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3 years ago

Solve −x16+1≥−5x2, for x without multiplying by a negative number. Then, solve by multiplying through by −16.

Mathematics

1 answer:

dedylja [7]3 years ago

7 0

Answer:

−1.13348172≤x≤1.13348172

Step-by-step explanation:

For this problem

Add −5x2 to both sides of the inequality.

−x16 + 1 + 5x2≥0

Convert inequality to an equation.

−x16 + 1 + 5x2 = 0

Factor −x16 + 1 + 5x2

Set the factor equal to 00.

x16−5x2−1 = 0

x≈ − 1.13348172,1.13348172

so,

x≤ − 1,13348172

−1.13348172≤x≤1.13348172

x≥1.13348172

−1.13348172≤x≤1.13348172

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3 years ago

a rectangular parking lot has a lenth that is 6 yard graeter than the width the area of the parking lot is 160 ssquar yardsd. fi

Lubov Fominskaja [6]

Answer:

<h3>Required Answer:-</h3>

Let

the breadth =x

the length =6x

$\sf Area=160yard^2$

As we know that in a rectangle

${\boxed{\sf Area=lb}}$

Substitute the values

${:}\longrightarrow$ $\sf x\times 6x=160$

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7 0

3 years ago

HURRY ASAP!!!

9966 [12]

The answer you’re looking for is 5.00 :)

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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

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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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Answer:

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Step-by-step explanation:

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