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4 years ago

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Simplify 8x^3-2x^2+x+3x^3

2 answers:

8_murik_8 [283]4 years ago

4 0

8x^3-2x^2+x+3x^3

combine like terms

8x^3+3x^3-2x^2+x

11x^3 -2x^2 +x

Dafna11 [192]4 years ago

4 0

For this case we have the following polynomial:

$P (x) = 8x ^ 3-2x ^ 2 + x + 3x ^ 3$

To simplify, we must add terms whose exponent is the same, that is:

Given:

$Q (x) = ax ^ n + bx ^ 3 + cx ^ n$

Where:

a, b, c are the coefficients of the terms of the polynomial

x: It is the variable

n, 3: Are the exponents associated with the terms

We can simplify the form:

$Q (x) = (a + c) x ^ n + bx ^ 3$

That said, we can simplify the given polynomial in the following way:

$P (x) = (8 + 3) x ^ 3-2x ^ 2 + x\\P (x) = 11x ^ 3-2x ^ 2 + x$

Answer:

$P (x) = 11x ^ 3-2x ^ 2 + x$

You might be interested in

A map has a scale of 1.5 in : 26 mi. If the distance between city A and City

alekssr [168]

Answer:

The real distance is 260 miles

Step-by-step explanation:

From the question;

1.5 inches in map is 26 miles

Thus;

15 inches will be x miles

To get the value of x, we do a cross-multiplication

we have this as:

15 * 26 = 1.5 * x

x = (15 * 26)/1.5

x = 10 * 26 = 260 miles

6 0

3 years ago

Tems11 [23]

QUESTION A

The given multiplication problem is

$\frac{39}{64} \times \frac{8}{13}$

Factor each term to obtain;

$\frac{13\times 3}{8\times8} \times \frac{8}{13}$

Cancel out the common factors to obtain;

$\frac{1\times 3}{8\times1} \times \frac{1}{1}$

Simplify to get;

$\frac{3}{8}$

QUESTION B

The given multiplication problem is

$\frac{2}{3}\times \frac{1}{5}\times \frac{4}{7}$

This the same as

$\frac{2\times 1\times 4}{3\times 5\times 7}$

This simplifies to;

$\frac{8}{105}$

QUESTION C

The given problem is

$\frac{3}{5}\times \frac{10}{12} \times \frac{1}{2}$

This is the same as

$\frac{3}{5}\times \frac{5}{6} \times \frac{1}{2}$

$=\frac{1}{1}\times \frac{1}{2} \times \frac{1}{2}$

This simplifies to

$=\frac{1}{4}$

QUESTION D.

The given expression is

$\frac{4}{9}\times 54$

Factor the 54 to obtain;

$\frac{4}{9}\times 9\times 6$

Cancel the common factors to get;

$\frac{4}{1}\times 1\times 6$

This simplifies to;

$=24$

QUESTION E

The given problem is

$20\times 3\frac{1}{5}$

Convert the mixed numbers to improper fraction to obtain;

$=20\times \frac{16}{5}$

$=4\times5 \times \frac{16}{5}$

Cancel the common factors to get;

$=4\times1 \times \frac{16}{1}$

$=64$

QUESTION F

The multiplication problem is

$11 \times 2 \frac{7}{11}$

Convert the mixed numbers to improper fractions to obtain;

$11 \times \frac{29}{11}$

Cancel out the common factors to get;

$=1 \times \frac{29}{1}$

Simplify;

$=29$

QUESTION G

The given problem is

$5\frac{1}{3}\times 5\frac{1}{8}$

Convert to improper fractions;

$=\frac{16}{3}\times \frac{41}{8}$

Cancel out the common factors to get;

$=\frac{2}{3}\times \frac{41}{1}$

$=\frac{82}{3}$

Convert back to mixed numbers

$=27\frac{1}{3}$

QUESTION H

The given expression is

$10\frac{2}{3} \times 1\frac{3}{8}$

Convert to improper fraction to get;

$\frac{32}{3} \times \frac{11}{8}$

Cancel common factors to get;

$=\frac{4}{3} \times \frac{11}{1}$

Simplify

$=\frac{44}{3}$

Convert back to mixed numbers;

$=14\frac{2}{3}$

7 0

3 years ago

Find the midpoint of the segment with the following endpoints.<br> (8,1) and (5,7)

Crazy boy [7]

The midpoint would be (13/2 , 4)

6 0

3 years ago

Read 2 more answers

If x>7, then |x|>7. |y|>7, so y=7<br><br><br> Valid or invalid?

Alex

$\text{if }x>7\text{, then }|x|>7$ is a valid argument

$|y|>7\text{, so }y=7$ is not a valid argument

For the first argument: $\text{if }x>7\text{, then }|x|>7$

From the definition of absolute value function

$|x|=x$ if $x\ge0$

That is every positive number is its own absolute value. Since

$x>7\implies x\ge0$ ,

we can argue that

$x>7\implies |x|>7$

so the first argument is valid

For the second argument: $|y|>7\text{, so }y=7$

From the definition of absolute value function

$|y|:=\left \{ {y\text{ if }y\ge0}\atop{-y\text{ if }y$

This means that

$|y|>7:=\left \{ {y>7\text{ if }y\ge0}\atop{-y>7\text{ if }y$

or

$|y|>7:=\left \{ {y>7\text{ if }y\ge0}\atop{y$

no part of the definition allow for the option $y=7$ . So the second argument is not valid.

Learn more here: brainly.com/question/11897796

7 0

3 years ago

Tennis player strikes a tennis ball from underneath with her racket. The ball is sent straight up with an initial velocity of 19

alekssr [168]

Answer:

18.42 m

Step-by-step explanation:

At that point, the ball will be static, with final velocity of 0.

The initial velocity is 19.

The acceleration due to gravity is -9.8 m/s^2.

We now have to use an equation of kinematics shown below:

$v_f^2=v_0^2+2as$

Where

$v_f$ is the final velocity

$v_0$ is the initial velocity

a is -9.8

s is the distance we are looking for

Lets substitute and find the answer:

$v_f^2=v_0^2+2as\\0=(19)^2+2(-9.8)s\\-361=-19.6s\\s=18.42$

7 0

4 years ago