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3 years ago

Is it right And why

Mathematics

1 answer:

antiseptic1488 [7]3 years ago

7 0

Yes it is indeed correct,
B. 7.47 is the correct answer!
Brainliest pls! thx! :)

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The Hyperbolic Sine (sinh(x)) and Hyperbolic Cosine (cosh(x)) functions are defined as such: sin h(x) = e^x - e^-x/2 cosh(x) = e

labwork [276]

Answer:

y-incercepts:

sinh(x):0, cosh(x)=1

Limits:

positive infinity: sinh(x): infinity, cosh(x): infinity

negative infinity: sinh(x): - infinity, cosh(x): infinity

Step-by-step explanation:

We are given that

$\sinh(x)=\frac{e^{x}-e^{-x}}{2}$

$\cosh(x)=\frac{e^{x}+e^{-x}}{2}$

To find out the y-incerpt of a function, we just need to replace x by 0. Recall that $e^{0}=1$ . Then,

$\sinh(0) = \frac{1-1}{2}=0$

$\cosh(0) = \frac{1+1}{2}=1$

For the end behavior, recall the following:

$\lim_{x\to \infty}e^{x} = \infty, \lim_{x\to \infty}e^{-x} = 0$

$\lim_{x\to -\infty}e^{x} = 0, \lim_{x\to -\infty}e^{-x} = \infty$

Using the properties of limits, we have that

$\lim_{x\to \infty} \sinh(x) =\frac{1}{2}(\lim_{x\to \infty}e^{x}-\lim_{x\to \infty}e^{-x})=(\infty -0) = \infty$

$\lim_{x\to \infty} \cosh(x) =\frac{1}{2}(\lim_{x\to \infty}e^{x}+\lim_{x\to \infty}e^{-x}) =(\infty -0)= \infty$

$\lim_{x\to -\infty} \sinh(x) =\frac{1}{2}(\lim_{x\to -\infty}e^{x}-\lim_{x\to -\infty}e^{-x}) = (0-\infty)=-\infty$

$\lim_{x\to -\infty} \cosh(x) =\frac{1}{2}(\lim_{x\to -\infty}e^{x}+\lim_{x\to -\infty}e^{-x}) =(0+\infty)= \infty$

8 0

3 years ago

Find the area of the surface. The part of the surface z = xy that lies within the cylinder x2 + y2 = 36.

rewona [7]

Answer:

Step-by-step explanation:

From the given information:

The domain D of integration in polar coordinates can be represented by:

D = {(r,θ)| 0 ≤ r ≤ 6, 0 ≤ θ ≤ 2π) &;

The partial derivates for z = xy can be expressed as:

$y =\dfrac{\partial z}{\partial x} , x = \dfrac{\partial z}{\partial y}$

Thus, the area of the surface is as follows:

$\iint_D \sqrt{(\dfrac{\partial z}{\partial x})^2+ (\dfrac{\partial z}{\partial y})^2 +1 }\ dA = \iint_D \sqrt{(y)^2+(x)^2+1 } \ dA$

$= \iint_D \sqrt{x^2 +y^2 +1 } \ dA$

$= \int^{2 \pi}_{0} \int^{6}_{0} \ r \sqrt{r^2 +1 } \ dr \ d \theta$

$=2 \pi \int^{6}_{0} \ r \sqrt{r^2 +1 } \ dr$

$= 2 \pi \begin {bmatrix} \dfrac{1}{3}(r^2 +1) ^{^\dfrac{3}{2}} \end {bmatrix}^6_0$

$= 2 \pi \times \dfrac{1}{3} \Bigg [ (37)^{3/2} - 1 \Bigg]$

$= \dfrac{2 \pi}{3} \Bigg [37 \sqrt{37} -1 \Bigg ]$

3 0

3 years ago

plz help I will mark brainliest if I can

MrRissso [65]

Figure out the density for one litre and then figure out how many litres in a gallon and calculate it like that:)

4 0

3 years ago

Ted creates a box plot using 14, 13, 21, 10, 28, 30, and 35 as the data. Which of the following box plots shows the data accurat

Slav-nsk [51]

<span> A box plot is drawn with end points at 10 and 35.The box extends from 13 to 30 and a vertical line is drawn inside the box at 21.</span>

8 0

4 years ago

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2(3 + x) – x(7 + x)<br> if x = 3<br> Pls someone answer and work

diamong [38]

Answer:

the answer would be -18

Step-by-step explanation:

3+3=6 2×6=12. 7+3=10 3×10=30 12-30= -18

7 0

3 years ago

Read 2 more answers