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asambeis [7]
3 years ago
13

For each pair of functions and below, find and .

Mathematics
1 answer:
Dima020 [189]3 years ago
5 0

Answer:

1- Inverses .

2 - Not inverses.  

Step-by-step explanation:

Notice that

1 -

f(g(x))= 3/(3/x) = (3x)/3  = x

Since their composition is the identity, then they are inverses.

2 -

f(g(x)) = ((-x+4)+4) = -x+8

Since their composition is not the identity, then they are not inverses.

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PLEASE HELP ASAP!! PLEASE ANSWER A AND B! GIVING AWAY 13 POINTS
telo118 [61]
According to google A. Would be 2.375 or as a fraction it would be 2 3/8
5 0
3 years ago
The endpoints of ab are a(1,4) and b(6,-1). If point c divides ab in the ratio 2:3, the coordinates of c are ?. If point d divid
aliya0001 [1]

Answer:

see explanation

Step-by-step explanation:

Find the coordinates of c using the section formula

x_{c} = \frac{(3(1)+(2(6)}{2+3} = \frac{3+12}{5} = 3

y_{c} = \frac{(3(4))+(2(-1))}{2+3} = \frac{12-2}{5} = 2

coordinates of c = (3, 2)

Similarly to find the coordinates of d

x_{d} = \frac{x(2(1))+(3(3))}{3+2} = \frac{2+9}{5} = \frac{11}{5}

y_{d} = \frac{(2(4))+(3(2))}{3+2} = \frac{8+6}{5} = \frac{14}{5}

coordinates of d = ( \frac{11}{5}, \frac{14}{5})


4 0
3 years ago
Read 2 more answers
I’ll mark u as brainlist if u get this right!!
WITCHER [35]

Answer:

A

Step-by-step explanation:

When X is less than -1, positive parabola.  

6 0
3 years ago
Read 2 more answers
When a breeding group of animals is introduced into a restricted area such as a wildlife reserve, the population can be expected
jasenka [17]

Answer:

A. Initially, there were 12 deer.

B. <em>N(10)</em> corresponds to the amount of deer after 10 years since the herd was introducted on the reserve.

C. After 15 years, there will be 410 deer.

D. The deer population incresed by 30 specimens.

Step-by-step explanation:

N=\frac{12.36}{0.03+0.55^t}

The amount of deer that were initally in the reserve corresponds to the value of N when t=0

N=\frac{12.36}{0.33+0.55^0}

N=\frac{12.36}{0.03+1} =\frac{12.36}{1.03} = 12

A. Initially, there were 12 deer.

B. N(10)=\frac{12.36}{0.03 + 0.55^t} =\frac{12.36}{0.03 + 0.0025}=\frac{12.36}{y}=380

B. <em>N(10)</em> corresponds to the amount of deer after 10 years since the herd was introducted on the reserve.

C. N(15)=\frac{12.36}{0.03+0.55^15}=\frac{12.36}{0.03 + 0.00013}=\frac{12.36}{0.03013}= 410

C. After 15 years, there will be 410 deer.

D. The variation on the amount of deer from the 10th year to the 15th year is given by the next expression:

ΔN=N(15)-N(10)

ΔN=410 deer - 380 deer

ΔN= 30 deer.

D. The deer population incresed by 30 specimens.

8 0
3 years ago
This is uhh due by 8 tommorow mornin please help me
Ira Lisetskai [31]

1 Simplify

1

3

x

3

1

​

x to

x

3

3

x

​

.

x

3

+

5

≤

−

4

3

x

​

+5≤−4

2 Subtract

5

5 from both sides.

x

3

≤

−

4

−

5

3

x

​

≤−4−5

3 Simplify

−

4

−

5

−4−5 to

−

9

−9.

x

3

≤

−

9

3

x

​

≤−9

4 Multiply both sides by

3

3.

x

≤

−

9

×

3

x≤−9×3

5 Simplify

9

×

3

9×3 to

27

27.

x

≤

−

27

x≤−27

I hope this help you

6 0
3 years ago
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