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solmaris [256]
4 years ago
6

Is the relation a function? Why or why not?

Mathematics
1 answer:
MAVERICK [17]4 years ago
3 0
The answer is C). Yes; only one range value exists for each domain value
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There are two numbers such that , if both of them are individually increased by 5 and then by the same percentage as they were i
QveST [7]
Let the two numbers be x and y.
Let m =  the fraction (or percentage) for increasing the numbers.
 
Increase x by 5, multiply (x+5) by (1+m), and set it qual to 36.
(x + 5)*(1 + m) = 36

Increase y by 5, multiply (y+5) by (1+m), and set it equal to 36.
(y+5)*(1+m) = 36

Therefore
(x+5)*(1+m) = (y+5)*(1+m)
x + 5 = y + 5
x = y
x - y = 0

Answer: 0.
The difference between the two numbers is zero.
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4 years ago
Find the distance between the points (2,8) and (-1,9).
Dima020 [189]

Answer:

need points sorry 38484343

Step-by-step explanation:

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3 years ago
London bought a bag containing 95 small candies. She starts eating the
Savatey [412]

Answer: 19

Step-by-step explanation: let the number of candies left in the bag be y and the number of minutes eating be x.

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8 0
2 years ago
A carpenter is building a rectangular shed with a fixed perimeter of 54 ft. What are the dimensions of the largest shed that can
lutik1710 [3]

Answer:

the Largest shed dimension is 13.5 ft by 13.5 ft

Largest Area is 182.25 ft²

Step-by-step explanation:

Given that;

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P = 2( L + B ) = 54ft

L + B = 54/2

L + B = 27 ft

B = 27 - L ------------Let this be equation 1

Area A = L × B

from equ 1, B = 27 - L

Area A = L × ( 27 - L)

A = 27L - L²

for Maxima or Minima

dA/dL = 0

27 - 2L = 0

27 = 2L

L = 13.5 ft

Now, d²A/dL² = -2 < 0

That is, area is maximum at L = 13.5 using second derivative test

B = 27 - L

we substitute vale of L

B = 27 - 13.5 = 13.5 ft

Therefore the Largest shed dimension = 13.5 ft by 13.5 ft

Largest Area = 13.5 × 13.5 = 182.25 ft²

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3 years ago
if a student does not like the mountains what is the probability that the student does also not like the beach
Yakvenalex [24]
<span>40.4% I would say but if you need the whole problem I can give it to you.</span>
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4 years ago
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