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3 years ago

A fish tank that holds 80 gallons of water is 55% full. How many gallons are

Mathematics

1 answer:

Amanda [17]3 years ago

5 0

<h2><u>Answer:</u></h2>

55% of the fish tank is <u>44 gallons.</u>

<h2><u>Step-by-step explanation:</u></h2>

To solve this, you turn the 55% into a decimal, by dividing it by 100.

55 ÷ 100 = 0.55

You then <u>multiply</u> the whole capacity (80) by 0.55

80 x 0.55 = <u>44 gallons</u>

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What is the perimeter of a polygon with vertices at (3, 2), (3, 9), (7, 12), (11, 9), and (11, 2)?

ICE Princess25 [194]

Answer:

32 units

Step-by-step explanation:

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2 years ago

Jim is 4 years older than mark if the sum of their ages is 44 find the age of each person

Nataliya [291]

Answer:

jim is 24 years old, Mark is 20 years old

Step-by-step explanation:

This prompt can be set up using this equation:

Let's represent Mark's age by variable m and Jim's age by variable J.

j + m = 44

m + 4 = j

m = j - 4

j + (j - 4) = 44

2j - 4 = 44

2j = 48

j = 24

So, jim is 24 years old, meaning mark must be 20 years old. 20 + 24 results in 44 years old together.

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<em>Let me know if you have any further questions! Spread the love.</em>

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3 years ago

Evaluate the limit with either L'Hôpital's rule or previously learned methods.lim Sin(x)- Tan(x)/ x^3x → 0

Vsevolod [243]

Answer:

$\dfrac{-1}{6}$

Step-by-step explanation:

Given the limit of a function expressed as $\lim_{ x\to \ 0} \dfrac{sin(x)-tan(x)}{x^3}$ , to evaluate the following steps must be carried out.

Step 1: substitute x = 0 into the function

$= \dfrac{sin(0)-tan(0)}{0^3}\\= \frac{0}{0} (indeterminate)$

Step 2: Apply L'Hôpital's rule, by differentiating the numerator and denominator of the function

$= \lim_{ x\to \ 0} \dfrac{\frac{d}{dx}[ sin(x)-tan(x)]}{\frac{d}{dx} (x^3)}\\= \lim_{ x\to \ 0} \dfrac{cos(x)-sec^2(x)}{3x^2}\\$

Step 3: substitute x = 0 into the resulting function

$= \dfrac{cos(0)-sec^2(0)}{3(0)^2}\\= \frac{1-1}{0}\\= \frac{0}{0} (ind)$

Step 4: Apply L'Hôpital's rule, by differentiating the numerator and denominator of the resulting function in step 2

$= \lim_{ x\to \ 0} \dfrac{\frac{d}{dx}[ cos(x)-sec^2(x)]}{\frac{d}{dx} (3x^2)}\\= \lim_{ x\to \ 0} \dfrac{-sin(x)-2sec^2(x)tan(x)}{6x}\\$

$= \dfrac{-sin(0)-2sec^2(0)tan(0)}{6(0)}\\= \frac{0}{0} (ind)$

Step 6: Apply L'Hôpital's rule, by differentiating the numerator and denominator of the resulting function in step 4

$= \lim_{ x\to \ 0} \dfrac{\frac{d}{dx}[ -sin(x)-2sec^2(x)tan(x)]}{\frac{d}{dx} (6x)}\\= \lim_{ x\to \ 0} \dfrac{[ -cos(x)-2(sec^2(x)sec^2(x)+2sec^2(x)tan(x)tan(x)]}{6}\\\\= \lim_{ x\to \ 0} \dfrac{[ -cos(x)-2(sec^4(x)+2sec^2(x)tan^2(x)]}{6}\\$

Step 7: substitute x = 0 into the resulting function in step 6

$= \dfrac{[ -cos(0)-2(sec^4(0)+2sec^2(0)tan^2(0)]}{6}\\\\= \dfrac{-1-2(0)}{6} \\= \dfrac{-1}{6}$

<em>Hence the limit of the function </em> $\lim_{ x\to \ 0} \dfrac{sin(x)-tan(x)}{x^3} \ is \ \dfrac{-1}{6}$ .

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3 years ago

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Answer:

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Step-by-step explanation:

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3 years ago

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Step-by-step explanation:

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3 years ago