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Colt1911 [192]
3 years ago
5

The sum of two numbers is 45. One number is 4 times and large as the other what are the numbers now

Mathematics
1 answer:
GarryVolchara [31]3 years ago
5 0

Answer:

The two numbers are 36 and 9

Step-by-step explanation:

Multiples of 4: 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48

We know that: n <44, 48

Let's try some values:

45 - 40 = 4

45- 36 = 9

45- 28 = 17

45- 24 = 21

Pattern: The lesser a natural numer's absolute value, the greater [larger number] it produces. (Bold = "larger numbers")

We recognize something: 36 + 9 = 45 and 36 is exactly 4 times greater than 9.

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17, 8, 16, 20, 13, 6. Find the mean and median
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Answer:

mean is 20

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Step-by-step explanation:

I hope this will help you ya wit yo quiz

5 0
3 years ago
40 points urgent really need answer
Kipish [7]

a_1=1\\a_2=1+4=5\\a_3=5+2\cdot4=5+8=13\\a_4=13+3\cdot4=13+12=25\\a_5=25+4\cdot4=25+16=41\\a_6=41+5\cdot4=41+20=61

Answer: d. 25 + 16 = 41

6 0
3 years ago
At what point does the curve have maximum curvature? y = 9 ln(x) (x, y) =
Andrews [41]

y = 9ln(x) 
<span>y' = 9x^-1 =9/x</span>
y'' = -9x^-2 =-9/x^2

curvature k = |y''| / (1 + (y')^2)^(3/2) 

<span>= |-9/x^2| / (1 + (9/x)^2)^(3/2) 
= (9/x^2) / (1 + 81/x^2)^(3/2) 
= (9/x^2) / [(1/x^3) (x^2 + 81)^(3/2)] 
= 9x(x^2 + 81)^(-3/2). 

To maximize the curvature, </span>

we find where k' = 0. <span>
k' = 9 * (x^2 + 81)^(-3/2) + 9x * -3x(x^2 + 81)^(-5/2) 
...= 9(x^2 + 81)^(-5/2) [(x^2 + 81) - 3x^2] 
...= 9(81 - 2x^2)/(x^2 + 81)^(5/2) 

Setting k' = 0 yields x = ±9/√2. 

Since k' < 0 for x < -9/√2 and k' > 0 for x > -9/√2 (and less than 9/√2), 
we have a minimum at x = -9/√2. 

Since k' > 0 for x < 9/√2 (and greater than 9/√2) and k' < 0 for x > 9/√2, 
we have a maximum at x = 9/√2. </span>

x=9/√2=6.36

<span>y=9 ln(x)=9ln(6.36)=16.66</span>  

the answer is
(x,y)=(6.36,16.66)
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4 years ago
A student has heard that spinning pennies on a table, rather than flipping them in the air, results in tails side up 65% of the
SVEN [57.7K]

so I believe it is C 0.8215

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3 years ago
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