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sdas [7]
3 years ago
11

DESPERATE HELP NEEDED!!! MATHS!!! WILL AWARD BRAINLIEST!!!

Mathematics
1 answer:
lapo4ka [179]3 years ago
6 0

Answer:

a) 24

b) f^-1 = \sqrt{x+1}

c) 3

Step-by-step explanation:

Just plug the numbers in for x:

(5)^2 -1 = 24

Switch x and y then solve for y and replace y with f^-1(x):

x = y^2 -1

sqrt(x +1) = y = f^-1(x)

Now plug in 8:

f^-1(x) = sqrt(8 +1)  = 3

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In direct variation, the variables change by a constant _______________.
jolli1 [7]
Direct variation is of the form y=kx   (inverse variation is of the form y=k/x)
5 0
3 years ago
In a chemical plant, 24 holding tanks are used for final product storage. Suppose that six of the tanks contain high viscosity m
andriy [413]

Answer:

The probability is 0.461

Step-by-step explanation:

Since six tanks contain high viscosity material, it means that 18 does not contain since the total number of tanks is 24.

Now, here, we know that we are selecting a total of 4 tanks.

out of the tanks with high viscosity materials, we will be selecting one only while out of the others without viscosity materials we shall need selecting three.

Thus,

the number of ways we can select one tank with viscosity material from a total of six tanks will be 6C1 = 6 ways

Now, we want to select three from the remaining 18 tanks

The number of ways we can do this is 18C3 = 816 ways

Finally, we are selecting 4 out of 24 tanks.

The number of ways we can do this is 24C4 = 10,626 ways

Now, we proceed to calculate the probability.

Kindly note that we are selecting 1 from 4 and 3 from 18

The term ‘and’ in probability means we are multiplying the probabilities of these events together and thus, that means that we are going to combine our numbers of selections.

So the total number of ways in which we can select 1 out of 3 and 3 out of 18 will be = 6 * 816 = 4,896 ways

The probability would now be 4896/10626 = 0.461

5 0
4 years ago
If a positive integer is equal to the following product: 25b3c425b3c4, where b and c are distinct prime numbers greater than 2,
Vlad [161]

Answer: 64 distinct even factors

Step-by-step explanation:

let the 2 distinct numbers be b and c and the integer is expected to be 25b3c425b3c4

since b, c > 2 and prime numbers,

potential options include 3, 5, and 7

hence the likelihoods are (b = 3, c = 5), (b = 5, c = 3), (b = 3, c = 7), (b = 7, c = 3), (b = 5, c = 7), (b = 7, c = 5)

Possibility 1 (b = 3, c = 5)

integer is now 253354253354

the distinct even factors = 2, 202, 262, 1934, 19802, 26462, 195334, 253354, 2000002, 2594062, 19148534, 25588754, 262000262, 1934001934, 2508457954, 253354253354

number of distinct even factors = 16

Possibility 2 (b = 5, c = 3)

integer is now 255334255334

the distinct even factors = 2, 86, 202, 5938, 8686, 19802, 253354, 599738, 851486, 2000002, 25788734, 58792138, 25588754, 86000086, 2528061934, 1934001934, 5938005938, 253354253354

number of distinct even factors = 18

Possibility 3 (b = 3, c = 7)

integer is now 253374253374

the distinct even factors = 2, 6, 22, 66, 202, 242, 606, 698, 726, 2094, 2222, 6666, 7678, 19802, 23034, 24442, 59406, 70498, 73326, 84458, 211494, 217822, 253374, 653466, 775478, 2000002, 2326434, 2396042, 6000006, 6910898, 7188126, 8530258, 20732694, 22000022, 25590774, 66000066, 76019878, 228059634, 242000242, 698000698, 726000726, 836218658, 2094002094, 2508655974, 7678007678, 23034023034, 84458084458,253354253354

number of distinct even factors = 48

Possibility 4 (b = 7, c = 3)

integer is now 257334257334

the distinct even factors = 2, 6, 14, 22, 42, 66, 154, 202, 462, 606, 1114, 1414, 2222, 3342, 4242, 6666, 7798, 12254, 15554, 19802, 23394, 36762, 46662, 59406, 85778, 112514, 138614, 217822, 257334, 337542, 415842, 653466, 787598, 1237654, 1524754, 2000002, 2362794, 3712962, 4574262, 6000006, 8663578, 11029714, 14000014, 22000022, 25990734, 33089142, 42000042, 66000066, 77207998, 121326854, 154000154, 231623994, 363980562, 462000462, 849287978, 1114001114, 2547863934, 3342003342, 7798007798, 12254012254, 23394023394, 36762036762, 85778085778, 257334257334

number of distinct even factors = 64

Possibility 5 (b = 5, c = 7)

integer is now 255374255374

the distinct even factors = 2, 14, 34, 58, 74, 202, 238, 406, 518, 986, 1258, 1414, 2146, 3434, 5858, 6902, 7474, 8806, 15022, 19802, 24038, 36482, 41006, 52318, 99586, 127058, 138614, 216746, 255374, 336634, 574258, 697102, 732674, 889406, 1517222, 2000002, 2356438, 3684682, 4019806, 5128718, 9762386, 12455458, 14000014, 21247546, 25792774, 34000034, 58000058, 68336702, 74000074, 87188206, 148732822, 238000238, 361208282, 406000406, 518000518, 986000986, 1258001258, 2146002146, 2528457974, 6902006902, 8806008806, 15022015022, 36482036482, 255374255374

number of distinct even factors = 64

Possibility 6 (b = 7, c = 5)

integer is now 257354257354

the distinct even factors = 2, 202, 19802, 257354, 2000002, 25992754, 2548061954, 257354257354

number of distinct even factors = 8

From the six possibilities, the highest number of likely distinct even factors is 64

7 0
4 years ago
H(n)=−10+12nh, left parenthesis, n, right parenthesis, equals, minus, 10, plus, 12, n
NARA [144]

Answer:

3

Step-by-step explanation:

3

3 0
4 years ago
What is the length of a kite string that is 6 meters above and 9 meters away from where it is being held? Round to the nearest t
ivann1987 [24]

Answer:

10.8 meters

Step-by-step explanation:

This situation forms a right triangle, where the length of the kite string is the hypotenuse, the distance from where it is held is the long leg, and the height is the short leg.

Use the pythagorean theorem to solve for c, the length of the kite string.

a² + b² = c²

6² + 9² = c²

36 + 81 = c²

117 = c²

10.8 = c

So, the length of the kite string is 10.8 meters

7 0
3 years ago
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