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Drupady [299]
3 years ago
15

To which set of numbers does the number-5 belong? select all that apply

Mathematics
1 answer:
Lelu [443]3 years ago
4 0
A, C, D are the correct answers.

-5 is neither a whole number nor a natural number. This is because negative numbers are not included in these. Hope this helps! Let me know:)
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What is the slope of line P / answers: -3, 1.5, 3
stepladder [879]
The slope of p is 3
the slope of q is -3
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3 years ago
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Solve for n: 2(n+5) = -2
Temka [501]
The answer should be n= -6
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Match the numerical expressions to their simplest forms.
Aloiza [94]

Answer:

(a^6b^1^2)^\frac{1}{3} = a^2b^4

\frac{(a^5b^3)^\frac{1}{2}}{(ab)^-^\frac{1}{2}} = a^3b^2

(\frac{a^5}{a^-^3b^-^4})^\frac{1}{4} = a^2b

(\frac{a^3}{ab^-^6})^\frac{1}{2} = ab^3

Step-by-step explanation:

Simplify each of the expressions:

1

(a^6b^1^2)^\frac{1}{3}

Distribute the exponent. Multiply the exponent of the term outside of the parenthesis by the exponents of the variable.

(a^6b^1^2)^\frac{1}{3}

a^6^*^\frac{1}{3}b^1^2^*^\frac{1}{3}

Simplify,

a^2b^4

2

Use a similar technique to solve this problem. Remember, a fractional exponent is the same as a radical, if the denominator is (2), then the operation is taking the square root of the number.

\frac{(a^5b^3)^\frac{1}{2}}{(ab)^-^\frac{1}{2}}

Rewrite as square roots:

\frac{\sqrt{a^5b^3}}{\sqrt{(ab)}^-^1}

A negative exponent indicates one needs to take the reciprocal of the number. Apply this here:

\frac{\sqrt{a^5b^3}}{\frac{1}{\sqrt{ab}}}

Simplify,

\sqrt{a^5b^3}*\sqrt{ab}

Since both numbers are under a radical, one can rewrite them such that they are under the same radical,

\sqrt{a^5b^3*ab}

Simplify,

\sqrt{a^6b^4}

Since this operation is taking the square root, divide the exponents in half to do this operation:

a^3b^2

3

(\frac{a^5}{a^-^3b^-^4})^\frac{1}{4}

Simplify, to simplify the expression in the numerator and the denominator, the base must be the same. Remember, the base is the number that is being raised to the exponent. One subtracts the exponent of the number in the denominator from the exponent of the like base in the numerator. This only works if all terms in both the numerator and the denominator have the operation of multiplication between them:

(\frac{a^8}{b^-^4})^\frac{1}{4}

Bring the negative exponent to the numerator. Change the sign of the exponent and rewrite it in the numerator,

(a^8b^4)^\frac{1}{4}

This expression to the power of the one forth. This is the same as taking the quartic root of the expression. Rewrite the expression with such,

\sqrt[4]{a^8b^4}

SImplify, divide the exponents by (4) to simulate taking the quartic root,

a^2b

4

(\frac{a^3}{ab^-^6})^\frac{1}{2}

Using all of the rules mentioned above, simplify the fraction. The only operation happening between the numbers in both the numerator and the denominator is multiplication. Therefore, one can subtract the exponents of the terms with the like base. The term in the denomaintor can be rewritten in the numerator with its exponent times negative (1).

(a^3^-^1b^(^-^6^*^(^-^1^)^))^\frac{1}{2}

(a^2b^6)^\frac{1}{2}

Rewrite to the half-power as a square root,

\sqrt{a^2b^6}

Simplify, divide all of the exponents by (2),

ab^3

7 0
3 years ago
There are approximately 3,000 bass in a lake. The population grows at a rate of 2% per year. Round answers to the nearest tenth.
Tamiku [17]

Answer:

years 1–4: 62.4 bass per year

years 5–8: 67.6 bass per year

Step-by-step explanation:

If the population in year n is ...

 p(n) = 3000·1.02^n

then the average rate of change from year 1 to year 4 is ...

 (p(4) -p(1))/(4 -1) = 3000(1.02^4 -1.02^1)/3 = 1020·(1.02^3 -1) ≈ 62.4

The average rate of change for years 1–4 is 62.4 bass per year.

For years 5–8, the rate of change is similarly computed:

 (p(8) -p(5))/(8 -5) = 3000(1.02^8 -1.02^5)/3 = 1000·1.02^5·(1.02^3 -1) ≈ 67.6

The average rate of change for years 5–8 is 67.6 bass per year.

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3 years ago
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postnew [5]
C = 2πr   Divide both sides by 2r
\frac{C}{2r} = π   Switch the sides to make it easier to read
π = <span>\frac{C}{2r}</span>
7 0
3 years ago
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