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STALIN [3.7K]
3 years ago
7

What is the unit rate for 5 pounds of lunch meat for $42.50?A. $5.31 per poundB. $6.50 per poundC. $7.25 per poundD. $8.50 per p

ound
Mathematics
2 answers:
Natalija [7]3 years ago
7 0

Answer:

D. $8.50 per pound

Step-by-step explanation:

We are asked to find the unit rate. In other words, we want to find the price per pound. Let's divide the cost by the pounds.

unit rate= cost/pounds

It costs $42.50 for 5 pounds. Divide $42.50 by 5.

unit rate= $42.50/5 pounds

unit rate= $8.50 / pound

It costs $8.50 per pound and the correct answer is D.

KengaRu [80]3 years ago
4 0

Hey there! I'm happy to help!

We want to find how much money the lunch meat costs for one pound. We see that it costs $42.50 for 5 pounds, so we simply divide by 5 to see how much it costs for 1 pound.

42.5/5=8.5

Therefore, the answer is D. $8.50 per pound.

Have a wonderful day! :D

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Answer:

Part 1) <em>Sphere</em> The surface area is equal to SA=196\pi\ m^{2} and the volume is equal to V=\frac{1,372}{3}\pi\ m^{3}

Part 2) <em>Cone</em> The surface area is equal to SA=(16+4\sqrt{65})\pi\ units^{2} and the volume is equal to V=\frac{112}{3}\pi\ units^{3}

Part 3) <em>Triangular Prism</em> The surface area is equal to SA=51.57\ mm^{2} and the volume is equal to V=17.388\ mm^{3}

Step-by-step explanation:

Part 1) The figure is a sphere

a) Find the surface area

The surface area of the sphere is equal to

SA=4\pi r^{2}

we have

r=14/2=7\ m ----> the radius is half the diameter

substitute

SA=4\pi (7)^{2}

SA=196\pi\ m^{2}

b) Find the volume

The volume of the sphere is equal to

V=\frac{4}{3}\pi r^{3}

we have

r=14/2=7\ m ----> the radius is half the diameter

substitute

V=\frac{4}{3}\pi (7)^{3}

V=\frac{1,372}{3}\pi\ m^{3}

Part 2) The figure is a cone

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The surface area of a cone is equal to

SA=\pi r^{2} +\pi rl

we have

r=4\ units

h=7\ units

Applying Pythagoras Theorem find the value of l (slant height)

l^{2}=r^{2} +h^{2}

substitute the values

l^{2}=4^{2} +7^{2}

l^{2}=65

l=\sqrt{65}\ units

so

SA=\pi (4)^{2} +\pi (4)(\sqrt{65})

SA=16\pi +4\sqrt{65}\pi

SA=(16+4\sqrt{65})\pi\ units^{2}

b) Find the volume

The volume of a cone is equal to

V=\frac{1}{3}\pi r^{2}h

we have

r=4\ units

h=7\ units

substitute

V=\frac{1}{3}\pi (4)^{2}(7)

V=\frac{112}{3}\pi\ units^{3}

Part 3) The figure is a triangular prism

a) The surface area of the triangular prism is equal to

SA=2B+PL

where

B is the area of the triangular base

P is the perimeter of the triangular base

L is the length of the prism

<em>Find the area of the base B</em>

B=\frac{1}{2} (2.7)(2.3)=3.105\ mm^{2}

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P=2.7*3=8.1\ mm

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L=5.6\ mm

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SA=2(3.105)+(8.1)(5.6)=51.57\ mm^{2}

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The volume of the triangular prism is equal to

V=BL

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we have

B=3.105\ mm^{2}

L=5.6\ mm

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V=(3.105)(5.6)=17.388\ mm^{3}

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