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scoundrel [369]
3 years ago
15

What's 955 ÷ 35 ?

align="absmiddle" class="latex-formula">
​
Mathematics
2 answers:
klasskru [66]3 years ago
5 0

Answer:

The answer is 27.28571429

If the answer helps you PLEASE mark my answer as brainliest

Stay Safe,Stay Healthy and Stay Happy

Maru [420]3 years ago
3 0

Answer:

Hello

Step-by-step explanation:

955÷35

27.2857142857

Or

35/955

35)955(27

-945

________

10

I hope this works♥️

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Find e^cos(2+3i) as a complex number expressed in Cartesian form.
ozzi

Answer:

The complex number e^{\cos(2+31)} = \exp(\cos(2+3i)) has Cartesian form

\exp\left(\cosh 3\cos 2\right)\cos(\sinh 3\sin 2)-i\exp\left(\cosh 3\cos 2\right)\sin(\sinh 3\sin 2).

Step-by-step explanation:

First, we need to recall the definition of \cos z when z is a complex number:

\cos z = \cos(x+iy) = \frac{e^{iz}+e^{-iz}}{2}.

Then,

\cos(2+3i) = \frac{e^{i(2+31)} + e^{-i(2+31)}}{2} = \frac{e^{2i-3}+e^{-2i+3}}{2}. (I)

Now, recall the definition of the complex exponential:

e^{z}=e^{x+iy} = e^x(\cos y +i\sin y).

So,

e^{2i-3} = e^{-3}(\cos 2+i\sin 2)

e^{-2i+3} = e^{3}(\cos 2-i\sin 2) (we use that \sin(-y)=-\sin y).

Thus,

e^{2i-3}+e^{-2i+3} = e^{-3}\cos 2+ie^{-3}\sin 2 + e^{3}\cos 2-ie^{3}\sin 2)

Now we group conveniently in the above expression:

e^{2i-3}+e^{-2i+3} = (e^{-3}+e^{3})\cos 2 + i(e^{-3}-e^{3})\sin 2.

Now, substituting this equality in (I) we get

\cos(2+3i) = \frac{e^{-3}+e^{3}}{2}\cos 2 -i\frac{e^{3}-e^{-3}}{2}\sin 2 = \cosh 3\cos 2-i\sinh 3\sin 2.

Thus,

\exp\left(\cos(2+3i)\right) = \exp\left(\cosh 3\cos 2-i\sinh 3\sin 2\right)

\exp\left(\cos(2+3i)\right) = \exp\left(\cosh 3\cos 2\right)\left[ \cos(\sinh 3\sin 2)-i\sin(\sinh 3\sin 2)\right].

5 0
2 years ago
Part of the graph of the function f(x) = (x – 1)(x + 7) is shown below.Which statements about the function are true? Select thre
sertanlavr [38]

The statements about the function that are true are

  • The vertex of the function is at (–3,–16).
  • The graph is increasing on the interval x > –3.
  • The graph is positive only on the intervals where x < –7 and where x > 1.

<h3>How to interpret the graph?</h3>

The equation of the graph is given as

f(x) = (x - 1)(x + 7)

The graph of the function is added as an attachment

On the graph, we have

Vertex = (-3, -16)

This is so because the graph has a minimum point of (-3, -16)

Also, the graph crosses the x-axis at x = 1 and x = -7

This means that the graph is positive at x >1 and x >-7

Because the vertex is a minimum, the graph is increasing at the left of its symmetry i.e. x > -3

Read more about quadratic functions at

brainly.com/question/14933288

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7 0
1 year ago
What are the endpoint coordinates for the midsegment of △BCD that is parallel to BC?
natulia [17]

Answer:

<em>The endpoint coordinates for the mid-segment are:  (-2,-1) and (1,0)</em>

Step-by-step explanation:

According to the given diagram, the coordinates of the vertices of \triangle BCD are:   B(-3,1), C(3,3) and D(-1,-3)

Now, the endpoints of the mid-segment of \triangle BCD which is parallel to BC will be the mid-points of sides BD and CD.

<u>Formula for the coordinate of mid-point</u> :   (\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}), where (x_{1}, y_{1}) and (x_{2}, y_{2}) are two endpoints.

So, the mid-point of BD will be:  (\frac{-3-1}{2},\frac{1-3}{2})=(\frac{-4}{2},\frac{-2}{2})=(-2,-1)

and the mid-point of CD will be:  (\frac{3-1}{2},\frac{3-3}{2})=(\frac{2}{2},\frac{0}{2})=(1,0)

Thus, the endpoint coordinates for the mid-segment of \triangle BCD that is parallel to BC are  (-2,-1) and (1,0)

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2 years ago
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Answer:

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2 years ago
You are at a restaurant and the check comes to a total of $75. If you want to leave an 18% tip, how much total money should you
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Answer:

93

Step-by-step explanation:

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2 years ago
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