3 years ago

Can someone help me with this equation n-6/3=-4

Mathematics

1 answer:

pentagon [3]3 years ago

8 0

Siplify
6/3=2
so
n-2=-4
ad 2 to both sides
n+2-2=-4+2
n+0=-2
n=-2

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X+78=3x-2 ..................

photoshop1234 [79]

Answer:

$x=40$

Step-by-step explanation:

$x+78=3x-2\\\mathrm{Subtract\:}78\mathrm{\:from\:both\:sides}\\x+78-78=3x-2-78\\x=3x-80\\\mathrm{Subtract\:}3x\mathrm{\:from\:both\:sides}\\x-3x=3x-80-3x\\\\$

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6 0

4 years ago

1+secA/sec A = sin^2 A / 1-cos A

Fofino [41]

Answer: see proof below

Step-by-step explanation:

$\dfrac{1+\sec A}{\sec A}=\dfrac{\sin^2 A}{1-\cos A}$

Use the following Identities:

sec Ф = 1/cos Ф

cos² Ф + sin² Ф = 1

Proof LHS → RHS

$\text{LHS:}\qquad \qquad \dfrac{1+\sec A}{\sec A}$

$\text{Identity:}\qquad \qquad \dfrac{1+\frac{1}{\cos A}}{\frac{1}{\cos A}}$

$\text{Simplify:}\qquad \qquad \dfrac{\frac{\cos A+1}{\cos A}}{\frac{1}{\cos A}}\\\\\\.\qquad \qquad \qquad =\dfrac{1+\cos A}{1}$

$\text{Multiply:}\qquad \qquad \dfrac{1+\cos A}{1}\cdot \bigg(\dfrac{1-\cos A}{1-\cos A}\bigg)\\\\\\.\qquad \qquad \qquad =\dfrac{1-\cos^2 A}{1-\cos A}$