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3 years ago

Which represents the solution set of the inequality 5x

1 answer:

8 0

<span>"Which represents the solution set of the inequality 5x" is not clear.
The inequation is 5<x, 2x-4>0, or x+y<3.......</span>

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What are my zeros (x^2 - 8x + 1)?

Pavel [41]

Answer:

x= 7.87298334621 x=.12701665379

Step-by-step explanation:

This is in standard form so we have to factor this into

x^2-8x + 16 + 1 = 16

(x-4)^2 = 15

(x-4) = $\sqrt{15\\}$

x-4= ± 3.87298334621

x-4 = 3.87298334621 x-4= -3.87298334621

x= 7.87298334621 x=.12701665379

7 0

3 years ago

My estimate was 2,808 is that a reasonable estimate and why

vova2212 [387]

<span>My estimate was 2,808 is that a reasonable estimate and why

312*9

9*2=18>>9*10=90>>9*300=2700

2700+18+90=>>2,808</span>

5 0

3 years ago

You spin a spinner, flip a coin, then spin the spinner again. Find the probability of spinning an odd number, flipping heads, th

vovangra [49]

Answer:

1/8

Step-by-step explanation:

you would multiply 5/10x1/2x5/1o. Since the spinner has 10 areas, 5 of them are odd and 5 are even and a coin has 2 sides, head or tails.

8 0

3 years ago

Given tan theta =9, use trigonometric identities to find the exact value of each of the following:_______

Ludmilka [50]

Answer:

$(a)\ \sec^2(\theta) = 82$

$(b)\ \cot(\theta) = \frac{1}{9}$

$(c)\ \cot(\frac{\pi}{2} - \theta) = 9$

$(d)\ \csc^2(\theta) = \frac{82}{81}$

Step-by-step explanation:

Given

$\tan(\theta) = 9$

Required

Solve (a) to (d)

Using tan formula, we have:

$\tan(\theta) = \frac{Opposite}{Adjacent}$

This gives:

$\frac{Opposite}{Adjacent} = 9$

Rewrite as:

$\frac{Opposite}{Adjacent} = \frac{9}{1}$

Using a unit ratio;

$Opposite = 9; Adjacent = 1$

Using Pythagoras theorem, we have:

$Hypotenuse^2 = Opposite^2 + Adjacent^2$

$Hypotenuse^2 = 9^2 + 1^2$

$Hypotenuse^2 = 81 + 1$

$Hypotenuse^2 = 82$

Take square roots of both sides

$Hypotenuse =\sqrt{82}$

So, we have:

$Opposite = 9; Adjacent = 1$

$Hypotenuse =\sqrt{82}$

Solving (a):

$\sec^2(\theta)$

This is calculated as:

$\sec^2(\theta) = (\sec(\theta))^2$

$\sec^2(\theta) = (\frac{1}{\cos(\theta)})^2$

Where:

$\cos(\theta) = \frac{Adjacent}{Hypotenuse}$

$\cos(\theta) = \frac{1}{\sqrt{82}}$

So:

$\sec^2(\theta) = (\frac{1}{\cos(\theta)})^2$

$\sec^2(\theta) = (\frac{1}{\frac{1}{\sqrt{82}}})^2$

$\sec^2(\theta) = (\sqrt{82})^2$

$\sec^2(\theta) = 82$

Solving (b):

$\cot(\theta)$

This is calculated as:

$\cot(\theta) = \frac{1}{\tan(\theta)}$

Where:

$\tan(\theta) = 9$ ---- given

So:

$\cot(\theta) = \frac{1}{\tan(\theta)}$

$\cot(\theta) = \frac{1}{9}$

Solving (c):

$\cot(\frac{\pi}{2} - \theta)$

In trigonometry:

$\cot(\frac{\pi}{2} - \theta) = \tan(\theta)$

Hence:

$\cot(\frac{\pi}{2} - \theta) = 9$

Solving (d):

$\csc^2(\theta)$

This is calculated as:

$\csc^2(\theta) = (\csc(\theta))^2$

$\csc^2(\theta) = (\frac{1}{\sin(\theta)})^2$

Where:

$\sin(\theta) = \frac{Opposite}{Hypotenuse}$

$\sin(\theta) = \frac{9}{\sqrt{82}}$

So:

$\csc^2(\theta) = (\frac{1}{\frac{9}{\sqrt{82}}})^2$

$\csc^2(\theta) = (\frac{\sqrt{82}}{9})^2$

$\csc^2(\theta) = \frac{82}{81}$

4 0

3 years ago

What is the sine value of 7 pi over 6? negative square root 3 over 2 square root 3 over 2 negative 1 over 2 1 over 2?

djverab [1.8K]

First thing to do is to change the radians to degrees so it's easier to determine our angle and where it lies in the coordinate plane. $\frac{7 \pi }{6} * \frac{180}{ \pi }=210$ . If we sweep out a 210 degree angle, we end up in the third quadrant, with a 30 degree angle. In this quadrant, x and y are both negative, but the hypotenuse, no matter where it is, will never ever be negative. So the side across from the 30 degree reference angle is -1, and the hypotenuse is 2, so the sine of this angle, opposite over hypotenuse, is -1/2

6 0

3 years ago

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