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3 years ago

Find the midpoint of the line segment

Mathematics

1 answer:

strojnjashka [21]3 years ago

5 0

Answer:

midpoint (5,4)

Step-by-step explanation:

The midpoint(M) of a segment with endpoints (x₁ , y₁) and ( x₂, y₂) is

where x₁ = 2 and x₂ = 8

y₁ = 0 and y₂ = 8

M = $\frac{x_1 + x_2}{2} ,\frac{y_1 + y_2}{2}$

M = $\frac{2 + 8}{2} ,\frac{0 + 8}{2}$

M = 5 , 4

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Someone please answer this lol xx

Tamiku [17]

The answer is 4 lol

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3 years ago

A store sells used DVDs for 8$ each and used videotapes for $6 each. the function rule 8d+6v can be used to represent the total

Nesterboy [21]

If you were to buy 5 DVDs and 3 videotapes, For 5 DVDs, it would be $40 for 5 DVDs, and $18 for 3 videotapes. I hope this helps :)

8 0

3 years ago

joe has a handful of dimes and quarters that values to $5.30. He has one fewer than twice as many dimes than quarters. How many

OlgaM077 [116]

Answer:

12 quarters 23 dimes

Step-by-step explanation:

x(0.25)+(2x-1)(0.10)=5.30

0.25x+0.2x-0.10=5.30

0.45x=5.40

x=12

12 quarters

12(2)=24-1=23 dimes

5 0

2 years ago

A solid is formed by adjoining two hemispheres to the ends of a right circular cylinder. An industrial tank of this shape must h

mestny [16]

Answer:

Radius =6.518 feet

Height = 26.074 feet

Step-by-step explanation:

The Volume of the Solid formed = Volume of the two Hemisphere + Volume of the Cylinder

Volume of a Hemisphere $=\frac{2}{3}\pi r^3$

Volume of a Cylinder $=\pi r^2 h$

Therefore:

The Volume of the Solid formed

$=2(\frac{2}{3}\pi r^3)+\pi r^2 h\\\frac{4}{3}\pi r^3+\pi r^2 h=4640\\\pi r^2(\frac{4r}{3}+ h)=4640\\\frac{4r}{3}+ h =\frac{4640}{\pi r^2} \\h=\frac{4640}{\pi r^2}-\frac{4r}{3}$

Area of the Hemisphere = $2\pi r^2$

Curved Surface Area of the Cylinder = $2\pi rh$

Total Surface Area=

$2\pi r^2+2\pi r^2+2\pi rh\\=4\pi r^2+2\pi rh$

Cost of the Hemispherical Ends = 2 X Cost of the surface area of the sides.

Therefore total Cost, C

$=2(4\pi r^2)+2\pi rh\\C=8\pi r^2+2\pi rh$

Recall: $h=\frac{4640}{\pi r^2}-\frac{4r}{3}$

Therefore:

$C=8\pi r^2+2\pi r(\frac{4640}{\pi r^2}-\frac{4r}{3})\\C=8\pi r^2+\frac{9280}{r}-\frac{8\pi r^2}{3}\\C=\frac{9280}{r}+\frac{24\pi r^2-8\pi r^2}{3}\\C=\frac{9280}{r}+\frac{16\pi r^2}{3}\\C=\frac{27840+16\pi r^3}{3r}$

The minimum cost occurs at the point where the derivative equals zero.

$C^{'}=\frac{-27840+32\pi r^3}{3r^2}$

$When \:C^{'}=0$

$-27840+32\pi r^3=0\\27840=32\pi r^3\\r^3=27840 \div 32\pi=276.9296\\r=\sqrt[3]{276.9296} =6.518$

Recall:

$h=\frac{4640}{\pi r^2}-\frac{4r}{3}\\h=\frac{4640}{\pi*6.518^2}-\frac{4*6.518}{3}\\h=26.074 feet$

Therefore, the dimensions that will minimize the cost are:

Radius =6.518 feet

Height = 26.074 feet

5 0

3 years ago

A rectangle has an area of 28 ft2. If the width of the rectangle is 5 ft, what is the length?

Jobisdone [24]

Area of a Rectangle = Length x Width
28 ft² = L x 5ft
28 ft² / 5ft = Length
5.6 ft = length. CHOICE A.

Perimeter of a Rectangle = 2(L + W)
20 ft = 2(L + 4)
20 ft = 2L + 8
20 - 8 = 2L
12 = 2L
12/2 = 2L/2
6 ft= Length CHOICE A.

5 0

3 years ago