1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Archy [21]
3 years ago
15

Please help me with 16-18 ty!!

Mathematics
1 answer:
jonny [76]3 years ago
3 0

16: B; irrational number, real

17: C; whole, integer, rational, real

18: B; real, rational, integer

You might be interested in
The common factors of 24 and 56 are
Tamiku [17]

Answer:

the answer would be B 1, 2, and 4

3 0
2 years ago
The students in math class use square tiles to make arrays. Celia says they can make more arrays with 8 tiles than with 9 tiles
Cerrena [4.2K]

The students in math class use square tiles to make arrays. If they

  • use 8 (8=1·2·2·2) tiles, then they can form arrays of the length in 1 tile, 2 tiles, 4 tiles and 8 tiles;
  • use 9 (9=1·3·3) tiles, they can form arrays of the length in 1 tile, 3 tiles and 9 tiles (one array less).

So, you can conclude that Celia is correct.

5 0
3 years ago
Read 2 more answers
Long division 40.8 divided by 4 <br> Plz I need to show my long division
VARVARA [1.3K]

Answer:

10.2

Step-by-step explanation:

6 0
2 years ago
Consider the given function and the given interval. f\(x\) = (x - 7)**2 text(, ) [5 text(, ) 11] (a) Find the average value fave
Jet001 [13]

a. f has an average value on [5, 11] of

f_{\rm ave}=\displaystyle\frac1{11-5}\int_5^{11}(x-7)^2\,\mathrm dx=\frac{(x-7)^3}{18}\bigg|_5^{11}=\frac{4^3-(-2)^3}{18}=4

b. The mean value theorem guarantees the existence of c\in(5,11) such that f(c)=f_{\rm ave}. This happens for

(c-7)^2=4\implies c-7=\pm2\implies c=9\text{ or }c=5

8 0
3 years ago
HELP <br><br>Determine whether the relation is a function<br><br> y=2w=2​
nexus9112 [7]

Answer:

Hi there!

I might be able to help you!

It is NOT a function.

<u>Determining whether a relation is a function on a graph is relatively easy by using the vertical line test. If a vertical line crosses the relation on the graph only once in all locations, the relation is a function. However, if a vertical line crosses the relation more than once, the relation is not a function</u>. <u>X = y2 would be a sideways parabola and therefore not a function.</u> Good test for function: Vertical Line test. If a vertical line passes through two points on the graph of a relation, it is <em>not </em>a function. A relation which is not a function. The x-intercept of a function is calculated by substituting the value of f(x) as zero. Similarly, the y-intercept of a function is calculated by substituting the value of x as zero. The slope of a linear function is calculated by rearranging the equation to its general form, f(x) = mx + c; where m is the slope.

A relation that is not a function

As we can see duplication in X-values with different y-values, then this relation is not a function.

A relation that is a function

As every value of X is different and is associated with only one value of y, this relation is a function.

Step-by-step explanation:

It's up there!

God bless you!

3 0
3 years ago
Other questions:
  • Choose the angle whose measure is greater than 90 degrees but less than 180 degrees.
    10·1 answer
  • How do I do this exercise?
    15·1 answer
  • Does anyone know the answer to this?​
    12·1 answer
  • Can someone please answer. There one problem. There's a picture. Thank you!
    9·2 answers
  • What is the pythagorean thereom? And how is it used to solve triangles
    14·1 answer
  • Joseph jumped off of a cliff into the ocean in Miami while vacationing with some friends. His height as a function of time is gi
    6·1 answer
  • Find the missing angle.
    8·1 answer
  • Please help with this one I’m almost done
    10·1 answer
  • Will give brainliest!
    12·1 answer
  • Thank you for answering this question.
    11·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!