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sertanlavr [38]
3 years ago
7

Is this linear or exponential ​

Mathematics
1 answer:
777dan777 [17]3 years ago
4 0
Exponential because the data is not increasing at a constant rate
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PLEASEEEEEE HELP IM RLLY STRUGGLING BRAINLIEST
Butoxors [25]

Answer:

see below

Step-by-step explanation:

1)  see attached image

2) <u> bearings from A to B</u>

Note: bearing is always clockwise from A and from North.

(i)

bearing 180° + 67°

247° from N

(ii)

bearing 148 + 180

328° from N

(iii)

bearing 180 - 21

159° from N

(iv)

39° from N 

(v)

bearing 90 - 40

50° from N

3) <u>back bearing from each diagram on no. 2</u>

<u />

Note: subtract 200 if bearing is more than 180

         add 180 if the bearing is less than 180

(i) 247 - 200 = 47°

(ii) 328 - 200 = 128°

(iii) 159 + 180 = 339°

(iv) 39 + 180 = 219°

(v) 50 + 180 = 230°

5 0
3 years ago
Solve for Y<br> P= 3y/q
kupik [55]
the answer is y=pq/3
4 0
3 years ago
Please solve I cannot solve this for the life of me
Paul [167]

Answer:

  • f(x) = x^5
  • g(x) = 2x -6 . . . . . and see below for other possible definitions

Step-by-step explanation:

You are given an expression for the composition H(x) = f(g(x)) and asked to decompose it into two functions. One way to do that is to look at what is being done to the variable in the function H(x):

  • the variable is multiplied by 2
  • 6 is subtracted from the product
  • the difference is raised to the 5th power.

One of the ways to create the functions g(x) and f(x) is to start at the top of this list and work your way down. Any subset of these transformations can be made into g(x). Then the rest of them are made into f(x).

__

For example, using

  g(x) = 2x

Then the rest of the list is f(x):

  f(x) = (x-6)^5

so when you put 2x as the argument for f(x), you get ...

  H(x) = f(g(x)) = f(2x) = (2x -6)^5 . . . . . the function we want.

__

We could also do the first two steps of our list as g(x):

  g(x) = 2x -6

  f(x) = x^5

so

  H(x) = f(g(x)) = f(2x -6) = (2x -6)^5

__

If we like, we can factor the expression for H(x):

  H(x) = (2(x -3))^5 = 32(x -3)^5

Using the methods above, we could write ...

  g(x) = x -3

  f(x) = 32x^5

so

  H(x) = f(g(x)) = f(x -3) = 32(x -3)^5

__

The multiplication and the addition can be made into several parts. We could choose ...

  g(x) = (1/4)x -1

  f(x) = (8x +2)^5

so

  H(x) = f(g(x)) = f(1/4x -1) = (8(1/4x -1) +2)^5 = (2x -6)^5

4 0
3 years ago
Can someone help on this?
bezimeni [28]

Step-by-step explanation:

this just means that the first function with input value x = 0 is calculated. that result becomes the input value x for the second function. and that result becomes the input value x for the 3rd function. and that result is then our result.

to answer a and b we only need to look at the main operation of each function :

the first one uses only basic multiplication and subtraction.

the second one uses squaring.

the third one is a 1/x operation.

a.

since the final result is a negative number (-31), the second function with the square operation cannot be last.

because a square operation always delivers a positive result.

+a × +a = +a²

-a × -a = +a²

remember, for a square operation we multiply the same number by itself. therefore we cannot mix signs like -a × +a, because then they would not be the same numbers anymore.

b.

since the initial input value is 0, the third function (1/x) cannot be the first, because 1/0 is not defined.

c.

because of the "large" -32 term in the first function I guess that this will be the last one to create such a low negative result of -31.

since 1/x cannot be first, (x - 2)² is then first. that makes 1/x second, and (4x - 32) third, as mentioned.

let's try :

x = 0

(x - 2)² = (0 - 2)² = (-2)² = 4

x = 4

1/x = 1/4

x = 1/4

4x - 32 = 4×1/4 - 32 = 1 - 32 = -31

hurrah ! we are correct !

3 0
1 year ago
Help ASAP<br><br> A)360<br><br> B)50<br><br> C)140
viktelen [127]
50

there’s no way x is 360 or 140
3 0
3 years ago
Read 2 more answers
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