Which inequality represents this statement?

The product of −6 and a number is no less than −12.


−6n≤−12

−6n>−12

−6n≥−12

−6n<−12

Answers

Answer 1

Answer:

−6n≥−12

Step-by-step explanation:

Answer 2

The correct inequality to show the given statement is,

⇒ - 6n ≥ - 12

Where, 'n' is any number.

What is Inequality?

A relation by which we can compare two or more mathematical expression is called an inequality.

Given that;

The statement is,

''The product of - 6 and a number is no less than - 12.''

Now, Let a number = n

So, We can formulate the correct inequality is,

⇒ - 6n ≥ - 12

Learn more about the inequality visit:

https://brainly.com/question/25944814

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8/15 is already a proper fraction, therefore it cannot be converted to a mixed number. Hope this helps!

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Part 1: Nouns
Directions: Write the plural form of each noun.
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Answers

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Determine which of the values is a solution of 4t+6 is less than or equal to 10.

Values:
1, -8, -4, 3, 0, 7

Answers

Answer:

1, 3, 7

Step-by-step explanation:

4t + 6≤10

4t≤4

t≤1

Therefore, t can either be equal to or greater than 1.

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The scale of a blueprint is 3 inches : 10 feet. On the blueprint, a house is 9 inches
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Step-by-step explanation:

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Answer:

A

Step-by-step explanation:

8/6 is also 1 1/3.

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Answer:

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Step-by-step explanation:

24 · 0.3 = 7.2

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Step-by-step explanation:

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Answers

Answer:

log14=1.146

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Step-by-step explanation:

Use log rules to break up the given logs, to be able to use the information give

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Step-by-step explanation:

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Answers

Answer:

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Step-by-step explanation:

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Answers

Answer:

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Answers

Answer:

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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Answers

Answer:

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Step-by-step explanation: Just simply multiply 14.9 by 100.

Answer:

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Answer:

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Step-by-step explanation:

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Answers

Answer:

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Step-by-step explanation:

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Answers

The equation in slope-intercept form that represents the situation is y = 3x + 12

A linear equation is on the form:

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Let x be the calf's age in month and y be the weight.

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The equation in slope-intercept form that represents the situation is y = 3x + 12

Find out more on linear equation at: brainly.com/question/14323743



1. What is the value of x that satisfies the equation? Must show all proper work/steps!

32^2x = 16^2x+8

Student's Work

Student's Answer

A. x = 4

B. x =- 4

C. x = 16

D. x = 1

Please explain all steps

Answers

[tex] {32}^{2x} = 16 ^{2x + 8} \\ = > {2}^{5(2x)} = {2}^{4(2x + 8)} \\ = > {2}^{10x} = {2}^{8x + 32} \\ = > 10x = 8x + 32 \\ = > 10x - 8x = 32 \\ = > 2x = 32 \\ = > x = \frac{32}{2} \\ = > x = 16[/tex]

Answer:

x = 16

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Answer:

Step-by-step explanation:

[tex]12000(1+\frac{.12}{4})^{4*45}=2454040.31463\\=2454040[/tex]


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Answers

Answer:

f-1(x) = 1/3x would be the answer

Step-by-step explanation:

Answer:

y= 1/3x

Step-by-step explanation:

inverse function perform the opposite.

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A finite geometric series is the sum of a sequence of numbers. Take the sequence
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previous number. So, a number in the sequence can be represented by the
function f(n) = 2^n-1. One way to write the sum of the sequence through the 5th
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This equation can also be written as S5 = 2^0+2^1+ 2^2+ 2^3+ 2^4. If we multiply this equation by 2. the equation becomes 2(S5) = 2^1+ 2^2+ 2^3+ 2^4+ 2^5

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Answers

Answer:

Step-by-step explanation:

2S₅ - S₅ = 2⁵ - 2⁰

        S₅ = 2⁵ - 1

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Answers

QuestionSimplify:

[tex]\displaystyle{\sf\:4\:\sqrt{48}\:-\:\dfrac{5}{2}\:\sqrt{\dfrac{1}{3}}\:+\:6\:\sqrt{3}}[/tex]

Answer :

[tex]\displaystyle{\boxed{\red{\sf\:4\:\sqrt{48}\:-\:\dfrac{5}{2}\:\sqrt{\dfrac{1}{3}}\:+\:6\:\sqrt{3}\:=\:\dfrac{127\:\sqrt{3}}{6}}}}[/tex]

Step-by-step-explanation:

We have given an expression.

We have to simplify the expression.

