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3 years ago

Which expressions are equivalent to the one below? Check all that apply.

Mathematics

2 answers:

V125BC [204]3 years ago

8 0

Answer:

Step-by-step explanation:

The given expression is $10x$ .

We have to check which of he given options on simplification gives the same given expression.

(A) The expression is:

= $\frac{50x}{5x}$

= $10$

which is not equivalent to the given expression, thus this option is incorrect.

(B) The expression is:

= $10(10x)-1$

= $100x-1$

which is not equivalent to the given expression, thus this option is incorrect.

= $10(10x)+1$

= $100x+1$

which is not equivalent to the given expression, thus this option is incorrect.

(D) The expression is:

= $\frac{50x}{5}$

= $10x$

which is equivalent to the given expression, thus this option is correct.

(E) The expression is:

= $(\frac{50}{5})x$

== $10x$

which is equivalent to the given expression, thus this option is correct.

(F) The expression is:

= $x^5$

which is not equivalent to the given expression, thus this option is incorrect.

Whitepunk [10]3 years ago

4 0

The correct options are ${\text{A} = {\left( {9 \cdot 9} \right)^x},C = {9^x} \cdot {9^x}{\text{ and }}F = {9^{2x}}.$

Further explanation:

The associative property is defined as a grouping of multiplication, addition, subtraction and division.

Always use the PEDMAS rule to solve the grouping of multiplication, addition, subtraction and division.

Here, P is parenthesis, E is exponents, M is multiplication, D is division, A is addition and S is subtraction.

Given:

The expression is $10x.$

The options are as follows,

(A). $\dfrac{{50x}}{{5x}}$

(B). $10 \cdot 10x - 1$

(C). $10 \cdot 10x + 1$

(D). $\dfrac{{50x}}{5}$

(E). $\left( {\dfrac{{50}}{5}} \right)x$

(F). ${x^5}$

Explanation:

The expression $10x$ can also be written as follows,

Solve option (A).

$\dfrac{{50x}}{{5x}} = 10$

Option (A) is not correct.

Solve option (B).

$10 \cdot 10x - 1 = 100x - 1$

Option (B) is not correct.

Solve option (C).

$10 \cdot 10x - 1 = 100x - 1$

Option (C) is not correct.

Solve option (D).

$\dfrac{{50x}}{5} = 10x$

Option (D) is correct.

Solve option (E).

$\left( {\dfrac{{50}}{5}} \right)x = 10x$

Option (E) is correct.

Solve option (F).

${x^5}$

The correct options are $D=\dfrac{{50x}}{5}{\text{ and }}E = \left( {\dfrac{{50}}{5}} \right)x.$

Option A, B, C and F are not correct.

Learn more:

1. Learn more about inverse of the functionhttps://brainly.com/question/1632445.

2. Learn more about equation of circle brainly.com/question/1506955.

3. Learn more about range and domain of the function brainly.com/question/3412497

Answer details:

Grade: Middle School

Subject: Mathematics

Chapter: Number System

Keywords: expression, equivalent, one below, 10x, exponents, addition.

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Match each equation with its solution set. Tiles a2 − 9a + 14 = 0 a2 + 9a + 14 = 0 a2 + 3a − 10 = 0 a2 + 5a − 14 = 0 a2 − 5a − 1

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We have that

N 1)
a² − 9a + 14 = 0

Group terms that contain the same variable, and move the constant to the opposite side of the equation

(a² − 9a)=-14

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(a² − 9a+20.25)=-14+20.25

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a1=4.5+√6.25-----> a1=7

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the solution problem N 1 is the pair {7, 2}

N 2)

a² + 9a + 14 = 0

Group terms that contain the same variable, and move the constant to the opposite side of the equation

(a² + 9a)=-14

Complete the square Remember to balance the equation by adding the same constants to each side

(a² +9a+20.25)=-14+20.25

Rewrite as perfect squares

(a+4.5)²=6.25--------> (a+4.5)=(+/-)√6.25

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N 3)

a² + 3a − 10 = 0

Group terms that contain the same variable, and move the constant to the opposite side of the equation

(a² + 3a)=10

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Rewrite as perfect squares

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a1=-1.5+√12.25-----> a1=2

a2=-1.5-√12.25-----> a2=-5

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N 4)

a² + 5a − 14 = 0

Group terms that contain the same variable, and move the constant to the opposite side of the equation

(a² + 5a) =14

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Rewrite as perfect squares

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a1=-2.5+√20.25-----> a1=2

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N 5)

a² − 5a − 14 = 0

Group terms that contain the same variable, and move the constant to the opposite side of the equation

(a² − 5a)=14

Complete the square Remember to balance the equation by adding the same constants to each side

(a² − 5a+6.25)=14+6.25

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a1=2.5+√20.25-----> a1=7

a2=2.5-√20.25-----> a2=-2

the solution problem N 5 is the pair {7, -2}

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