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2 years ago

Stephen says that the numbers 38 and 40 are relatively prime. Explain why he is incorrect in making this statement.

Mathematics

2 answers:

Jet001 [13]2 years ago

6 0

Answer:

Relatively prime numbers have a GCF of 1. The numbers 38 and 40 have a GCF of 2, so they are not relatively prime.

Step-by-step explanation:

When you say that two numbers are relatively prime, it means that both numbers do not have a common factor except 1. Stephen is incorrect because 38 and 40 are both multiples of 2.

inna [77]2 years ago

6 0

Step-by-step explanation:

Relatively prime numbers have a GCF of 1. The numbers 38 and 40 have a GCF of 2, so they are not relatively prime.

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An open-top rectangular box is being constructed to hold a volume of 150 in3. The base of the box is made from a material costin

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Answer:

minimum cost of construction box are x = x = 3.4199 in and y = 10.2598 in and z = 4.2750 in

Step-by-step explanation:

given data

volume = 150 in³

base material costing = 5 cents/in²

front cost = 10 cents/in²

remainder sides cost = 2 cents/in²

to find out

the dimensions that will minimize the cost

solution

we consider here length = x and breadth = y and height = z

and

area of base = xy

area of front = xz

and area of remaining side = xz + 2yz .....................1

cost of base will be = 5xy

cost of front = 10xz

cost of remaining side = 2 ( xz+ 2yz)

and

total cost will be

total cost TC = 5xy + 10xz + 2 ( xz+ 2yz)

total cost TC = 5xy + 10xz + 2xz+ 4yz

total cost TC = 5xy + 12xz + 4yz ....................2

and total volume will be = xyz

150 = xyz

z = $\frac{150}{xy}$ .......................3

now put z value in equation 2

total cost TC = 5xy + 12xz + 4yz

total cost TC = 5xy + 12x $\frac{150}{xy}$ + 4y $\frac{150}{xy}$

total cost TC = 5xy + $\frac{1800}{y}$ + $\frac{600}{x}$ ...........4

now differentiate TC w.r.t x and y

TC (x) = 5y - $\frac{600}{x^2}$

TC (x) = 5x - $\frac{1800}{y^2}$

now equating with 0 these

5y - $\frac{600}{x^2}$ = 0

x² = $\frac{120}{y}$

and

5x - $\frac{1800}{y^2}$ = 0

x = $\frac{360}{y^2}$

solve we get

y = 10.2598

x = 3.4199

now put x and y value in equation 3

z = $\frac{150}{xy}$

z = $\frac{150}{3.4199*10.2598}$

z = 4.2750

so minimum cost of construction box are x = x = 3.4199 in and y = 10.2598 in and z = 4.2750 in

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