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vlabodo [156]
3 years ago
11

Evaluate the expression.

Mathematics
1 answer:
chubhunter [2.5K]3 years ago
4 0

um is this all one equation?

like -2.2×(-2)÷(-14)×5-2.2×(-2)÷(-41)×5?

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No explanation required, just a quick answer please!
hram777 [196]
The answer to the question

4 0
3 years ago
Yup there's more pls help
Artemon [7]
Top half:
D (x-2, y+3) E (x-3, y(+\-) 0) F (x+1, y-2)
P (x+1, y+2) Q (x+3, y+4) R (x+5, y+2)
7 0
3 years ago
24y - 22 = 4 ( 6y - 6 )
Pie
I don’t think their is a solution to this equation
because if you expand the second half it is= 24y-24 which would make the equation
- 24y-22=24y-24
and because the number next to the y is the same on both sides, no matter what y is if we subtract different numbers from each side we will never get the same value for each side of the =
5 0
3 years ago
The school that Michael goes to is selling tickets to a dance. On the first day of ticket sales the school sold 3 staff tickets
Minchanka [31]

The price of a staff ticket and the price of a student ticket is $8 and $14

Given:

Day 1:

Number of staff tickets sold = 3

Number of students tickets sold = 1

Total revenue day 1 = $38

Day 2:

<em>Number of staff tickets sold</em> = 3

<em>Number of students tickets sold</em> = 2

<em>Total revenue day</em> 2 = $52

let

<em>cost of staff tickets</em> = x

<em>cost of students tickets</em> = y

The equation:

<em>3x + y = 38 (1)</em>

<em>3x + y = 38 (1)3x + 2y = 52 (2)</em>

subtract (1) from (2)

2y - y = 52 - 38

y = 14

substitute y = 14 into (1)

3x + y = 38 (1)

3x + 14 = 38

3x = 38 - 14

3x = 24

x = 24/3

x = 8

Therefore,

cost of staff tickets = x

= $8

cost of students tickets = y

= $14

Read more:

brainly.com/question/22940808

8 0
3 years ago
Math:
Dahasolnce [82]

Answer:

a) x_{1} = \frac{\pi}{3}\pm 2\pi \cdot i, \forall i \in \mathbb{N}_{O}, x_{2} = \frac{5\pi}{6}\pm 2\pi\cdot i, \forall i \in \mathbb{N}_{O}, b) x_{1} = \frac{\pi}{3}\pm 2\pi \cdot i, \forall i \in \mathbb{N}_{O}, x_{2} = \frac{5\pi}{3}\pm 2\pi\cdot i, \forall i \in \mathbb{N}_{O}

Step-by-step explanation:

a) The equation must be rearranged into a form with one fundamental trigonometric function first:

\sqrt{3}\cdot \csc x - 2 = 0

\sqrt{3} \cdot \left(\frac{1}{\sin x} \right) - 2 = 0

\sqrt{3} - 2\cdot \sin x = 0

\sin x = \frac{\sqrt{3}}{2}

x = \sin^{-1} \frac{\sqrt{3}}{2}

Value of x is contained in the following sets of solutions:

x_{1} = \frac{\pi}{3}\pm 2\pi \cdot i, \forall i \in \mathbb{N}_{O}

x_{2} = \frac{5\pi}{6}\pm 2\pi\cdot i, \forall i \in \mathbb{N}_{O}

b) The equation must be simplified first:

\cos x + 1 = - \cos x

2\cdot \cos x = -1

\cos x = -\frac{1}{2}

x = \cos^{-1} \left(-\frac{1}{2} \right)

Value of x is contained in the following sets of solutions:

x_{1} = \frac{\pi}{3}\pm 2\pi \cdot i, \forall i \in \mathbb{N}_{O}

x_{2} = \frac{5\pi}{3}\pm 2\pi\cdot i, \forall i \in \mathbb{N}_{O}

7 0
4 years ago
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