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Dmitrij [34]
3 years ago
13

Please find the general limit of the following function:

Mathematics
2 answers:
valentinak56 [21]3 years ago
8 0

Answer:

The general limit exists at <em>x</em> = 9 and is equal to 300.

Step-by-step explanation:

We want to find the general limit of the function:

\displaystyle \lim_{x \to 9}(x^2+2^7+(9.1\times 10))

By definition, a general limit exists at a point if the two one-sided limits exist and are equivalent to each other.

So, let's find each one-sided limit: the left-hand side and the right-hand side.

The left-hand limit is given by:

<h3>\displaystyle \lim_{x \to 9^-}(x^2+2^7+(9.1 \times 10))</h3>

Since the given function is a polynomial, we can use direct substitution. This yields:

=(9)^2+2^7+(9.1\times 10)

Evaluate:

300

Therefore:

\displaystyle \lim_{x \to 9^-}(x^2+2^7+(9.1 \times 10))=300

The right-hand limit is given by:

\displaystyle \lim_{x \to 9^+}(x^2+2^7+(9.1\times 10))

Again, since the function is a polynomial, we can use direct substitution. This yields:

=(9)^2+2^7+(9.1\times 10)

Evaluate:

=300

Therefore:

\displaystyle \lim_{x \to 9^+}(x^2+2^7+(9.1\times 10))=300

Thus, we can see that:

\displaystyle \lim_{x \to 9^-}(x^2+2^7+(9.1\times 10))=\displaystyle \lim_{x \to 9^+}(x^2+2^7+(9.1\times 10))=300

Since the two-sided limits exist and are equivalent, the general limit of the function does exist at <em>x</em> = 9 and is equal to 300.

creativ13 [48]3 years ago
5 0

Step-by-step explanation:

Hey there!

Please look your required answer in picture.

Note: In left hand limit always take a smaller near number of the approaching number. For example as in the solution I took the 8.99,8.999 as it is smaller than 9 but very near to it.

And in right hand limit always take a smaller and just greater near number than the approaching number. For example, I took 9.01,9.001 which a just greater but very near to 9.

<u>Hope</u><u> it</u><u> helps</u><u>!</u>

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Hello!

The figure is made up of a cone and a hemisphere. To the nearest whole number, what is the approximate volume of this figure? Use 3.14 to approximate π . Enter your answer in the box. cm³

Data: (Cone)

h (height) = 12 cm

r (radius) = 4 cm (The diameter is 8 being twice the radius)

Adopting: \pi \approx 3.14

V (volume) = ?

Solving: (Cone volume)

V = \dfrac{ \pi *r^2*h}{3}

V = \dfrac{ 3.14 *4^2*\diagup\!\!\!\!\!12^4}{\diagup\!\!\!\!3}

V = 3.14*16*4

\boxed{V = 200.96\:cm^3}

Note: Now, let's find the volume of a hemisphere.

Data: (hemisphere volume)

V (volume) = ?

r (radius) = 4 cm

Adopting: \pi \approx 3.14

If: We know that the volume of a sphere is V = 4* \pi * \dfrac{r^3}{3} , but we have a hemisphere, so the formula will be half the volume of the hemisphere V = \dfrac{1}{2}* 4* \pi * \dfrac{r^3}{3} \to \boxed{V = 2* \pi * \dfrac{r^3}{3}}

Formula: (Volume of the hemisphere)

V = 2* \pi * \dfrac{r^3}{3}

Solving:

V = 2* \pi * \dfrac{r^3}{3}

V = 2*3.14 * \dfrac{4^3}{3}

V = 2*3.14 * \dfrac{64}{3}

V = \dfrac{401.92}{3}

\boxed{ V_{hemisphere} \approx 133.97\:cm^3}

What is the approximate volume of this figure?

Now, to find the total volume of the figure, add the values: (cone volume + hemisphere volume)

Volume of the figure = cone volume + hemisphere volume

Volume of the figure = 200.96 cm³ + 133.97 cm³

\boxed{\boxed{\boxed{V = 334.93\:cm^3 \to Volume\:of\:the\:figure \approx 335\:cm^3 }}}\end{array}}\qquad\quad\checkmark

Answer:

The volume of the figure is approximately 335 cm³

_______________________

I Hope this helps, greetings ... Dexteright02! =)

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