Collect like terms and you'll get -2 - 3(a - b)
Sin(B) = opp/hyp = 32/40 = 0.8
cos(B) = adj/hyp = 24/40 = 0.6
tan(B) = opp/adj = 32/24 = 1.33 (repeating)
Answer:
114°
Step-by-step explanation:
The exterior angle is the sum of the remote interior angles.
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<h3>setup</h3>
(11x +15)° = 60° +6x°
<h3>solution</h3>
5x = 45 . . . . . . . . . divide by °, subtract 15+6x
x = 9 . . . . . . . . . . divide by 5
The measure of exterior angle KMN is ...
m∠KMN = (11(9) +15)° = 114°
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<em>Additional comment</em>
Both the sum of interior angles and the sum of angles of a linear pair are 180°. If M represents the interior angle at vertex M, then we have ...
60° +6x° +M = 180°
(11x +15)° +M = 180°
Equating these expressions for 180° and subtracting M gives the relation we used above:
(11x +15)° +M = 60° +6x° +M . . . . . equate the two expressions for 180°
(11x +15)° = 60° +6x° . . . . . . . . . . . subtract M
This is also described by "supplements to the same angle are equal."
<em>*To solve an inequality, it's the same as solving an equation; isolate the variable to one side.</em>
For this, all you need to do is divide both sides by -4. And since we are dividing by a <u>negative</u>, flip the inequality sign and your inequality is 
Now to graph this. Since x can be "equal to" 1/4 ( ≤ = less than or equal to), you will have a <u>closed circle on 1/4.</u> And since x can also be "less than" 1/4, <u>the arrow will be going to the left of 1/4.</u>
Answer:
For Lin's answer
Step-by-step explanation:
When you have a triangle, you can flip it along a side and join that side with the original triangle, so in this case the triangle has been flipped along the longest side and that longest side is now common in both triangles. Now since these are the same triangle the area remains the same.
Now the two triangles form a quadrilateral, which we can prove is a parallelogram by finding out that the opposite sides of the parallelogram are equal since the two triangles are the same(congruent), and they are also parallel as the alternate interior angles of quadrilateral are the same. So the quadrilaral is a paralllelogram, therefore the area of a parallelogram is bh which id 7 * 4 = 7*2=28 sq units.
Since we already established that the triangles in the parallelogram are the same, therefore their areas are also the same, and that the area of the parallelogram is 28 sq units, we can say that A(Q)+A(Q)=28 sq units, therefore 2A(Q)=28 sq units, therefore A(Q)=14 sq units, where A(Q), is the area of triangle Q.