The given expression is

[tex]\displaystyle{\sf\:4\:\sqrt{48}\:-\:\dfrac{5}{2}\:\sqrt{\dfrac{1}{3}}\:+\:6\:\sqrt{3}}[/tex]

[tex]\displaystyle{\implies\sf\:4\:\sqrt{16\:\times\:3}\:-\:\dfrac{5}{2}\:\sqrt{\dfrac{1}{3}}\:+\:6\:\sqrt{3}}[/tex]

We know that,

[tex]\displaystyle{\boxed{\pink{\sf\:\sqrt{a\:\times\:b}\:=\:\sqrt{a}\:\times\:\sqrt{b}\:}}\:\cdots\sf\:a\:,\:b\:\geq\:0}[/tex]

[tex]\displaystyle{\implies\sf\:4\:\sqrt{16}\:\times\:\sqrt{3}\:-\:\dfrac{5}{2}\:\sqrt{\dfrac{1}{3}}\:+\:6\:\sqrt{3}}[/tex]

[tex]\displaystyle{\implies\sf\:4\:\times\:4\:\sqrt{3}\:-\:\dfrac{5}{2}\:\sqrt{\dfrac{1}{3}}\:+\:6\:\sqrt{3}}[/tex]

We know that,

[tex]\displaystyle{\boxed{\blue{\sf\:\sqrt{\dfrac{a}{b}}\:=\:\dfrac{\sqrt{a}}{\sqrt{b}}\:}}\:\cdots\sf\:b\: > \:0}[/tex]

[tex]\displaystyle{\implies\sf\:16\:\sqrt{3}\:-\:\dfrac{5}{2}\:\times\:\dfrac{\sqrt{1}}{\sqrt{3}}\:+\:6\:\sqrt{3}}[/tex]

[tex]\displaystyle{\implies\sf\:16\:\sqrt{3}\:-\:\dfrac{5}{2}\:\times\:\dfrac{1}{\sqrt{3}}\:+\:6\:\sqrt{3}}[/tex]

[tex]\displaystyle{\implies\sf\:\dfrac{16\:\sqrt{3}\:\times\:2\:\sqrt{3}\:-\:5}{2\:\sqrt{3}}\:+\:6\:\sqrt{3}}[/tex]

[tex]\displaystyle{\implies\sf\:\dfrac{16\:\times\:2\:\sqrt{3}\:\times\:\sqrt{3}\:-\:5}{2\:\sqrt{3}}\:+\:6\:\sqrt{3}} \\ \\\displaystyle{\implies\sf\:\dfrac{32\:\times\:3\:-\:5}{2\:\sqrt{3}}\:+\:6\:\sqrt{3}}[/tex]

[tex]\displaystyle{\implies\sf\:\dfrac{8\:\times\:4\:\times\:3\:-\:5\:+\:(\:6\:\sqrt{3}\:\times\:2\:\sqrt{3}\:)}{2\:\sqrt{3}}} \\ \\ \\\displaystyle{\implies\sf\:\dfrac{8\:\times\:12\:-\:5\:+\:(\:6\:\times\:2\:\times\:\sqrt{3}\:\times\:\sqrt{3}\:)}{2\:\sqrt{3}}} \\ \\ \\\displaystyle{\implies\sf\:\dfrac{8\:\times\:12\:-\:5\:+\:(\:12\:\times\:3\:)}{2\:\sqrt{3}}} \\ \\ \\\displaystyle{\implies\sf\:\dfrac{8\:\times\:12\:-\:5\:+\:12\:\times\:3}{2\:\sqrt{3}}} \\ \\ \\ \displaystyle{\implies\sf\:\dfrac{8\:\times\:12\:+\:12\:\times\:3\:-\:5}{2\:\sqrt{3}}} \\ \\ \\\displaystyle{\implies\sf\:\dfrac{12\:(\:8\:+\:3\:)\:-\:5}{2\:\sqrt{3}}} \\ \\ \\\displaystyle{\implies\sf\:\dfrac{12\:\times\:11\:-\:5}{2\:\sqrt{3}}} \\ \\ \\\displaystyle{\implies\sf\:\dfrac{132\:-\:5}{2\:\sqrt{3}}} \\ \\ \\\displaystyle{\implies\sf\:\dfrac{127}{2\:\sqrt{3}}} \\ \\ \\ \displaystyle{\implies\sf\:\dfrac{127}{2\:\sqrt{3}}\:\times\:\dfrac{\sqrt{3}}{\sqrt{3}}} \\ \\ \\\displaystyle{\implies\sf\:\dfrac{127\:\sqrt{3}}{2\:\times\:3}} \\ \\ \\\displaystyle{\implies\sf\:\dfrac{127\:\sqrt{3}}{6}} \\ \\ \\ \displaystyle{\therefore\:\underline{\boxed{\red{\sf\:4\:\sqrt{48}\:-\:\dfrac{5}{2}\:\sqrt{\dfrac{1}{3}}\:+\:6\:\sqrt{3}\:=\:\dfrac{127\:\sqrt{3}}{6}}}}}[/tex]

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