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3 years ago

Solve the inequality. 4x-8<-2x+1

Mathematics

1 answer:

lord [1]3 years ago

3 0

Answer:

x< 9/2 or

x< 4 1/2

Step-by-step explanation:

act like th <is a = for a moment

add 2x to both sides

add 8 to both sides

2x=(<) 9

devide both sides by 2

x < 9/2 or

x< 4 1/2

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Verify cot x sec^4x=cotx +2tanx +tan^3x

Tanzania [10]

Answer:

See explanation

Step-by-step explanation:

We want to verify that:

$\cot(x) \: { \sec}^{4} x = \cot(x) + 2 \tan(x) + { \tan}^{3} x$

Verifying from left, we have

$\cot(x) \: { \sec}^{4} x = \cot(x) \: ( 1 + { \tan}^{2} x )^{2}$

Expand the perfect square in the right:

$\cot(x) \: { \sec}^{4} x = \cot(x) \: ( 1 + { 2\tan}^{2} x + { \tan}^{4} x)$

We expand to get:

$\cot(x) \: { \sec}^{4} x = \cot(x) \: + \cot(x){ 2\tan}^{2} x +\cot(x) { \tan}^{4} x$

We simplify to get:

$\cot(x) \: { \sec}^{4} x = \cot(x) \: + 2 \frac{ \cos(x) }{\sin(x) ) } \times \frac{{ \sin}^{2} x}{{ \cos}^{2} x} +\frac{ \cos(x) }{\sin(x) ) } \times \frac{{ \sin}^{4} x}{{ \cos}^{4} x}$

Cancel common factors:

$\cot(x) \: { \sec}^{4} x = \cot(x) \: + 2 \frac{{ \sin}x}{{ \cos}x} +\frac{{ \sin}^{3} x}{{ \cos}^{3} x}$

This finally gives:

$\cot(x) \: { \sec}^{4} x = \cot(x) + 2 \tan(x) + { \tan}^{3} x$

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4 years ago

Ax + z = aw - y<br> Solve for a

olga2289 [7]

Ax=aw-z-y
divide by x
a=awx-z-y
a=awx-zy

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