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3 years ago

What the solution 2(4+2x)≥5x+5

Mathematics

2 answers:

Ber [7]3 years ago

6 0

Answer:

3 ≥ x

Step-by-step explanation:

2(4+2x)≥5x+5 | Given Expression

8 + 4x ≥ 5x + 5 | Distributive Property

^(multiply the terms on the inside of the parenthesis by 2)

8 ≥ 1x + 5 | Subtraction Property of Equality

^(subtract 4x from both sides)

3 ≥ x | Subtraction PoE

^(subtract 5 from both sides)

olga nikolaevna [1]3 years ago

4 0

Answer:

x ≤ 3

Step-by-step explanation:

2(4+2x) ≥ 5x+5 (by PEDMAS, expand parentheses first)

4(2) + 2x(2) ≥ 5x + 5

8 + 4x ≥ 5x + 5 (subtract 8 from each side)

4x ≥ 5x + 5 - 8

4x ≥ 5x - 3 (subtract 5x from both sides)

4x -5x ≥ - 3

-x ≥ - 3 (multiply both sides by -1, remember to flip the inequality when multiplying both sides by a negative number)

x ≤ 3

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3 years ago

Which equation has infinite solutions?

AnnyKZ [126]

The equation that has an infinite number of solutions is $2x + 3 = \frac{1}{2}(4x + 2) + 2$

<h3>How to determine the equation?</h3>

An equation that has an infinite number of solutions would be in the form

a = a

This means that both sides of the equation would be the same

Start by simplifying the options

3(x – 1) = x + 2(x + 1) + 1

3x - 3 = x + 3x + 2 + 1

3x - 3 = 4x + 3

Evaluate

x = 6 ----- one solution

x – 4(x + 1) = –3(x + 1) + 1

x - 4x - 4 = -3x - 3 + 1

-3x - 4 = -3x - 2

-4 = -2 ---- no solution

$2x + 3 = \frac{1}{2}(4x + 2) + 2$

2x + 3 = 2x + 1 + 2

2x + 3 = 2x + 3

Subtract 2x

3 = 3 ---- infinite solution

Hence, the equation that has an infinite number of solutions is $2x + 3 = \frac{1}{2}(4x + 2) + 2$

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brainly.com/question/15349799

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<u>Complete question</u>

Which equation has infinite solutions?

3(x – 1) = x + 2(x + 1) + 1

x – 4(x + 1) = –3(x + 1) + 1

$2x + 3 = \frac{1}{2}(4x + 2) + 2$

$\frac 13(6x - 3) = 3(x + 1) - x - 2$

5 0

2 years ago

First question, thanks. I believe there should be 3 answers

zysi [14]

Given: The following functions

$A)cos^2\theta=sin^2\theta-1$ $B)sin\theta=\frac{1}{csc\theta}$ $\begin{gathered} C)sec\theta=\frac{1}{cot\theta} \\ D)cot\theta=\frac{cos\theta}{sin\theta} \\ E)1+cot^2\theta=csc^2\theta \end{gathered}$

To Determine: The trigonometry identities given in the functions

Solution

Verify each of the given function

$\begin{gathered} cos^2\theta=sin^2\theta-1 \\ Note\text{ that} \\ sin^2\theta+cos^2\theta=1 \\ cos^2\theta=1-sin^2\theta \\ Therefore \\ cos^2\theta sin^2\theta-1,NOT\text{ }IDENTITIES \end{gathered}$

$\begin{gathered} sin\theta=\frac{1}{csc\theta} \\ Note\text{ that} \\ csc\theta=\frac{1}{sin\theta} \\ sin\theta\times csc\theta=1 \\ sin\theta=\frac{1}{csc\theta} \\ Therefore \\ sin\theta=\frac{1}{csc\theta},is\text{ an identities} \end{gathered}$

$\begin{gathered} sec\theta=\frac{1}{cot\theta} \\ note\text{ that} \\ cot\theta=\frac{1}{tan\theta} \\ tan\theta cot\theta=1 \\ tan\theta=\frac{1}{cot\theta} \\ Therefore, \\ sec\theta\ne\frac{1}{cot\theta},NOT\text{ IDENTITY} \end{gathered}$

$\begin{gathered} cot\theta=\frac{cos\theta}{sin\theta} \\ Note\text{ that} \\ cot\theta=\frac{1}{tan\theta} \\ cot\theta=1\div tan\theta \\ tan\theta=\frac{sin\theta}{cos\theta} \\ So, \\ cot\theta=1\div\frac{sin\theta}{cos\theta} \\ cot\theta=1\times\frac{cos\theta}{sin\theta} \\ cot\theta=\frac{cos\theta}{sin\theta} \\ Therefore \\ cot\theta=\frac{cos\theta}{sin\theta},is\text{ an Identity} \end{gathered}$

$\begin{gathered} 1+cot^2\theta=csc^2\theta \\ csc^2\theta-cot^2\theta=1 \\ csc^2\theta=\frac{1}{sin^2\theta} \\ cot^2\theta=\frac{cos^2\theta}{sin^2\theta} \\ So, \\ \frac{1}{sin^2\theta}-\frac{cos^2\theta}{sin^2\theta} \\ \frac{1-cos^2\theta}{sin^2\theta} \\ Note, \\ cos^2\theta+sin^2\theta=1 \\ sin^2\theta=1-cos^2\theta \\ So, \\ \frac{1-cos^2\theta}{sin^2\theta}=\frac{sin^2\theta}{sin^2\theta}=1 \\ Therefore \\ 1+cot^2\theta=csc^2\theta,\text{ is an Identity} \end{gathered}$

Hence, the following are identities

$\begin{gathered} B)sin\theta=\frac{1}{csc\theta} \\ D)cot\theta=\frac{cos\theta}{sin\theta} \\ E)1+cot^2\theta=csc^2\theta \end{gathered}$

The marked are the trigonometric identities

3 0

2 years ago