Answer: 18
Step-by-step explanation:
See the photo for work shown
-22 - x = 5 + 6x + 9 Combine like terms (5 and 9)
-22 - x = 14 + 6x Subtract 6x from both sides
-22 - 7x = 14 Add 22 to both sides
-7x = 36 Divide both sides by -7
x = 5
Heres a tip---- if it has a square its 90 degrees if it doesnt its 180 and if its in an x shape theyre equal
see the attached figure with the letters
1) find m(x) in the interval A,BA (0,100) B(50,40) -------------- > p=(y2-y1(/(x2-x1)=(40-100)/(50-0)=-6/5
m=px+b---------- > 100=(-6/5)*0 +b------------- > b=100
mAB=(-6/5)x+100
2) find m(x) in the interval B,CB(50,40) C(100,100) -------------- > p=(y2-y1(/(x2-x1)=(100-40)/(100-50)=6/5
m=px+b---------- > 40=(6/5)*50 +b------------- > b=-20
mBC=(6/5)x-20
3)
find n(x) in the interval A,BA (0,0) B(50,60) -------------- > p=(y2-y1(/(x2-x1)=(60)/(50)=6/5
n=px+b---------- > 0=(6/5)*0 +b------------- > b=0
nAB=(6/5)x
4) find n(x) in the interval B,CB(50,60) C(100,90) -------------- > p=(y2-y1(/(x2-x1)=(90-60)/(100-50)=3/5
n=px+b---------- > 60=(3/5)*50 +b------------- > b=30
nBC=(3/5)x+30
5) find h(x) = n(m(x)) in the interval A,B
mAB=(-6/5)x+100
nAB=(6/5)x
then
n(m(x))=(6/5)*[(-6/5)x+100]=(-36/25)x+120
h(x)=(-36/25)x+120
find <span>h'(x)
</span>h'(x)=-36/25=-1.44
6) find h(x) = n(m(x)) in the interval B,C
mBC=(6/5)x-20
nBC=(3/5)x+30
then
n(m(x))=(3/5)*[(6/5)x-20]+30 =(18/25)x-12+30=(18/25)x+18
h(x)=(18/25)x+18
find h'(x)
h'(x)=18/25=0.72
for the interval (A,B) h'(x)=-1.44
for the interval (B,C) h'(x)= 0.72
<span> h'(x) = 1.44 ------------ > not exist</span>
9514 1404 393
Answer:
38.2°
Step-by-step explanation:
The law of sines tells you ...
sin(x)/15 = sin(27°)/11
sin(x) = (15/11)sin(27°) . . . . . multiply by 15
x = arcsin((15/11)sin(27°)) ≈ arcsin(0.619078) ≈ 38.2488°
x ≈ 38.2°
_____
<em>Additional comment</em>
In "law of sines" problems, you need to identify a side and opposite angle that you know both values of. Then, you need to identify whether you're looking for an angle or a side, and whether its opposite side or angle is known. If two angles are known, you can always figure the third from the sum of angles in a triangle.
Here, we have angle 27° opposite side 11. We are looking for an angle, and we know its opposite side. This lets us use the ratio formula directly. Since the angle is the unknown, it is useful to write the equation with sines on top and sides on the bottom.
The given angle is opposite the shorter of the given sides, so this triangle has two solutions. We assume that we want the solution that is an acute angle (141.8° is the other solution). That assumption is based on the drawing. Usually, you're cautioned not to take the drawings at face value.