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2 years ago

What is the inverse of H(x)=1/2x+3

Mathematics

2 answers:

storchak [24]2 years ago

6 0

Answer:

x=-6

Step-by-step explanation:

dalvyx [7]2 years ago

4 0

Answer:

$h^{-1}$ (x) = 2x - 6

Step-by-step explanation:

let y = h(x) and rearrange making x the subject

y = $\frac{1}{2}$ x + 3 ( multiply through by 2 to clear the fraction )

2y = x + 6 ( subtract 6 from both sides )

2y - 6 = x

Change y back into terms of x with x = $h^{-1}$ (x)

$h^{-1}$ (x) = 2x - 6

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Please help solve for x

Ludmilka [50]

Answer:

8.49

Step-by-step explanation:

there is a little formula related to the famous formula of Pythagoras.

it says that the height of a triangle is the square root of the product of both segments of the baseline (the segments the height splits the baseline into).

so, x is actuality the height of the triangle.

x = sqrt(3×24) = sqrt(72) = 8.49

7 0

3 years ago

What is 3 copies of 1/3

jarptica [38.1K]

Answer: 1

Step-by-step explanation:

3 copies of 1/3 would be 1/3+1/3+1/3= 1

6 0

2 years ago

Can I get help solving this graph please?

Daniel [21]

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

2 years ago

Which of the following describes the end behavior of the

slava [35]

B . Approaches -4
I’m pretty sure it’s -4

6 0

1 year ago

I WILL AWARD BRAINLIEST!! PLEASE HELP!!! The figure below shows the movement of a pedestrian from point B to point E. Using the

jeyben [28]

Answer:

Step-by-step explanation:

A) What is the speed of the pedestrian BC, CD, and DE?

Speed from B to C = distance/time = (40 - 20) / 4 = 20/4

= 5 km/h

Speed from C to D = distance/time = 0 / 2

= 0 km/h

Speed from D to E = distance/time = (20 - 0) / (10 - 6) = 20/4

= 5 km/h

B) After what time since the stop did he arrive at point E?

Since the stop at D, he arrived at E after (10 - 6) = 4 h

C) Write the formulas for function d(t) for sections BC, CD, and DE

For BC, d = 40 when t = 0 and d = 20 when t = 4

So d(t) = 40 - 5t

For CD, d = 20 when t = 4 and t = 6

So d(t) = 20

For DE, d = 20 when t = 6 and d = 0 when t = 10

So d(t) = 5 * (10 - t) or d(t) = 50 - 5t

4 0

2 years ago

Read 2 more answers