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3 years ago

50 POINTS AND BRAINLIEST

Mathematics

1 answer:

algol [13]3 years ago

8 0

From the left $(x\to2^-)$ the function is clearly approaching 5, while from the right $(x\to2^+)$ it is approaching -2.

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Select the correct answer.

ad-work [718]

Answer:

A' (-1, 2)

Step-by-step explanation:

(x, y) -> (-y, x)

A' (-1, 2)

B' (1, -2)

C' (2, -2)

D' (0, -2)

3 0

3 years ago

A) Let f(x) = 4x + 2 and g(x) = 2x^2-4. Find the formula for the composition function gof

ollegr [7]

Answer:

g o f = $g(f(x)) = 32x^2 + 32x + 4$

Step-by-step explanation:

Given

$f(x) = 4x +2$

$g(x) =2x^2 - 4$

Required:

Find g o f

This is calculated as:

$gof = g(f(x))$

$g(x) =2x^2 - 4$

So:

$g(f(x)) = 2(4x+2)^2 - 4$

$g(f(x)) = 2(4x+2)(4x+2) - 4$

$g(f(x)) = 2[ 16x^2 + 16x + 4)] - 4$

$g(f(x)) = 32x^2 + 32x + 8 - 4$

$g(f(x)) = 32x^2 + 32x + 4$

5 0

3 years ago

Answer this question please; number 5... show all work thank you

My name is Ann [436]

Answer:

rate of the plane in still air is 33 miles per hour and the rate of the wind is 11 miles per hour

Step-by-step explanation:

We will make a table of the trip there and back using the formula distance = rate x time

d = r x t

there

back

The distance there and back is 264 miles, so we can split that in half and put each half under d:

d = r x t

there 132

back 132

It tells us that the trip there is with the wind and the trip back is against the wind:

d = r x t

there 132 = (r + w)

back 132 = (r - w)

Finally, the trip there took 3 hours and the trip back took 6:

d = r * t

there 132 = (r + w) * 3

back 132 = (r - w) * 6

There's the table. Using the distance formula we have 2 equations that result from that info:

132 = 3(r + w) and 132 = 6(r - w)

We are looking to solve for both r and w. We have 2 equations with 2 unknowns, so we will solve the first equation for r, sub that value for r into the second equation to solve for w:

132 = 3r + 3w and

132 - 3w = 3r so

44 - w = r. Subbing that into the second equation:

132 = 6(44 - w) - 6w and

132 = 264 - 6w - 6w and

-132 = -12w so

w = 11

Subbing w in to solve for r:

132 = 3r + 3(11) and

132 = 3r + 33 so

99 = 3r and

r = 33

7 0

3 years ago

Double the difference of a number and seven

KengaRu [80]

Answer:

2x-14

Step-by-step explanation:

2(x-7)

distribute

2x-14

4 0

3 years ago

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

3 years ago