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Mila [183]
3 years ago
12

PLS HELP. AND SHOW WORK !!! NO BOTS PLS IM TIRED

Mathematics
1 answer:
Sauron [17]3 years ago
5 0

It had four sides, ultimately making it a quadrilateral. Plot all four lines, and it could be a square, rectangle, rhombus, trapezoid, parallelogram, and kite.

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Please help me quickly the question is on the image <br><br>​
Paul [167]
I’m not 100% sure but I think it’s y=5x+20
8 0
3 years ago
Help me asap .no links! imma fail please help​
Dovator [93]
Bear with my working out lol,

8x3 = 24in^2
9x12 = 108in^2
6x7 = 42in^2
(11x4) / 2 = 22in^2

Total = 196in^2

Try that, I couldn’t tell where the tip on the triangle fell at. So it could be wrong but that is what I got and what I would put :)
4 0
3 years ago
Read 2 more answers
The table represents the function (x) = 2x+1<br> Which value goes in the empty cell?
frozen [14]

Answer:

8 maybe

Step-by-step explanation:

My guess but if it is not Sorry

7 0
3 years ago
A librarian has 4 identical copies of Hamlet, 3 identical copies of Macbeth, 2 identical copies of Romeo and Juliet, and one cop
lesantik [10]

Answer:

The number of distinct arrangements is <em>12600</em><em>.</em>

Step-by-step explanation:

This is a permutation type of question and therefore the number of distinguishable permutations is:

n!/(n₁! n₂! n₃! ... nₓ!)

where

  • n₁, n₂, n₃ ... is the number of arrangements for each object
  • n is the number of objects
  • nₓ is the number of arrangements for the last object

In this case

  • n₁ is the identical copies of Hamlet
  • n₂ is the identical copies of Macbeth
  • n₃ is the identical copies of Romeo and Juliet
  • nₓ = n₄ is the one copy of Midsummer's Night Dream

Therefore,

<em>Number of distinct arrangements =  10!/(4! × 3! × 2! × 1!)</em>

<em>                                                         = </em><em>12600 ways</em>

<em />

Thus, the number of distinct arrangements is <em>12600</em><em>.</em>

4 0
3 years ago
Pleasseee hellp for brainleist answer and extra points if you correct the 2 first top rows
andrew-mc [135]
So close! But instead of adding the multiplication sign after the number you need to add them before. Good job though! :)
7 0
3 years ago
Read 2 more answers
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