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4 years ago

9

Someone solve this please like NOW 4(5x-7)+12=16x+24

1 answer:

Bingel [31]4 years ago

7 0

20x-28+12=16x+24 (i distributed the 4 to get this)

20x-16=16x+24 (added -28 & 12)

4x=40 (Subtracted 16x from both sides and added 16 to both sides)

(divide by 4 on both sides)

x=10

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3 years ago

1. Construct a table of values of the following functions using the interval of 5

Morgarella [4.7K]

Complete Question:

Construct a table of values of the following functions using the interval of -5 to 5.

$g(x) = \frac{x^3 + 3x - 5}{x^2}$

Answer:

See Explanation

Step-by-step explanation:

Required

Construct a table with the given interval

When $x = -5$

$g(x) = \frac{x^3 + 3x - 5}{x^2}$ becomes

$g(-5) = \frac{-5^3 + 3(-5) - 5}{-5^2}$

$g(-5) = \frac{-125 -15 - 5}{25}$

$g(-5) = \frac{-145}{25}$

$g(-5) = -5.8$

When $x = -4$

$g(x) = \frac{x^3 + 3x - 5}{x^2}$ becomes

$g(-4) = \frac{-4^3 + 3(-4) - 5}{-4^2}$

$g(-4) = \frac{-64 -12 - 5}{16}$

$g(-4) = \frac{-81}{16}$

$g(-4) = -5.0625$

When $x = -3$

$g(x) = \frac{x^3 + 3x - 5}{x^2}$ becomes

$g(-3) = \frac{-3^3 + 3(-3) - 5}{-3^2}$

$g(-3) = \frac{-27 -9 - 5}{9}$

$g(-3) = \frac{-41}{9}$

$g(-3) = -4.56$

When $x = -2$

$g(x) = \frac{x^3 + 3x - 5}{x^2}$ becomes

$g(-2) = \frac{-2^3 + 3(-2) - 5}{-2^2}$

$g(-2) = \frac{-8 -6 - 5}{4}$

$g(-2) = \frac{-19}{4}$

$g(-2) = -4.75$

When $x = -1$

$g(x) = \frac{x^3 + 3x - 5}{x^2}$ becomes

$g(-1) = \frac{-1^3 + 3(-1) - 5}{-1^2}$

$g(-1) = \frac{-1 + 3 - 5}{1}$

$g(-1) = \frac{-3}{1}$

$g(-1) = -3$

When $x = 0$

$g(x) = \frac{x^3 + 3x - 5}{x^2}$ becomes

$g(0) = \frac{0^3 + 3(0) - 5}{0^2}$

$g(0) = \frac{0 + 0 - 5}{0}$

$g(0) = \frac{- 5}{0}$

<em>g(0) = undefined</em>

When $x = 1$

$g(x) = \frac{x^3 + 3x - 5}{x^2}$ becomes

$g(1) = \frac{1^3 + 3(1) - 5}{1^2}$

$g(1) = \frac{1 + 3 - 5}{1}$

$g(1) = \frac{-1}{1}$

$g(1) = 1$

When $x = 2$

$g(x) = \frac{x^3 + 3x - 5}{x^2}$ becomes

$g(2) = \frac{2^3 + 3(2) - 5}{2^2}$

$g(2) = \frac{8 + 6 - 5}{4}$

$g(2) = \frac{9}{4}$

$g(2) = 2.25$

When $x = 3$

$g(x) = \frac{x^3 + 3x - 5}{x^2}$ becomes

$g(3) = \frac{3^3 + 3(3) - 5}{3^2}$

$g(3) = \frac{27 + 9 - 5}{9}$

$g(3) = \frac{31}{9}$

$g(3) = 3.44$

When $x = 4$

$g(x) = \frac{x^3 + 3x - 5}{x^2}$ becomes

$g(4) = \frac{4^3 + 3(4) - 5}{4^2}$

$g(4) = \frac{64 + 12 - 5}{16}$

$g(4) = \frac{71}{16}$

$g(4) = 4.4375$

When $x = 5$

$g(x) = \frac{x^3 + 3x - 5}{x^2}$ becomes

$g(5) = \frac{5^3 + 3(5) - 5}{5^2}$

$g(5) = \frac{125 + 15 - 5}{25}$

$g(5) = \frac{135}{25}$

$g(5) = 5.4$

<em>Hence, the complete table is:</em>

x ---- g(x)

-5 --- -5.8

-4 --- -5.0625

-3 --- -4.56

-2 --- -4.75

-1 --- -3

0 -- Undefined

1 --- 1

2 -- 2.25

3 --- 3.44

4 --- 4.4375

5 --- 5.4

7 0

3 years ago

What do you need to find

Yuki888 [10]

Please type in a question if you did please type in the hole comment to get full help
Thank You :)

8 0

3 years ago