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3 years ago

Solve for x: 3 < x + 3 < 6

Mathematics

2 answers:

nirvana33 [79]3 years ago

5 0

Answer:

0 < x < 3

Step-by-step explanation:

3 < x + 3 < 6

Subtract 3 from all sides

3-3 < x + 3-3 < 6-3

0 < x < 3

agasfer [191]3 years ago

3 0

Steps to solve:

3 < x + 3 < 6

~Subtract 3 to all sides

3 - 3 < x + 3 - 3 < 6 - 3

~Simplify

0 < x < 3

Best of Luck!

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Find exact values for sin θ and tan θ if cos θ = -4/9 and tan θ > 0.

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Answer:

Part 1) $sin(\theta)=-\frac{\sqrt{65}}{9}$

Part 2) $tan(\theta)=\frac{\sqrt{65}}{4}$

Step-by-step explanation:

we have that

The cosine of angle theta is negative and the tangent of angle theta is positive

That means that the sine of angle theta is negative

step 1

Find $sin(\theta)$

we know that

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we have

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substitute

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$sin^{2}(\theta)=1-\frac{16}{81}$

$sin^{2}(\theta)=\frac{65}{81}$

square root both sides

$sin(\theta)=\pm\frac{\sqrt{65}}{9}$

Remember that

In this problem the sine of angle theta is negative

$sin(\theta)=-\frac{\sqrt{65}}{9}$

step 2

Find $tan(\theta)$

we know that

$tan(\theta)=\frac{sin(\theta)}{cos(\theta)}$

we have

$sin(\theta)=-\frac{\sqrt{65}}{9}$

$cos(\theta)=-\frac{4}{9}$

substitute the given values

$tan(\theta)=-\frac{\sqrt{65}}{9}:-\frac{4}{9}=\frac{\sqrt{65}}{4}$

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