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2 years ago

PLEASE HELP!!! Which graph shows a function whose inverse is also a function?

Mathematics

2 answers:

RideAnS [48]2 years ago

4 0

Answer:graph 3

Step-by-step explanation:

Vika [28.1K]2 years ago

3 0

Answer:

Graph 3

Step-by-step explanation:

A function is a relation between to sets, Domain and Range, such that for every element in the domain there is one and only one image in the range. This definition implies basically that, given the x's in the domain, any value x on the domain must have only one image. Thus, lets analyze the graphs and identify which has a f^(-1)(x) that fulfills this condition.

In the first we can see that, for example, x=-3 has two images one of about -0.3 and other of about 0.5. We notice that because the curve passes trough x=-3 more than one time. So, this is not a function.

In the second. Again, we can see that there are values of x with two images. More precisely, every x < 1 has to images, one positive and one negative. If you draw a vertical line in every x<1 you can see it will cut the curve twice, while a function only does it once.

In the third we do have a function. Pick f^-1 and try to draw a vertical line in every x, you will see that it only cuts f^1 once, so every x has only one image. This is because we are working with linear functions. Linear functions (with slope different from 0) always have inverse functions in its domain.

Finally, the fourth graph does not has a inverse function. Similar to what happened in graph 2, take every x > 1 and see that those x will have 2 images by tracing a vertical line in every x.

So, only graph 3 has an inverse function.

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So increased weight = 92-80 = 12

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so his weight is increased by 15℅ !!

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2 years ago

I need help on #3 I don't understand systems of linear inequalities since I wasn't there when my teacher explained it

photoshop1234 [79]

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3 years ago

A certain cold remedy has an 88% rate of success of reducing symptoms within 24 hours. Find the probability that in a random sam

Hitman42 [59]

Answer:

The probability of cured people in who took the remedy is 8/9.

Step-by-step explanation:

Success rate of the cold remedy = 88%

The number of people who took the remedy = 45

Now, 88% of 45 = $\frac{88}{100} \times 45 = 39.6$

and 39.6 ≈ 40

So, out of 45 people, the remedy worked on total 40 people.

Now, let E: Event of people being cured by cold remedy

Favorable outcomes = 40

$\textrm{Probability of Event E} = \frac{\textrm{Total number of favorable outcomes}}{\textrm{Total outcomes}}$

or, $\textrm{Probability of people getting cured} = \frac{\textrm{40}}{\textrm{45}}$ = $\frac{8}{9}$

Hence, the probability of cured people in who took the remedy is 8/9.

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2 years ago

The number line shows the locations of points P 0. R. and S 털 HR & -7 -6 = Which points have a distance of 5 units between t

Svetradugi [14.3K]

Answer: point P and point S

Step-by-step explanation:

took the test

6 0

2 years ago

Which of the following is not one of the 8th roots of unity?

Anika [276]

Answer:

1+i

Step-by-step explanation:

To find the 8th roots of unity, you have to find the trigonometric form of unity.

1. Since $z=1=1+0\cdot i,$ then

$Rez=1,\\ \\Im z=0$

and

$|z|=\sqrt{1^2+0^2}=1,\\ \\\\\cos\varphi =\dfrac{Rez}{|z|}=\dfrac{1}{1}=1,\\ \\\sin\varphi =\dfrac{Imz}{|z|}=\dfrac{0}{1}=0.$

This gives you $\varphi=0.$

Thus,

$z=1\cdot(\cos 0+i\sin 0).$

2. The 8th roots can be calculated using following formula:

$\sqrt[8]{z}=\{\sqrt[8]{|z|} (\cos\dfrac{\varphi+2\pi k}{8}+i\sin \dfrac{\varphi+2\pi k}{8}), k=0,\ 1,\dots,7\}.$

Now

at k=0, $z_0=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 0}{8}+i\sin \dfrac{0+2\pi \cdot 0}{8})=1\cdot (1+0\cdot i)=1;$

at k=1, $z_1=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 1}{8}+i\sin \dfrac{0+2\pi \cdot 1}{8})=1\cdot (\dfrac{\sqrt{2}}{2}+i\dfrac{\sqrt{2}}{2})=\dfrac{\sqrt{2}}{2}+i\dfrac{\sqrt{2}}{2};$

at k=2, $z_2=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 2}{8}+i\sin \dfrac{0+2\pi \cdot 2}{8})=1\cdot (0+1\cdot i)=i;$

at k=3, $z_3=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 3}{8}+i\sin \dfrac{0+2\pi \cdot 3}{8})=1\cdot (-\dfrac{\sqrt{2}}{2}+i\dfrac{\sqrt{2}}{2})=-\dfrac{\sqrt{2}}{2}+i\dfrac{\sqrt{2}}{2};$

at k=4, $z_4=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 4}{8}+i\sin \dfrac{0+2\pi \cdot 4}{8})=1\cdot (-1+0\cdot i)=-1;$

at k=5, $z_5=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 5}{8}+i\sin \dfrac{0+2\pi \cdot 5}{8})=1\cdot (-\dfrac{\sqrt{2}}{2}-i\dfrac{\sqrt{2}}{2})=-\dfrac{\sqrt{2}}{2}-i\dfrac{\sqrt{2}}{2};$

at k=6, $z_6=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 6}{8}+i\sin \dfrac{0+2\pi \cdot 6}{8})=1\cdot (0-1\cdot i)=-i;$

at k=7, $z_7=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 7}{8}+i\sin \dfrac{0+2\pi \cdot 7}{8})=1\cdot (\dfrac{\sqrt{2}}{2}-i\dfrac{\sqrt{2}}{2})=\dfrac{\sqrt{2}}{2}-i\dfrac{\sqrt{2}}{2};$

The 8th roots are

$\{1,\ \dfrac{\sqrt{2}}{2}+i\dfrac{\sqrt{2}}{2},\ i, -\dfrac{\sqrt{2}}{2}+i\dfrac{\sqrt{2}}{2},\ -1, -\dfrac{\sqrt{2}}{2}-i\dfrac{\sqrt{2}}{2},\ -i,\ \dfrac{\sqrt{2}}{2}-i\dfrac{\sqrt{2}}{2}\}.$

Option C is icncorrect.

5 0

2 years ago