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Dafna11 [192]
3 years ago
14

You are a member of a darts team and they have won £1750 prize money in a competition. £70 is to be saved for a celebration part

y; the remainder is to be divided equally between the 14 team members. How much will you get?
Mathematics
2 answers:
grandymaker [24]3 years ago
7 0
Most of the information's that are required for solving this question is already given in the question. It is important to check them minutely before trying to solve the problem in hand.
Amount of prize money won by the darts team = 1750 pounds
The amount of money that needs to be saved = 70 pounds
Then
Amount of money remaining after savings = (1750 - 70) pounds
                                                                     = 1680 pounds
The number of  team members in the darts team = 14
So
The amount of money
that each members of the darts team gets = 1680/14
                                                                     = 70 pounds
From the above deduction we can easily conclude that each member of the darts teams will get an amount of 70 pounds.
SashulF [63]3 years ago
7 0

Answer: £120

Step-by-step explanation:

We are given that : You are a member of a darts team and they have won £1750 prize money in a competition.

If £70 is to be saved for a celebration party.

Then, the Money remained : Prize Money-Money saved of party

=£1750-£70 = £1680

Now, if the remainder money is to be divided equally between the 14 team members (including you).

Then,. the amount received by each member :  Money remained ÷ 14

=£1680÷ 14

=£120

Hence, the money you will get = £120

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To solve this equation you would divide both sides by 5 to get 7. s=7 (35/5=7)

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mx = np \\  \\  x \:  =  \frac{np}{m}

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Step-by-step explanation:

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3 years ago
Use the given information to find the exact value of the trigonometric function
eimsori [14]
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Half Angle Formula

\cos \frac{\theta}{2}=\pm\sqrt[\square]{\frac{1+\cos\theta}{2}}\tan \theta>0\text{ and csc}\theta\text{ is negative in the third quadrant}\begin{gathered} \csc \theta=-\frac{6}{5}=\frac{r}{y} \\ x^2+y^2=r^2 \\ x=\pm\sqrt[\square]{r^2-y^2} \\ x=\pm\sqrt[\square]{6^2-(-5)^2} \\ x=\pm\sqrt[\square]{36-25} \\ x=\pm\sqrt[\square]{11} \\ \text{x is negative since the angle is on the 3rd quadrant} \end{gathered}\begin{gathered} \cos \theta=\frac{x}{r}=\frac{-\sqrt[\square]{11}}{6} \\ \cos \frac{\theta}{2}=\pm\sqrt[\square]{\frac{1+\cos\theta}{2}} \\ \cos \frac{\theta}{2}is\text{ also negative in the 3rd quadrant} \\ \cos \frac{\theta}{2}=-\sqrt[\square]{\frac{1+\frac{-\sqrt[\square]{11}}{6}}{2}} \\ \cos \frac{\theta}{2}=-\sqrt[\square]{\frac{\frac{6-\sqrt[\square]{11}}{6}}{2}} \\ \cos \frac{\theta}{2}=-\sqrt[\square]{\frac{6-\sqrt[\square]{11}}{12}} \\  \\  \end{gathered}

Answer:

\cos \frac{\theta}{2}=-\sqrt[\square]{\frac{6-\sqrt[\square]{11}}{12}}

Checking:

\begin{gathered} \frac{\theta}{2}=\cos ^{-1}(-\sqrt[\square]{\frac{6-\sqrt[\square]{11}}{12}}) \\ \frac{\theta}{2}=118.22^{\circ} \\ \theta=236.44^{\circ}\text{  (3rd quadrant)} \end{gathered}

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\csc \theta=\frac{1}{\sin\theta}=\frac{1}{\sin (236.44)}=-\frac{6}{5}\text{ QED}

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1 year ago
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7 0
3 years ago
Read 2 more answers
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