All categories

2 years ago

How to solve the inequality

Mathematics

1 answer:

GaryK [48]2 years ago

4 0

The problem is not shown clearly therefore i am not sure if my answer will be correct, but, this inequality is already solved. all this means is that y is more than or equal to 4.5 (meaning that y can be 4.5, 5, 6, 7, 8 and so own but never a number less than 4.5), the "I I" only mean that y will never be a negative.

You might be interested in

What is the equation of the line represented by the table below? X -2 -1 0 1 2 - y 16 11 6 1 -4 O A. y = -5x+9 OB. y = -5x+6 O C

Nina [5.8K]

Option (c) is the correct answer

The equation y = -5x + 6 is represented by the given table i.e.

x -2 -1 0 1 2

y 16 11 6 1 -4

Given, a table

x -2 -1 0 1 2

y 16 11 6 1 -4

Now, we have to find the equation of the line, satisfying the given values

As, the equation

y = -5x + 6 satisfying the given set of values,

So, the equation y = -5x + 6 is represented by the given table.

Hence, the equation y = -5x + 6 is represented by the given table.

Learn more about Linear Equations here here brainly.com/question/2030026

#SPJ9

5 0

1 year ago

PLEASE HELP WILL MARK BRAINLIEST!!

noname [10]

Answer:

m∠A = 91°

m∠B = 146°

m∠C = 89°

m∠D = 34°

Step-by-step explanation:

If the four vertices of a quadrilateral lie on the edge of a circle, then this quadrilateral is called cyclic quadrilateral
In the cyclic quadrilateral each two opposite angles are supplementary (means the sum of their measures is 180°)
The sum of the measures of the interior angles of any quadrilateral is 360°

In quadrilateral ABCD

∵ A, B, C, And D lie on the circumference of the circle

∴ ABCD is a cyclic quadrilateral

∴ The sum of the measures of each opposite angles is 180°

∵ ∠A and ∠C are opposite angle in the cyclic quadrilateral ABCD

∴ m∠A + m∠C = 180°

∵ m∠A = (2x + 3)°

∵ m∠C = (2x + 1)°

- Add them and equate the answer by 180

∴ (2x + 3) + (2x + 1) = 180

- Add the like terms in the left hand side

∴ 4x + 4 = 180

- Subtract 4 from both sides

∴ 4x = 176

- Divide both sides by 4

∴ x = 44

Substitute the value of x in the expressions of angle A, C, D

∵ m∠A = 2(44) + 3 = 88 + 3

∴ m∠A = 91°

∵ m∠C = 2(44) + 1 = 88 + 1

∴ m∠C = 89°

∵ m∠D = x - 10

∴ m∠D = 44 - 10

∴ m∠D = 34°

- ∠B and ∠D are opposite angles in the cyclic quadrilateral ABCD

∴ m∠B + m∠D = 180°

∴ m∠B + 34 = 180

- Subtract 34 from both sides

∴ m∠B = 146°

5 0

2 years ago

F(×)= 5×-10. What is the vaule of f^-1 (-3)

Katena32 [7]

If I’m not mistaken I think this is what you’re asking for?

4 0

2 years ago

On Tuesday Katy ran a mile in 8.34 minutes. She ran another mile in 7.89 minutes on Wednesday. What is the total time for both m

miv72 [106K]

She ran a total time of 16.23 minutes

3 0

3 years ago

The given matrix is the augmented matrix for a linear system. Use technology to perform the row operations needed to transform t

shtirl [24]

Answer:

$x_{1} = \frac{176}{127} + \frac{71}{127}x_{4}\\\\ x_{2} = \frac{284}{127} + \frac{131}{254}x_{4}\\\\x_{3} = \frac{845}{127} + \frac{663}{254}x_{4}\\$

Step-by-step explanation:

As the given Augmented matrix is

$\left[\begin{array}{ccccc}9&-2&0&-4&:8\\0&7&-1&-1&:9\\8&12&-6&5&:-2\end{array}\right]$

Step 1 :

$r_{1}$ ↔ $r_{1} - r_{2}$

$\left[\begin{array}{ccccc}1&-14&6&-9&:10\\0&7&-1&-1&:9\\8&12&-6&5&:-2\end{array}\right]$

Step 2 :

$r_{3}$ ↔ $r_{3} - 8r_{1}$

$\left[\begin{array}{ccccc}1&-14&6&-9&:10\\0&7&-1&-1&:9\\0&124&-54&77&:-82\end{array}\right]$

Step 3 :

$r_{2}$ ↔ $\frac{r_{2}}{7}$

$\left[\begin{array}{ccccc}1&-14&6&-9&:10\\0&1&-\frac{1}{7} &-\frac{1}{7} &:\frac{9}{7} \\0&124&-54&77&:-82\end{array}\right]$

Step 4 :

$r_{1}$ ↔ $r_{1} + 14r_{2}$ , $r_{3}$ ↔ $r_{3} - 124r_{2}$

$\left[\begin{array}{ccccc}1&0&4&-11&:-8\\0&1&-\frac{1}{7} &-\frac{1}{7} &:\frac{9}{7} \\0&0&- \frac{254}{7} &\frac{663}{7} &:-\frac{1690}{7} \end{array}\right]$

Step 5 :

$r_{3}$ ↔ $\frac{r_{3}. 7}{254}$

$\left[\begin{array}{ccccc}1&0&4&-11&:-8\\0&1&-\frac{1}{7} &-\frac{1}{7} &:\frac{9}{7} \\0&0&1&-\frac{663}{254} &:-\frac{1690}{254} \end{array}\right]$

Step 6 :

$r_{1}$ ↔ $r_{1} - 4r_{3}$ , $r_{2}$ ↔ $r_{2} + \frac{1}{7} r_{3}$

$\left[\begin{array}{ccccc}1&0&0&-\frac{71}{127} &:\frac{176}{127} \\0&1&0&-\frac{131}{254} &:\frac{284}{127} \\0&0&1&-\frac{663}{254} &:\frac{845}{127} \end{array}\right]$

∴ we get

$x_{1} = \frac{176}{127} + \frac{71}{127}x_{4}\\\\ x_{2} = \frac{284}{127} + \frac{131}{254}x_{4}\\\\x_{3} = \frac{845}{127} + \frac{663}{254}x_{4}\\$

6 0

2 years ago