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3 years ago

What are the domain and range of the function f(x)=logx-5?

Mathematics

1 answer:

lora16 [44]3 years ago

7 0

<h2>Answer</h2>

Domain is (5, ∞), {x| x is greater than 5} for all integer n

Range is (-∞, ∞), {y| y belongs to Reals numbers}

<h2>Explanation</h2>

$F(x) = log (x - 5)$

Set up the equation for x

$x - 5 > 0$

or,

x > 5

So the domain is all values of x greater than 5 to infinity.

Range is set of all correct or valid values of y.

Range = (-∞, ∞), {y| y belongs to Reals numbers}

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2 years ago

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

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3 years ago

Help Please!!!<br> Find the volume of the sphere

yuradex [85]

Answer:

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Step-by-step explanation:

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2 years ago

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Y = -2x + 8<br> y = x^2 - 9x + 18<br><br> solve the system of nonlinear equation

ryzh [129]

Substitution

-2x+8=x^2-9x+18

x^2-7x+10=0

Factoring, we get

(x-5)(x-2)=0

x=5, x=2

If x equals 5,

y=-2(5)+8=

-10+8=

-2

Therefore, we derive our first solution:

Now we solve for our second x

If x=2,

y=-4+8=

Therefore, we derive our second solution:

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2 years ago

I need some help please

Mama L [17]

Answer:

109

Step-by-step explanation:

Formula

<ZHG + <IHZ = <IHG

Givens

<ZHF = 11x - 1

<IHZ = 24

<IHG = 12x + 13

Solution

11x - 1 + 24 = 12x + 13 Combine terms on the left

11x + 23 = 12x + 13 Subtract 13 from both sides

11x + 10 = 12x Subtract 11x from both sides

10 = x

Answer

11x - 1 = 11*10 - 1 = 110 - 1 = 109

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2 years ago