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3 years ago

What function is the inverse of the exponential function y = 3x?

Mathematics

2 answers:

lukranit [14]3 years ago

7 0

Answer:I dont know not gonna lie

Step-by-step explanation:

madam [21]3 years ago

4 0

Answer:

y = $\frac{(x + 2)}{3}$

Step-by-step explanation:

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Solve the following proportion for x. 3/13 = x/5 Round your answer to the nearest tenth.

anastassius [24]

To solve, isolate the x. Cross multiply

3/13(13)(5) = x/5(5)(13)

3(5) = (13)x

Simplify

13x = 15

Isolate the x. Divide by 13.

13x/13 = 15/13

x = 15/13

x = 1.2

1.2 is your answer

hope this helps

8 0

2 years ago

What is the sum of 1/4 and 3/8

kupik [55]

$\dfrac{1}{4}+\dfrac{3}{8}=(*)\\Find\ LCD\ (Least\ Common\ Denominator)\\list\ the\ multiples\\of\ 4:\ \ 0;\ 4;\ \fbox8;\ 12;...\\of\ 8:\ \ 0;\ \fbox8;\ 16;\ 24;\ ...\\LCD\left(\dfrac{1}{4};\ \dfrac{3}{8}\right)=8\\\\(*)=\dfrac{1\cdot2}{4\cdot2}+\dfrac{3}{8}=\dfrac{2}{8}+\dfrac{3}{8}=\dfrac{2+3}{8}=\dfrac{5}{8}$

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3 years ago

Answer quickkkkkkkkkkk

Hoochie [10]

Answer:

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Step-by-step explanation:

6 0

2 years ago

Read 2 more answers

Whats is 2400000 in standard form

Ghella [55]

The best answer would be
2.4 x 10 ^6
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3 years ago

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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

3 years ago