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2 years ago

-8 ( y + 4 ) = 10y + 4

2 answers:

8 0

-8y-32 = 10y+4 Subract 10y on both sides.
-8y-10-32 = 4 Add 32 on both sides.
-18y = 4+32
-18y = 36 Divide 18 on both sides.
-18/y = 36/-18
y = -2

maks197457 [2]2 years ago

7 0

Let's solve your equation step-by-step.−8(y+4)=10y+4

Step 1: Simplify both sides of the equation.−8(y+4)=10y+4

Simplify: (−8)(y)+(−8)(4)=10y+4(Distribute)−8y+−32=10y+4−8y−32=10y+4

Step 2: Subtract 10y from both sides.−8y−32−10y=10y+4−10y−18y−32=4

Step 3: Add 32 to both sides.−18y−32+32=4+32−18y=36

Step 4: Divide both sides by -18.−18y−18=36−18y=−2

Answer:y=−2

Good luck mate :P

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2 years ago

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Answer:

Step-by-step explanation:

We are to show that $\sqrt{3} cosec140^{0} - sec140^{0} = 4\\$

Proof:

From trigonometry identity;

$cosec \theta = \frac{1}{sin\theta} \\sec\theta = \frac{1}{cos\theta}$

$\sqrt{3} cosec140^{0} - sec140^{0} \\= \frac{\sqrt{3} }{sin140} - \frac{1}{cos140} \\= \frac{\sqrt{3}cos140-sin140 }{sin140cos140} \\$

From trigonometry, 2sinAcosA = Sin2A

$= \frac{\sqrt{3}cos140-sin140 }{sin140cos140} \\\\= \frac{\sqrt{3}cos140-sin140 }{sin280/2}\\= \frac{4(\sqrt{3}/2cos140-1/2sin140) }{2sin280}\\\\= \frac{4(\sqrt{3}/2cos140-1/2sin140) }{sin280}\\since sin420 = \sqrt{3}/2 \ and \ cos420 = 1/2 \\ then\\\frac{4(sin420cos140-cos420sin140) }{sin280}$

Also note that sin(B-C) = sinBcosC - cosBsinC

sin420cos140 - cos420sin140 = sin(420-140)

The resulting equation becomes;

$\frac{4(sin(420-140)) }{sin280}$

= $\frac{4sin280}{sin280}\\ = 4 \ Proved!$

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