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liq [111]
3 years ago
14

The length of a rectangle is 5 in. More than 3 times its width. The quadratic function f(x) = x(3x + 5) represents the area of t

he rectangle in terms of its width. What is a reasonable domain of the function?
Mathematics
1 answer:
True [87]3 years ago
6 0

Answer: Domain of function is R^{ + } or (0, \infty )

Step-by-step explanation:

Assuming x as width of rectangle

so, Width = x

Question says, Length is 5in more than 3 times id width.

we write as Length = 3x+5

Therefore, Area of rectangle will be given function f(x)

f(x) = x(3x+5)

Here, It is important to notice that width and length of rectangle will always be positive values and also, area of rectangle is always positive.

we can write in equation as

x > 0

f(x) > 0

Thus, Domain of function is R^{ + } or (0, \infty )

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Lena [83]

put a 1 under the 13, so it is 13/1, and 5 5/6, turn it into an improper fraction by multiplying 5x6=30. then 30+5, which is 35. so just keep the denominator, which would be 35/6. so you would have 13/1 - 35/6. you need the same denominator, so see what both 1 and 6 would fit into. which is 6. so 13x6 which is 78, so 78/6 - 35/6, so 78-35 is 43. so 43/6, then simplify, which is 7 1/6

6 0
3 years ago
What is 1042 divided by 45?
Daniel [21]
1042 / 45 
=23.15

You should be able to do that with a calculator though.
4 0
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A student has some $1 and $5 bills in his wallet. he has a total of 15 bills that are worth $51. How much of each type of bill d
snow_tiger [21]

Answer:

# of 5 dollar bills: 9

# of 1 dollar bills: 6

Step-by-step explanation:

9•5 = 45

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How many dimes equal the value of 6 quarters.show you step​
4vir4ik [10]

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4 0
2 years ago
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How many positive integers $n$ from 1 to 5000 satisfy the congruence $n \equiv 5 \pmod{12}$?
irga5000 [103]
The equivalence n \equiv 5 \pmod{12}

means that n-5 is a multiple of 12.

that is

n-5=12k, for some integer k

and so

n=12k+5


for k=-1, n=-12+5=-7

for k= 0, n=0+5=5 (the first positive integer n, is for k=0)


we solve 5000=12k+5 to find the last k

12k=5000-5=4995

k=4995/12=416.25

so check k = 415, 416, 417 to be sure we have the right k:

n=12k+5=12*415+5=4985

n=12k+5=12*416+5=4997

n=12k+5=12*417+5=5009


The last k which produces n<5000 is 416


For all k∈{0, 1, 2, 3, ....416}, n is a positive integer from 1 to 5000,

thus there are 417 integers n satisfying the congruence.


Answer: 417

6 0
3 years ago
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