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notsponge [240]
3 years ago
5

300 L= mL I NEED THE ANSWER RIGHT NOW

Mathematics
2 answers:
vodka [1.7K]3 years ago
5 0

Explanation + Answer: First, 1 Liter equals 1,000 Milliliters.

If you need 300 liters to Milliliters, we need to do multiplication.

You need to think, 300 x 1000= ?

Well 3 x 1 = 3 and add 5 zeroes, 300000 mL.

Hope this helps.

sergiy2304 [10]3 years ago
4 0

Answer:

300000ml

Step-by-step explanation:

Litres to ml = * 1000

300L * 1000 = 300000ml

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s344n2d4d5 [400]

Expanded Notation:

A. 654.362 = (6x100) + (5x10) + (4x1) + (3x0.1) + (6x0.01) + (2x0.001)

B. 125.384 = (1x100) + (2x10) + (5x1) + (3x0.1) + (4x0.01) + (8x0.001)

8 0
3 years ago
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Which pair of funtions is not a pair of inverse functions? please help!!
antiseptic1488 [7]

Answer:

f(x)=\frac{x}{x+20} , g(x)=\frac{20x}{x-1}

Step-by-step explanation:

we know that

To find the inverse of a function, exchange variables x for y and y for x. Then clear the y-variable to get the inverse function.

we will proceed to verify each case to determine the solution of the problem

<u>case A)</u> f(x)=\frac{x+1}{6} , g(x)=6x-1

Find the inverse of f(x)

Let

y=f(x)

Exchange variables x for y and y for x

x=\frac{y+1}{6}

Isolate the variable y

6x=y+1

y=6x-1

Let

f^{-1}(x)=y

f^{-1}(x)=6x-1

therefore

f(x) and g(x) are inverse functions

<u>case B)</u> f(x)=\frac{x-4}{19} , g(x)=19x+4

Find the inverse of f(x)

Let

y=f(x)

Exchange variables x for y and y for x

x=\frac{y-4}{19}

Isolate the variable y

19x=y-4

y=19x+4

Let

f^{-1}(x)=y

f^{-1}(x)=19x+4

therefore

f(x) and g(x) are inverse functions

<u>case C)</u> f(x)=x^{5}, g(x)=\sqrt[5]{x}

Find the inverse of f(x)

Let

y=f(x)

Exchange variables x for y and y for x

x=y^{5}

Isolate the variable y

fifth root both members

y=\sqrt[5]{x}

Let

f^{-1}(x)=y

f^{-1}(x)=\sqrt[5]{x}

therefore

f(x) and g(x) are inverse functions

<u>case D)</u> f(x)=\frac{x}{x+20} , g(x)=\frac{20x}{x-1}

Find the inverse of f(x)

Let

y=f(x)

Exchange variables x for y and y for x

x=\frac{y}{y+20}

Isolate the variable y

x(y+20)=y

xy+20x=y

y-xy=20x

y(1-x)=20x

y=20x/(1-x)

Let

f^{-1}(x)=y

f^{-1}(x)=20x/(1-x)

\frac{20x}{1-x}\neq \frac{20x}{x-1}

therefore

f(x) and g(x) is not a pair of inverse functions

7 0
3 years ago
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The answer is 21.

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The correct answer would be C. Hope that helps.
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the answers are there

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