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Effectus [21]
3 years ago
12

Asap need help fast pls!!!!!!

Mathematics
1 answer:
anastassius [24]3 years ago
6 0

Answer:

i think its 20 too, tho good luck

Step-by-step explanation:

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Write the set of real numbers x less than 6 in set builder notation
maksim [4K]

Answer:

{x ∈ R: x<6}

Step-by-step explanation:

Given

  • x is a real number
  • x is less than 6

Required

Write the set using set builder notation

The very first thing to do is to list out the range of x, using inequalities;

x is less than 6 implies that -infiniti < x < 6

The next step is to translate this to set builder. This is done as follows

x ∈ R - > This means that x is a real number

x < 6 -> where x is less than 6.

Bringing these two together, it gives:

{x ∈ R: x<6}

Hence, the set of real numbers x less than 6 is equivalent to {x ∈ R: x<6} using set builder notation

6 0
3 years ago
PLEASE PLEASEE HELP ITS PAST DUE AND ITS A TEST ILL GIVE BRAINLIEST
agasfer [191]

Answer:

The length will be letter C, 50 m

5 0
3 years ago
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My teacher has 91 apples, and ate 82, and added 10000000000000000 more apples. How many does she have
kondor19780726 [428]

Answer:

10000000000000009

OML that's a lot of apples lol.

Step-by-step explanation:

Hope this helps ;)

3 0
3 years ago
Read 2 more answers
A customer pays $3.27 for oranges and $4.76 for pears. How many pounds of fruit does the customer buy?
ruslelena [56]
Take 3.27 and divide that by 1.09  (for the oranges)
3.27/1.09=3lbs. of oranges
then take 4.76 and divide that by 1.19       (for the pears)
4.76/1.19=4lbs. of pears
add 4+3 to get how many lbs. in all
4+3=7

3 0
3 years ago
The function f(x)=(x-5)^2 +2 is not one-to-one. Identify a restricted domain that makes the function one-to-one, and find the in
Anon25 [30]

We have been given a quadratic function f(x)=(x-5)^{2} +2 and we need to restrict the domain such that it becomes a one to one function.

We know that vertex of this quadratic function occurs at (5,2).

Further, we know that range of this function is [2,\infty).

If we restrict the domain of this function to either (-\infty,5] or [5,\infty), it will become one to one function.

Let us know find its inverse.

y=(x-5)^{2}+2

Upon interchanging x and y, we get:

x=(y-5)^{2}+2

Let us now solve this function for y.

(y-5)^{2}=x-2\\&#10;y-5=\pm \sqrt{x-2}\\&#10;y=5\pm \sqrt{x-2}\\

Hence, the inverse function would be f^{-1}(x)=5+\sqrt{x-2} if we restrict the domain of original function to [5,\infty) and the inverse function would be f^{-1}(x)=5-\sqrt{x-2} if we restrict the domain to (-\infty,5].

8 0
3 years ago
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