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Aleksandr [31]
3 years ago
12

What number is 4 times as many as 25

Mathematics
2 answers:
Oduvanchick [21]3 years ago
6 0
This is a standard relationship/systems of equations question.  Here is how you attack it.  Firstly, set up equations to represent the relationships. Ed = 4*Kim That shows that Ed has four times as many pins as Kim.  Next, we see that the two have 25 together.  So: Ed + Kim = 25 Now we have our two equations.  In order to solve, we need to get one of the equations down to one variable.  We can achieve this via substitution.  The first equation tells us that we can substitute 4*Kim for Ed (in the second equation).  So, let's do just that: 4*Kim + Kim = 255*Kim = 25Kim = 5 Using the first equation again, we can solve for Ed: Ed = 4*KimEd = 4*(5)<span>Ed = 20</span>
Debora [2.8K]3 years ago
3 0
Just multiply 4 by 25. You get 100.
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Answer:

See Explanation

Step-by-step explanation:

\frac{ \sin5A}{\sin A}  -  \frac{ \cos5A}{\cos A}  = 4\cos2A \\  \\ LHS = \frac{ \sin5A}{\sin A}  -  \frac{ \cos5A}{\cos A}   \\  \\  =  \frac{ \sin5A \:\cos A -  \cos5A \:  \sin A}{\sin A \:\cos A }  \\  \\  =  \frac{ \sin(5A -A )}{\sin A \:\cos A}  \\  \\ =  \frac{ \sin 4A}{\sin A \:\cos A}  \\  \\ =  \frac{ 2\sin 2A \: \cos 2A}{\sin A \:\cos A}  \\  \\ =  \frac{ 2 \times 2\sin A \: \cos A \: \cos 2A}{\sin A \:\cos A}  \\  \\ =  \frac{ 4\sin A \: \cos A \: \cos 2A}{\sin A \:\cos A}  \\  \\ =4\cos 2A \\  \\  = RHS \\  \\ thus \\  \\  \frac{ \sin5A}{\sin A}  -  \frac{ \cos5A}{\cos A}  = 4\cos2A \\  \\ hence \: proved

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