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2 years ago

MARKING AS BRAINLIEST!!

Mathematics

1 answer:

devlian [24]2 years ago

3 0

Answer: They are inverses of each other

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Explanation:

We need to check two things:

f(g(x)) = x
g(f(x)) = x

If both of those equations are true, then f and g are inverses of each other.

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Part 1

$f(x) = -3x-12\\\\f(g(x)) = -3( g(x) )-12\\\\f(g(x)) = -3\left( \frac{-12-x}{3} \right)-12\\\\f(g(x)) = 12+x-12\\\\f(g(x)) = x\\\\$

So far, so good. Now we need to check the other equation mentioned.

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Part 2

$g(x) = \frac{-12-x}{3}\\\\g(f(x)) = \frac{-12-( f(x) )}{3}\\\\g(f(x)) = \frac{-12-( -3x-12 )}{3}\\\\g(f(x)) = \frac{-12+3x+12}{3}\\\\g(f(x)) = \frac{3x}{3}\\\\g(f(x)) = x\\\\$

That works out as well. Both equations mentioned are true, which confirms the two original functions are inverses of one another.

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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>

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A baker has 3/4 a pound of sugar 2/5 of a pound of cinnamon and 4/5 of a pound of flour. She uses 1/5 of a pound of each. How mu

Komok [63]

First, make all the denominators the same.

In this case, all the denominators are of the value of either 4 or 5. So, now what is the biggest number we can multiply by, using four AND five.

Answer is 20 because 5*4 is 20 and 4*5 is 20.

So now for all the ingredients, we have to :

-multiply the whole fraction by 4 if the denominator is 5

- Multiply the whole fraction by 5 if the denominator is 4

So for sugar, you would do 3/4 ( times 3 by 5, times 4 by 5)

Which is 15/20

For cinnamon, you would do 2/5 ( times 2 by 4, times 5 by 4)

Which is 8/20

For flour, you would do 4/5 ( times 4 by 4, times 5 by 4)

Which is 16/20

But, for your problem, you need to work out how much the baker has left over after using 1/5 of pound of all the ingredients. So, you would do 1/5 ( times 1 by 4, times 5 by 4)

Which is 4/20

Now, we can subtract it by the ingredients For sugar:

15/20 - 4/20

Now the baker has 11/20 pound of sugar

For cinnamon:

8/20 - 4/20

Now the baker has 4/20 pound of cinnamon ( simplified fraction is 1/5)

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16/20 - 4/20

Now the baker has 12/20 pound of flour ( simplified fraction is 3/5)

Hope this helps;)

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