Step-by-step explanation:
In,
Question 2 I have already answered,
Q. 3)
<h3>
<em><u>Given</u></em><em><u> </u></em><em><u>Question</u></em><em><u>:</u></em></h3>
<em><u>Given</u></em><em><u> </u></em><em><u>values</u></em><em><u>:</u></em><em><u>-</u></em>
c = 6 , d = 3
<em><u>By</u></em><em><u> </u></em><em><u>putting</u></em><em><u> </u></em><em><u>the</u></em><em><u> </u></em><em><u>values</u></em><em><u> </u></em><em><u>in</u></em><em><u> </u></em><em><u>the</u></em><em><u> </u></em><em><u>question</u></em><em><u>, </u></em>
= <em><u>12 (Ans)</u></em>
Answer:
30=3x+6
Step-by-step explanation:
Answer:
The solution is x = 10 and x = -6. This quadratic could be represented in factored form as (x - 10)(x + 6) or on a graph with x-intercepts at (10,0) and (-6,0).
Step-by-step explanation:
To solve the quadratic, write the quadratic in standard form and factor the equation.
x² - 4x = 60
x² - 4x - 60 = 0
(x - 10)(x + 6) = 0
x = 10 and x = -6
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>
Pi is the same as
private investigator
