^2 + (3 X + X) + (2 + 5)
3 X + X = 4 X:
7 X^3 + 6 X^2 + 4 X + (2 + 5)
2 + 5 = 7:Answer: 7 X^3 + 6 X^2 + 4 X + 7

You might be interested in

(WORTH 20 POINTS ANSWER ASAP)

Volgvan

The answer is B i think

8 0

3 years ago

What is the probability of rolling a number less than 4 on a number cube labeled 1 through 6?

cluponka [151]

Answer:

The answer is B. 1/3

Step-by-step explanation:

Hope this helps! :)

-------------------------------------

3 0

3 years ago

Find the solution set of this inequality|10x+20| ≤10

Pavlova-9 [17]

Answer:

solution is

[-3,-1]

Step-by-step explanation:

we are given

$|10x+20|\leq 10$

Firstly, we will find critical values

so, let's assume it is equal

$|10x+20|= 10$

now, we can break absolute sign

For $|10x+20|= -(10x+20)$ :

$-(10x+20)= 10$

we can solve for x

$-10x-20= 10$

Add both sides by 20

$-10x-20+20= 10+20$

$-10x= 30$

Divide both sides by -10

and we get

$x=-3$

For $|10x+20|= (10x+20)$ :

$(10x+20)= 10$

we can solve for x

$10x+20= 10$

Subtract both sides by 20

$10x+20-20= 10-20$

$10x= -10$

Divide both sides by 10

and we get

$x=-1$

so, critical values are

$x=-3$

$x=-1$

now, we can draw a number line and locate these values

and then we can check inequality on each intervals

For $(-\infty,-3)$ :

We can select any random value from this interval and plug that in inequality

and we get

we can plug x=-5

$|10\times -5+20|\leq 10$

$|-50+20|\leq 10$

$30\leq 10$

so, this is FALSE

For $[-3,-1]$ :

We can select any random value from this interval and plug that in inequality

and we get

we can plug x=-2

$|10\times -2+20|\leq 10$

$|-20+20|\leq 10$

$0\leq 10$

so, this is TRUE

For $(-1,\infty)$ :

We can select any random value from this interval and plug that in inequality

and we get

we can plug x=0

$|10\times 0+20|\leq 10$

$|0+20|\leq 10$

$20\leq 10$

so, this is FALSE

so, solution is

[-3,-1]

7 0

3 years ago

Find the length of side X in simplest radical form with a rational denominator.<br> X

german

Answer:

given in the figure an isosceles right triangle

hence

x = √3

√3 is a <u>surd</u> it can't be simplified further

7 0

2 years ago

A bakery offers a sale price of $3.40 for 4 muffins. What is the price pe dozen?

maxonik [38]

Answer:

10.2

Step-by-step explanation:

3.40 * 3 = 10.2

5 0

3 years ago