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3 years ago

Help me on number problem 4 please.

Mathematics

2 answers:

VashaNatasha [74]3 years ago

6 0

J =jaylens score

T = trevors score

J = T-25

J+T = 485

Substitute into above equation

(T-25) +T = 485

combine like terms

2T-25 = 485

add 25 to each side

2T = 510

divide by 2

T = 255

Trever scored 255

J = 255 -25

J= 230

Scores : Trever:255 Jaylen:230

Tresset [83]3 years ago

3 0

The answer is 460 because you have to subtract

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GrogVix [38]

Answer:

$\frac{\cos\left(2x\right)-\cos\left(4x\right)}{\sin\left(2x\right)+\sin\left(4x\right)}=\tan\left(x\right)$

Step-by-step explanation:

$\frac{\cos\left(2x\right)-\cos\left(4x\right)}{\sin\left(2x\right)+\sin\left(4x\right)}$

Apply formula:

$\cos\left(A\right)-\cos\left(B\right)=-2\cdot\sin\left(\frac{A+B}{2}\right)\cdot\sin\left(\frac{A-B}{2}\right)$ and

$\sin\left(A\right)+\sin\left(B\right)=2\cdot\sin\left(\frac{A+B}{2}\right)\cdot\sin\left(\frac{A-B}{2}\right)$

We get:

$=\frac{-2\cdot\sin\left(\frac{2x+4x}{2}\right)\cdot\sin\left(\frac{2x-4x}{2}\right)}{2\cdot\sin\left(\frac{2x+4x}{2}\right)\cdot\cos\left(\frac{2x-4x}{2}\right)}$

$=\frac{-\sin\left(\frac{2x-4x}{2}\right)}{\cos\left(\frac{2x-4x}{2}\right)}$

$=\frac{-\sin\left(\frac{-2x}{2}\right)}{\cos\left(\frac{-2x}{2}\right)}$

$=\frac{-\sin\left(-x\right)}{\cos\left(-x\right)}$

$=\frac{-\cdot-\sin\left(x\right)}{\cos\left(x\right)}$

$=\frac{\sin\left(x\right)}{\cos\left(x\right)}$

$=\tan\left(x\right)$

Hence final answer is

$\frac{\cos\left(2x\right)-\cos\left(4x\right)}{\sin\left(2x\right)+\sin\left(4x\right)}=\tan\left(x\right)$

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Its b
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