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3 years ago

6

Solve the equation. 3x + 2(4x − 6) = 8x + 1 What is the value for x? x =

2 answers:

Aleonysh [2.5K]3 years ago

8 0

Answer: x= 13/3

Step-by-step explanation: solve x

by simplifying both sides of the equation, then isolating the variable.

Ivahew [28]3 years ago

3 0

Answer:

$x=\frac{13}{3}$

Step-by-step explanation:

The explanation on how to solve it is from the attached picture.

Hope this helps! :)

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What are the characteristics of the graph of the inequality x < 4.5?

lyudmila [28]

Answer:

see explanation

Step-by-step explanation:

Given x < 5

The graph will have an open circle at 5 indicating that x is not equal to 5, which would be a closed circle.

The ray will move to the left since values less than 5 are to the left of 5

4 0

3 years ago

Read 2 more answers

Select the correct answer.

andrew-mc [135]

Answer:

OPTION C

Step-by-step explanation:

We have the following rules:

$\sqrt[a]{x} = x^{\frac{1}{a}$ and

$x^a . x^b = x^{a + b}$

OPTION A: $x \sqrt[7]{x}$

= $x . x^{\frac{1}{7}$

= $x^{1 + \frac{1}{7}$

$= x^{\frac{8}{7}$ $\ne x^{\frac{9}{7}$

So, it can be eliminated.

OPTION B: $\bigg ( \frac{1}{\sqrt[7]{x}} \bigg )^9$

= $\bigg ( \frac{1}{x^{\frac{1}{7}}} \bigg )^9$

= $\bigg ( x^ {\frac{-1}{7}} \bigg )^9$

= $x^{\frac{-9}{7}}$

So, this option is also eliminated.

OPTION C: $x . \sqrt[7]{x}$

= $x . x^{\frac{1}{7}}$

= $x^{1 + \frac{1}{7}}$

= $x^{\frac{9}{7}}$

So, OPTION C is the answer.

5 0

3 years ago

Hep plz plz plz plz desperate help

Vikki [24]

Answer:

The equation is: 120 - 4x = 92. There are 7 incorrect answers.

Step-by-step explanation:

5 0

3 years ago

which of the following is equivalent to 3 sqrt 32x^3y^6 / 3 sqrt 2x^9y^2 where x is greater than or equal to 0 and y is greater

Nutka1998 [239]

Answer:

$\frac{\sqrt[3]{16y^4}}{x^2}$

Step-by-step explanation:

The options are missing; However, I'll simplify the given expression.

Given

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$

Required

Write Equivalent Expression

To solve this expression, we'll make use of laws of indices throughout.

From laws of indices $\sqrt[n]{a} = a^{\frac{1}{n}}$

So,

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$ gives

$\frac{(32x^3y^6)^{\frac{1}{3}}}{(2x^9y^2)^\frac{1}{3}}$

Also from laws of indices

$(ab)^n = a^nb^n$

So, the above expression can be further simplified to

$\frac{(32^\frac{1}{3}x^{3*\frac{1}{3}}y^{6*\frac{1}{3}})}{(2^\frac{1}{3}x^{9*\frac{1}{3}}y^{2*\frac{1}{3}})}$

Multiply the exponents gives

$\frac{(32^\frac{1}{3}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

Substitute $2^5$ for 32

$\frac{(2^{5*\frac{1}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

$\frac{(2^{\frac{5}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

From laws of indices

$\frac{a^m}{a^n} = a^{m-n}$

This law can be applied to the expression above;

$\frac{(2^{\frac{5}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$ becomes

$2^{\frac{5}{3}-\frac{1}{3}}x^{1-3}*y^{2-\frac{2}{3}}$

Solve exponents

$2^{\frac{5-1}{3}}*x^{-2}*y^{\frac{6-2}{3}}$

$2^{\frac{4}{3}}*x^{-2}*y^{\frac{4}{3}}$

From laws of indices,

$a^{-n} = \frac{1}{a^n}$ ; So,

$2^{\frac{4}{3}}*x^{-2}*y^{\frac{4}{3}}$ gives

$\frac{2^{\frac{4}{3}}*y^{\frac{4}{3}}}{x^2}$

The expression at the numerator can be combined to give

$\frac{(2y)^{\frac{4}{3}}}{x^2}$

Lastly, From laws of indices,

$a^{\frac{m}{n} = \sqrt[n]{a^m}$ ; So,

$\frac{(2y)^{\frac{4}{3}}}{x^2}$ becomes

$\frac{\sqrt[3]{(2y)}^{4}}{x^2}$

$\frac{\sqrt[3]{16y^4}}{x^2}$

Hence,

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$ is equivalent to $\frac{\sqrt[3]{16y^4}}{x^2}$

8 0

3 years ago

10 POINTS ARE AWARDED! :(

amid [387]

Answer:

Step-by-step explanation:

An index number is a number which is raised to a power. The power, also known as the index, tells you how many times you have to multiply the number by itself. For example, 2^5 means that you have to multiply 2 by itself five times = 2×2×2×2×2 = 32. There are a number of important rules of index numbers: y^a × y^b = y.

3 0

3 years ago