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3 years ago

Simplify. Your answer should only contain positive exponents.

Mathematics

1 answer:

natali 33 [55]3 years ago

8 0

Answer:

Step-by-step explanation:

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(Distribute , then simplify the remaining expression. Final answer must be in standard form) Please help !!

Sergeu [11.5K]

Problem 22
$4x^2(2x^2+x-5) - x(x^3+5x^2-3) + 17$
$4x^2(2x^2)+4x^2(x)+4x^2(-5) - x(x^3)-x(5x^2)-x(-3) + 17$
$8x^4+4x^3-20x^2 - x^4-5x^3+3x + 17$
$7x^4-x^3-20x^2+3x + 17$

So the final simplified answer to problem 22 is
$7x^4-x^3-20x^2+3x + 17$

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

Problem 23
$-2x(x^3-6x^2+6)+4x^3-(5x^4+10x)$
$-2x(x^3)-2x(-6x^2)-2x(6)+4x^3-(5x^4)-(10x)$
$-2x^4+12x^3-12x+4x^3-5x^4-10x$
$-7x^4+16x^3-22x$

So problem 23 simplifies to
$-7x^4+16x^3-22x$

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

Problem 24
$4b[2a^2-5(3ab-2b^2)]+29ab^2$
$4b[2a^2-5(3ab)-5(-2b^2)]+29ab^2$
$4b[2a^2-15ab+10b^2]+29ab^2$
$4b[2a^2]+4b[-15ab]+4b[10b^2]+29ab^2$
$8a^2b-60ab^2+40b^3+29ab^2$
$8a^2b-31ab^2+40b^3$

So problem 24 simplifies to
$8a^2b-31ab^2+40b^3$

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

Problem 25
$2y(6x-4x^2y)+13x^2y^2-6$
$2y(6x)+2y(-4x^2y)+13x^2y^2-6$
$12xy-8x^2y^2+13x^2y^2-6$
$12xy+5x^2y^2-6$

The final answer for problem 25 is
$12xy+5x^2y^2-6$

8 0

4 years ago

What is 0.5x+1/5y-3.5x-4/5y

Virty [35]

Answer:

−3x+ −3/5y

Step-by-step explanation:

5 0

3 years ago

What is the measure of x in triangle ABC? A. 41° B. 61° C. 31° D. 51°

iogann1982 [59]

Answer:D.51

Step-by-step explanation:180-68-61=51

4 0

3 years ago

Read 2 more answers

Which of the following best describes the expression 7(y + 5)?

WINSTONCH [101]

Answer:

d The product of a constant factor of seven and a factor with the sum of two terms

Step-by-step explanation:

It's a product, and one of the factors is 7.

Answer:

d The product of a constant factor of seven and a factor with the sum of two terms

8 0

3 years ago

What is the range of the function y = x 2?

Leno4ka [110]

Answer:

$\mathrm{Range\:of\:}x^2:\quad \begin{bmatrix}\mathrm{Solution:}\:&\:f\left(x\right)\ge \:0\:\\ \:\mathrm{Interval\:Notation:}&\:[0,\:\infty \:)\end{bmatrix}$

The graph is also attached below.

Step-by-step explanation:

Given the function

$y=x^2$

We know that the range of a function is the set of values of the dependent variable for which a function is defined.

$\mathrm{For\:a\:parabola}\:ax^2+bx+c\:\mathrm{with\:Vertex}\:\left(x_v,\:y_v\right)$

$\mathrm{If}\:a$

$\mathrm{If}\:a>0\:\mathrm{the\:range\:is}\:f\left(x\right)\ge \:y_v$

$a=1,\:\mathrm{Vertex}\:\left(x_v,\:y_v\right)=\left(0,\:0\right)$

$f\left(x\right)\ge \:0$

Thus,

$\mathrm{Range\:of\:}x^2:\quad \begin{bmatrix}\mathrm{Solution:}\:&\:f\left(x\right)\ge \:0\:\\ \:\mathrm{Interval\:Notation:}&\:[0,\:\infty \:)\end{bmatrix}$

The graph is also attached below.

5 0

3 years ago