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alexira [117]
3 years ago
6

Need help with all my question if anyone can help I would be very grateful

Mathematics
1 answer:
Step2247 [10]3 years ago
8 0
Is it B -4.44 lucky
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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
Is 45 a multiple of 4? Explain your answer Lmk ASAP
arsen [322]
No 45 is a multiple for 5
6 0
3 years ago
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Determine the intervals on which the function is increasing, decreasing, and constant
max2010maxim [7]
I believe it's increasing on all fronts, because if you start from the right, you see that the y values always increase, hence they are increasing. They do it for when x<0 and when x>0. So, it should be increasing on all real numbers
3 0
3 years ago
Read 2 more answers
Let O be an angle in quadrant III such that cos 0 = -2/5 Find the exact values of csco and tan 0.​
vivado [14]

well, we know that θ is in the III Quadrant, where the sine is negative and the cosine is negative as well, or if you wish, where "x" as well as "y" are both negative, now, the hypotenuse or radius of the circle is just a distance amount, so is never negative, so in the equation of cos(θ) = - (2/5), the negative must be the adjacent side, thus

cos(\theta)=\cfrac{\stackrel{adjacent}{-2}}{\underset{hypotenuse}{5}}\qquad \textit{let's find the \underline{opposite side}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies \sqrt{c^2-a^2}=b \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ \pm\sqrt{5^2 - (-2)^2}=b\implies \pm\sqrt{25-4}\implies \pm\sqrt{21}=b\implies \stackrel{III~Quadrant}{-\sqrt{21}=b}

\dotfill\\\\ csc(\theta)\implies \cfrac{\stackrel{hypotenuse}{5}}{\underset{opposite}{-\sqrt{21}}}\implies \stackrel{\textit{rationalizing the denominator}}{-\cfrac{5}{\sqrt{21}}\cdot \cfrac{\sqrt{21}}{\sqrt{21}}\implies -\cfrac{5\sqrt{21}}{21}} \\\\\\ tan(\theta)=\cfrac{\stackrel{opposite}{-\sqrt{21}}}{\underset{adjacent}{-2}}\implies tan(\theta)=\cfrac{\sqrt{21}}{2}

4 0
2 years ago
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