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zavuch27 [327]
2 years ago
6

The box plot measures the amount of people in Chess Club the past month. What was the range number of students in Chess Club las

t month?
Mathematics
2 answers:
frez [133]2 years ago
7 0
Sorry not rlly sure about this one
g100num [7]2 years ago
5 0
Huh? Who goes to the chess club
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How do you describe the end behavior of a graph?
timofeeve [1]

Answer:

The end behavior of a function [f] describes the behavior of the graph of the function [f] at the ends of the x-axis.

Step-by-step explanation:

I've done this last year and this is what I can come up with

7 0
1 year ago
If (x-7) is a factor of 2x^2-11x+k, what is the value of k
kondaur [170]

If x-7 =0

Then x =7

Substitute x=7 into f(x):

2(7)^2 - 11(7) + k

= (98-77)+k

= 21+k

Therefore k=-21 because 21+(-21)=0

4 0
3 years ago
Can someone please be generous and help me
Ugo [173]

Answer:

The point slope form is y-y_{1} =m(x-x_{1} )

The slope(m) can be calculated using \frac{y-y_{1} }{x-x_{1}}:

(x, y)=(4,-3)\\(x_{1} ,y_{1} )=(5,0)\\\\\frac{0-(-3)}{5-4} =\frac{0+3}{1} =\frac{3}{1} =3

Using Point One, the point-slope form is determined as:

y-(-3)=3(x-4)\\y+3=3(x-4)

The y-intercept(b) can be calculated as:

(5,0)\\y=3x+b\\0=3(5)+b\\0=15+b\\b=-15\\(0,-15)

7 0
3 years ago
Use Euclid's division algorithm to find the HCF of 441, 567, 693
Illusion [34]
Let a = 693, b = 567 and c = 441

Now first we will find HCF of 693 and 567 by using Euclid’s division algorithm as under

693 = 567 x 1 + 126
567 = 126 x 4 + 63
126 = 63 x 2 + 0
Hence, HCF of 693 and 567 is 63

Now we will find HCF of third number i.e., 441 with 63 So by Euclid’s division alogorithm for 441 and 63

441 = 63 x 7+0
=> HCF of 441 and 63 is 63.

Hence, HCF of 441, 567 and 693 is 63.
6 0
3 years ago
Selected-Response
maks197457 [2]

Answer: 6 pints

Step-by-step explanation:

32 times 3 in 96

so you must multiply 2 times 3

5 0
3 years ago
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