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Anit [1.1K]
3 years ago
14

3.

Mathematics
1 answer:
larisa [96]3 years ago
8 0

Answer:

a. 16 gallons

b. 32

Step-by-step explanation:

Let the full capacity of the tank is x gallons.

a. It is given that 12 gallons of water fill a tank to \frac{3}{4} capacity.

Hence, we can write \frac{3x}{4} = 12

⇒ x = \frac{4 \times 12}{3} = 16 gallons.

b. If the tank is filled to capacity, then there will be 16 gallons of water, with which \frac{16}{\frac{1}{2} } = 32 numbers of half-gallon bottles of be filled with water. (Answer)

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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}.

3 0
3 years ago
Write a number in which value of the digit 3 is 10 times as much as the value of the digit 3 in 942381 show your thinking.
choli [55]

Answer:

943,281

Step-by-step explanation:

Just move the 3 up a value

If it's in the ten's place, then move it to the hundred's place value.

If it's in the hundred's place, then move it to thousand's place value.

So on and so on...

8 0
4 years ago
Read 2 more answers
Can y'all help me with these too please
Bess [88]

Answer:

6. Z= -11

7. N= 15.25

Step-by-step explanation:

6. Subtract 10 from each side to get Z+3= -8

Then Subtract 3 from each side to get Z= -11

7. Subtract 4 from each side to get 3/4 n= 16

Then Subtract 3/4 to get N= 15.25

8 0
3 years ago
Read 2 more answers
On a number line what is the distance between -12 and 20
djverab [1.8K]

Answer:

The distance is 32

Step-by-step explanation:

One way is to subtract -12 and 20, which equals 32.

Another way is to add 12 and 20 since distance on a number line is always positive, which equals 32.

7 0
3 years ago
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If u can answer these please do
enot [183]

Answer:

1) 8.5x10^8

2) 5.3x10^-3

3) 9.95x10^12

Step-by-step explanation:

Since they have the same exponents, you just add or subtract and leave the rest the same.

1) 8.5x10^8

2) 5.3x10^-3

3) 9.95x10^12

8 0
3 years ago
Read 2 more answers
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