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Romashka-Z-Leto [24]
3 years ago
14

3. 'a' and 'b' are the intercepts made

Mathematics
1 answer:
Julli [10]3 years ago
5 0

Given:

'a' and 'b' are the intercepts made  by a straight-line with the co- ordinate axes.

3a = b and the line  pass through the point (1, 3).

To find:

The equation of the line.

Solution:

The intercept form of a line is

\dfrac{x}{a}+\dfrac{y}{b}=1         ...(i)

where, a is x-intercept and b is y-intercept.

We have, 3a=b.

\dfrac{x}{a}+\dfrac{y}{3a}=1           ...(ii)

The line  pass through the point (1, 3). So, putting x=1 and y=3, we get

\dfrac{1}{a}+\dfrac{3}{3a}=1

\dfrac{1}{a}+\dfrac{1}{a}=1

\dfrac{2}{a}=1

Multiply both sides by a.

2=a

The value of a is 2. So, x-intercept is 2.

Putting a=2 in b=3a, we get

b=3(2)

b=6

The value of b is 6. So, y-intercept is 6.

Putting a=2 and b=6 in (i), we get

\dfrac{x}{2}+\dfrac{y}{6}=1

Therefore, the equation of the required line in intercept form is \dfrac{x}{2}+\dfrac{y}{6}=1.

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see the attached figure with the letters

1) find m(x) in the interval A,B
A (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,C
B(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,B
A (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,C
B(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>

8 0
3 years ago
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