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ollegr [7]
4 years ago
11

Jeremy is a snowboarder. One of his best tricks is catching big air, jumping off the ramp while holding on to the snowboard. Jer

emy did this trick 20 times, the heights (in meters) of which were {18, 23, 15, 16, 15, 17, 22, 16, 19, 11, 24, 14, 12, 16, 20, 18, 14, 13, 21, 11}. What is the median value of Jeremy’s big air trick?
Mathematics
1 answer:
OLga [1]4 years ago
3 0
Write down all of the #'s on a piece of paper
18,23,15,16,15,17,22,16,19,11,24,14,12,16,20,18,14,13,21,11

Put them in order
11,11,12,13,1414,15,15,16,16,16,17,18,18,19,20,21,22
Cross em off on eah side until you reach the middle 
16, 16
Since there's 2 instead of 1 middle # Add up the 2 (16+16 = 32) 
Then divide by 2 (32/2 = 16)

You're answer is 16
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Answer:

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7 0
3 years ago
PLEASE HELP ME, LIKE ACTUALLY! The sum of two numbers is 193. One of them in 71 more than the other. Find the larger of these tw
3241004551 [841]

Answer:

132

Step-by-step explanation:

X is smaller number

x + 71 is larger number.

x+ (x+71)=193

2x = 193-71

2x=122

x=61

x+71=largest number

61+71=132 largest numher

check answer

x+(x+71)=193

61+132=193

5 0
3 years ago
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What is the slope of the line that passes through the points (4, -7) and (9, 1)?
nekit [7.7K]
The slope is 8/5 if you use the formula
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6 0
4 years ago
Verify that:
Lelu [443]

Answer:

See Below.

Step-by-step explanation:

Problem 1)

We want to verify that:

\displaystyle \left(\cos(x)\right)\left(\cot(x)\right)=\csc(x)-\sin(x)

Note that cot(x) = cos(x) / sin(x). Hence:

\displaystyle \left(\cos(x)\right)\left(\frac{\cos(x)}{\sin(x)}\right)=\csc(x)-\sin(x)

Multiply:

\displaystyle \frac{\cos^2(x)}{\sin(x)}=\csc(x)-\sin(x)

Recall that Pythagorean Identity: sin²(x) + cos²(x) = 1 or cos²(x) = 1 - sin²(x). Substitute:

\displaystyle \frac{1-\sin^2(x)}{\sin(x)}=\csc(x)-\sin(x)

Split:

\displaystyle \frac{1}{\sin(x)}-\frac{\sin^2(x)}{\sin(x)}=\csc(x)-\sin(x)

Simplify:

\csc(x)-\sin(x)=\csc(x)-\sin(x)

Problem 2)

We want to verify that:

\displaystyle (\csc(x)-\cot(x))^2=\frac{1-\cos(x)}{1+\cos(x)}

Square:

\displaystyle \csc^2(x)-2\csc(x)\cot(x)+\cot^2(x)=\frac{1-\cos(x)}{1+\cos(x)}

Convert csc(x) to 1 / sin(x) and cot(x) to cos(x) / sin(x). Thus:

\displaystyle \frac{1}{\sin^2(x)}-\frac{2\cos(x)}{\sin^2(x)}+\frac{\cos^2(x)}{\sin^2(x)}=\frac{1-\cos(x)}{1+\cos(x)}

Factor out the sin²(x) from the denominator:

\displaystyle \frac{1}{\sin^2(x)}\left(1-2\cos(x)+\cos^2(x)\right)=\frac{1-\cos(x)}{1+\cos(x)}

Factor (perfect square trinomial):

\displaystyle \frac{1}{\sin^2(x)}\left((\cos(x)-1)^2\right)=\frac{1-\cos(x)}{1+\cos(x)}

Using the Pythagorean Identity, we know that sin²(x) = 1 - cos²(x). Hence:

\displaystyle \frac{(\cos(x)-1)^2}{1-\cos^2(x)}=\frac{1-\cos(x)}{1+\cos(x)}

Factor (difference of two squares):

\displaystyle \frac{(\cos(x)-1)^2}{(1-\cos(x))(1+\cos(x))}=\frac{1-\cos(x)}{1+\cos(x)}

Factor out a negative from the first factor in the denominator:

\displaystyle \frac{(\cos(x)-1)^2}{-(\cos(x)-1)(1+\cos(x))}=\frac{1-\cos(x)}{1+\cos(x)}

Cancel:

\displaystyle \frac{\cos(x)-1}{-(1+\cos(x))}=\frac{1-\cos(x)}{1+\cos(x)}

Distribute the negative into the numerator. Therefore:

\displaystyle \frac{1-\cos(x)}{1+\cos(x)}=\displaystyle \frac{1-\cos(x)}{1+\cos(x)}

3 0
3 years ago
How many of the numbers from 10 through 97 have the sum of their digits equal to a perfect​ square?
Anarel [89]
Perfect squares are:
1,4,9,16,25,36,49,64,81,100,....

the sum of the digits of our biggest number is 16 so any perfect square bigger than 16 doesn't work for us
1-
1+0=1 so any number containing the digits will work(keep in mind we only will look into whole numbers because digits can't be negative or have fractions or be irrational)
thereful 10 works for our category
2-
0+4=4
1+3=4
2+2=4
22 13 31 and 40 will work two
3-
0+9
1+8
2+7
3+6
4+5
90 18 81 27 72 36 63 45 54
4-
0+16
1+15
2+14
3+13
4+12
5+11
6+10
7+9
8+8
79 97 88

so our set of numbers contain:
10 22 13 31 40 90 18 81 27 72 36 63 45 54 79 97 88
3 0
4 years ago
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