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3 years ago

8

Cuantas decenas hay 47

1 answer:

Scorpion4ik [409]3 years ago

5 0

Hay 4 decenas en 47, y 7 unidades.

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Solving radical equations

My name is Ann [436]

Answer:

top: no solution
bottom: x = 7

Step-by-step explanation:

I find a graphing calculator very helpful to solving radical equations. It helps avoid extraneous solutions.

<u>top</u>

√(-14x +2) = x -3

-14x +2 = x² -6x +9 . . . . . square both sides

x² +8x +7 = 0 . . . . . . . . . .subtract the left side

Descartes' rule of signs tells you this has no positive real solutions. (There are no sign changes among the coefficients.) The value of the radical will be non-negative, so we must have x-3≥0, or x≥3. Since there are no positive real solutions to the quadratic, there are no values of x≥3 that will satisfy the equation. NO SOLUTION

__

<u>bottom</u>

4√(x -3) = 8 . . . . . given

√(x -3) = 2 . . . . . . divide by 4

x -3 = 4 . . . . . . . . . square both sides

x = 7 . . . . . . . . . . . add 3

3 0

4 years ago

What are the next three triangular numbers, 1, 3, 6, 10

lord [1]

It’s 15, 21, 28 because it’s going up 1,2,3 etc and adding one each time it goes up

7 0

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

Write and solve an equation to find the value of x and a missing angle in each triangle below. Use the number bank to help you c

jolli1 [7]

Answer:

1. x = 30

REC = 130

Step-by-step explanation:

2x + 70 = 4x + 10

60 = 2x

x = 30

you try the others

Good luck :D

5 0

2 years ago

Plz solve!!<br> 10-2^2/-5 <br> Plz show your work for brainly.

vfiekz [6]

Gujhhijftujjajwkciqje

7 0

3 years ago