Answer:
x ≥ - 91 / 2
Step-by-step explanation:
In this sample problem, the first thing we want to do is expand the part in parenthesis through the distributive property. This will make the simplification process easier. Another approach would be to divide either side by x + 8, but let's try the first.
Approach 1 :
- so we have this simplified expression. We now want to isolate x, so let's combine common terms here. Start by subtracting 6x from either side,
- now add 43 to either side,
- remember that dividing or multiplying a negative value changed the inequality sign. Dividing - 2 on either side, the sign changes to greater than or equal to, with respect to x,
, or in other words
. This is our solution.
Answer:
Step-by-step explanation:
Given
Let the angle be
is co-terminal with -5
Required
Find
Start by dividing 5 by 360 (ignore the negative)
The nearest integer greater than 0.0138 is 1.
<em>This implies that if you were to draw for this problem, the circle you would draw to show -5 degrees would go around 1 time</em>
So, is calculated as:
Answer:
Step-by-step explanation:
First, let's determine which quadrant our angle θ lies in.
Remember ASTC, where:
Everything is positive in QI,
Only sine (and cosecant) is positive in QII,
Only tangent (and cotangent) is positive in QIII,
And only cosine (and secant) is positive in QIV.
Since our tangent is negative, and our cosine is positive, this means that our θ <em>must</em> be in QIV.
In QIV, sine is negative, tangent is negative, and cosine is positive.
With that, let's figure out the remaining trig ratios.
We know that:
Remember that tangent is the ratio of the opposite side to the adjacent side.
Let's figure out our hypotenuse using the Pythagorean Theorem:
Substitute 1 for a and 2 for b (we can ignore the negative since we're squaring anyways). This yields:
Square:
Add:
Take the square root:
So, our square root is √5.
So, our three sides are: Opposite=1, Adjacent=2, and Hypotenuse=√5.
Sine and Cosecant:
Remember that:
Substitute 1 for the opposite and √5 for the hypotenuse. This yields:
Rationalize:
And since our angle is in QIV, we add a negative:
Cosecant is simply the reciprocal of sine. So:
Cosine and Secant:
Remember that:
Substitute 2 for the adjacent and √5 for the hypotenuse. This yields:
Rationalize:
Since our angle is in QIV, cosine stays positive.
Secant is the reciprocal of cosine. So:
Tangent and Cotangent:
We were given that:
To find cotangent, flip:
And we're done!