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4 years ago

Part one out of six thx

Mathematics

1 answer:

Ulleksa [173]4 years ago

5 0

Factors of 16:
1,2,4,8,16

factors of 40:
1,2,4,5,8,10,20,40

highest common denominator of 16 and 40:
8
Hope this helps!

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150m - 100m + 47.750 = 50.500- 200m

coldgirl [10]

Answer:

m=11÷600

Step-by-step explanation:

150m-100m+47.750=50.5000-200m

collect the like terms

-50m+47.75=50.5-200m

move variable to the left side and change its sign

-50m+200m+47.75=50.5

Move constant to the right side and change its sign

50m+200m=50.5-47.75

collect the like terms(again)

150m=50.5-47.75

subtract the numbers

150m=2.75

divied both sides of the equation by 150

m=11÷600

solution

m=11÷600

alternative form

m=0.0183

5 0

3 years ago

Find the equation of a plane that passes through (6,3,2) and is perpendicular to <-2,1,5>

My name is Ann [436]

no rurio greob tryo yeu

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3 years ago

HELP ME PLZ WILL MARK U AS BRAINLIEST

natulia [17]

Answer:

Step-by-step explanation:

The mean for ground floor is 20.67

The mean for balcony is 15.8

the difference between the means is ground floor - balcony

20.67 - 15.8 = 4.87

You can talk about it yourself for part c

3 0

3 years ago

The radius of a sphere whose volume is 12 π cubic cm is??

dimaraw [331]

Given a radius $r$ , the volume of a sphere is given by

$V=\dfrac{4}{3}\pi r^3$

We can solve this formula for the radius, given the volume:

$r=\sqrt[3]{\dfrac{3V}{4\pi}}$

We can plug our value for the volume to get the radius:

$r=\sqrt[3]{\dfrac{3\cdot 12\pi}{4\pi}}=\sqrt[3]{9}$

8 0

4 years ago

Read 2 more answers

A closed box with a square base is to have a volume of 171 comma 500 cm cubed. The material for the top and bottom of the box co

Zepler [3.9K]

Answer:

$C(x)=\dfrac{20x^3+1715000}{x}\\$Minimum cost, C(35)=\$29,400$

The dimensions that will lead to minimum cost of the box are a base length of 35 cm and a height of 140 cm.

Step-by-step explanation:

Volume of the Square-Based box=171,500 cubic cm

Let the length of a side of the base=x cm

Volume $=x^2h$

$x^2h=171,500\\h=\dfrac{171500}{x^2}$

The material for the top and bottom of the box costs $10.00 per square centimeter.

Surface Area of the Top and Bottom $=2x^2$

Therefore, Cost of the Top and Bottom $=\$10X2x^2=20x^2$

The material for the sides costs $2.50 per square centimeter.

Surface Area of the Sides=4xh

Cost of the sides=$2.50 X 4xh =10xh

$\text{Substitute h}$=\dfrac{171500}{x^2} $into 10xh\\Cost of the sides=10x(\dfrac{171500}{x^2})=\dfrac{1715000}{x}$

Therefore, total Cost of the box

$= 20x^2+\dfrac{1715000}{x}\\C(x)=\dfrac{20x^3+1715000}{x}$

To find the minimum total cost, we solve for the critical points of C(x). This is obtained by equating its derivative to zero and solving for x.

$C'(x)=\dfrac{40x^3-1715000}{x^2}\\\dfrac{40x^3-1715000}{x^2}=0\\40x^3-1715000=0\\40x^3=1715000\\x^3=1715000\div 40\\x^3=42875\\x=\sqrt[3]{42875}=35$

Recall that:

$h=\dfrac{171500}{x^2}\\Therefore:\\h=\dfrac{171500}{35^2}=140cm$

The dimensions that will lead to minimum costs are base length of 35cm and height of 140cm.

Therefore, the minimum total cost, at x=35cm

$C(35)=\dfrac{20(35)^3+1715000}{35}=\$29,400$

8 0

3 years ago