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3 years ago

Solve each equation and match the solution. e +8= 31

Mathematics

1 answer:

Leokris [45]3 years ago

8 0

Answer:e=23

Step-by-step explanation:

e+8=31

8-31=23

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Evaluate the expression e^3

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Answer:

The answer is 20.0855

Step-by-step explanation:

Calculate the approximate value

e³ = 20.0855

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The temperature at 7:00 a.m. was -5°F. By 4:00 p.m., the temperature had risen 9

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Answer:

The answer is -7 degrees. :)

Step-by-step explanation:

-5+9-11=7

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4 years ago

Where the points (0,0) Where the x-axis and the y-axis intersect is also called a

I am Lyosha [343]

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3 years ago

PLEASE HELP! URGENT WILL GIVE BRAINLEST! PLEASE ANSWER ALL QUESTIONS!!! IF YOU DON'T I WILL REPORT.

Nesterboy [21]

Answer:

1. b) 135°, 225°

2. c) 0°, 180°

3. a) π/2

4. b) 30°, 150°, 210°, 330°

5. a) 1.25 radians and 4.39 radians

Step-by-step explanation:

Question 1

$\begin{aligned}\cos \theta & = -\dfrac{\sqrt{2}}{2}\\\implies \theta & = \cos^{-1}\left(-\dfrac{\sqrt{2}}{2}\right)\\\\& = 135^{\circ} \pm 360^{\circ}n, 225^{\circ} \pm 360^{\circ}\\\\& = 135^{\circ}, 225^{\circ}\quad \textsf{for}\:0^{\circ}\leq \theta < 360^{\circ}\end{aligned}$

Question 2

$\begin{aligned}\cos^2 \theta-1 & = 0\\\implies \cos^2 \theta & = 1\\\cos \theta & = \pm1\\\theta & = \cos^{-1}(\pm1)\\& = 0^{\circ} \pm 360^{\circ}n, 180^{\circ} \pm360^{\circ}n\\& = 0^{\circ}, 180^{\circ}\quad \textsf{for}\:0^{\circ}\leq \theta < 360^{\circ}\end{aligned}$

Question 3

$\begin{aligned}\sin \theta & = 1\\\implies \theta & = \sin^{-1}(1)\\\theta & = \dfrac{\pi}{2}\pm2\pi n\\\theta & = \dfrac{\pi}{2}\quad \textsf{for}\:0 \leq \theta < 2 \pi \end{aligned}$

Question 4

$\begin{aligned}3 \tan^2 \theta-1 & = 0\\\implies \tan^2 \theta & = \dfrac{1}{3}\\\tan \theta & = \pm\dfrac{1}{\sqrt{3}}\\\theta & = \tan^{-1}\left(\pm\dfrac{1}{\sqrt{3}}\right)\\& = 30^{\circ}\pm180^{\circ}n, 150^{\circ}\pm180^{\circ}n\\& = 30^{\circ}, 150^{\circ}, 210^{\circ}, 330^{\circ}\quad \textsf{for}\:0^{\circ}\leq \theta < 360^{\circ}\end{aligned}$

Question 5

$\begin{aligned}2 \tan \theta -6 & = 0\\\implies \tan \theta & = 3\\\theta & = \tan^{-1}(3)\\\theta & = 1.25\pm \pi n\\\theta & = 1.25, 4.39 \quad \textsf{for}\:0^{\circ}\leq \theta < 2 \pi \end{aligned}$

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2 years ago

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