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algol [13]
3 years ago
7

Find the range 19, 21, 18, 17, 18, 22, 46

Mathematics
1 answer:
rewona [7]3 years ago
7 0

Answer:

46-17=29

Step-by-step explanation:

LARGEST VALUE:46

SMALLEST VALUE: 17

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Solve 4x2-25&lt;0<br><br> in words 4x squared -minus 25 &lt; 0
muminat
<h3>I'll teach you how to solve 4x^2-25<0</h3>

-------------------------------------------------

4x^2-25<0

Add 25 to both sides:

4x^2-25+25 < 0+25

Simplify:

4x^2 < 25

Divide both sides by 4:

4x^2/4 < 25/4

Simplify:

x^2 < 25/4

whole sqrt -25/4 < x < whole sqrt 25/4

whole sqrt 25/4= 5/2

-5/2 < x < 5/2

Your Answer Is -5/2 < x < 5/2

plz mark me as brainliest if this helped :)

4 0
3 years ago
F(1)=6;f(n)=f(n−1)−2
ch4aika [34]

Answer: f(7) = f(4) + 3(3) = 15 + 9 = 24

Step by Step Explanation:

f(1) = 6

f(n) = f(n-1) + 3

 

f(2) = f(1) + 3 = 6+3 = 9

f(3) = f(2) + 3 = 12

f(4) = f(3) + 3 = 15

 

Each term is 3 more than the previous one  

Therefore to get from the 4th term to the 7th term you will add 3

For a total of 3 more times.

 

f(7) = f(4) + 3(3) = 15 + 9 = 24

Hope this helps

5 0
3 years ago
What is the length of a diagonal of a square with a side length of 6?
natka813 [3]
8.5? i’m pretty sure
3 0
3 years ago
Read 2 more answers
Question 3: In les Miserables Monsier, Thenardier, and Madame Thenardier are married and during act two, Cosette and Marius beco
Natasha2012 [34]
Answer: 14,515,200

Note: this is a single number (not an ordered triple or a collection of three different numbers) roughly equal to about 14.5 million if you round to the nearest hundred thousand.

----------------------------------------------

Explanation:

There are 13 people. Let's call them person A, person B, person C, ... all the way up to person M. The first four people are given who we'll call A through D. The rest (E through M) aren't really important since they aren't named. 

A = Monsier Thenardier
B = Madame Thenardier
C = Cosette
D = Marius
Peron's E through M = remaining 9 people

------------------

A and B must stick together. Because of this, we can consider "AB" as one "person".
So we go from 13 people to 13-2+1 = 12 "people". 

Likewise, C and D must stick together. We can consider "CD" as one "person". So we go from 12 "people" to 12-2+1 = 11 "people"

------------------

The question is now: how many ways can we arrange these 11 "people" around a circular table? The answer is (n-1)! ways where n = 11 in this case

So, (n-1)! = (11-1)! = 10! = 10*9*8*7*6*5*4*3*2*1 = 3,628,800

We're almost at the answer. We need to do two adjustments. 

First off, for any single permutation, there are two ways to arrange "AB". The first is "AB" itself and the second is the reverse of that "BA". So we will multiply 3,628,800 by 2 to get 2*3,628,800 = 7,257,600

Using similar logic for "CD", we double 7,257,600 to get 2*7,257,600 = 14,515,200

The final answer is 14,515,200


7 0
3 years ago
Help quick, please !!!
Vedmedyk [2.9K]

Let the third side be x

using Pythagoras theoram

x²+4²=7²

x²= 49-16

x²= 33

x=√33

4 0
3 years ago
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