Answer: x = 5π/6Explanation:1) Given function: 
2) x-intercept are the roots of the function, i.e. the solution to
y = 03) to find when y = 0, you can either solve the equation or look at the graph.
4) Solving the equation you get:
y = 0 ⇒ tan(x - 5π/6) = 0 ⇒ x - 5π/6 = arctan(0)arctan(0) is the angle whose tangent is zero,so this is 0
⇒ x - 5π/6 = 0 ⇒ x = 5π/6.Then, one example of an x-intercept is x = 5π/6, which means that when x = 5π/6, the value of the function is 0.
Since, the tangent function is a periodic function, there are infinite x-intecepts, that is why the questions asks for one example and not all the values.
You can
verify by replacing the value x = 5π/6 in the given function:
y = tan (5π/6 - 5π/6) = tan(0) = 0.
The answer is z = -13.9
In order to solve this, you need to use inverse operations. The inverse operation of addition is subtraction. So, this is the function we'll use
4.9 + z = -9 ----> subtract 4.9 from both sides
z = -13.9
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>
Answer: 4x - 6 = 563
Suma 6 a ambos lados para aislar la variante 4x.
4x - 6 = 563
+6. +6
4x = 569
Divide ambos lados por 4
4x/4 = 569/4
X = 142.5
Step-by-step explanation: