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4 years ago

642 ÷ 5 standard algorithm

Mathematics

2 answers:

damaskus [11]4 years ago

7 0

642/5= 128.4

i hope this helps

Serhud [2]4 years ago

3 0

The answer to that is 128.4

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3 out of 5 sophomore study integrated math course. how many study this course in a sophomore class of 300?

ollegr [7]

3/5 = ?/300
We multiply 5 by 60 to get 300
So you do the same to 3
3x60= 180

180

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4 years ago

What is an intermediate good?

cluponka [151]

A product used to produce a final good or finished product

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3 years ago

Simplify 3r(4r-5s)+2r(-6r-9s)

katovenus [111]

Answer:

-33rs

Step-by-step explanation:

First, distribute in the first part of the equation.

3r(4r-5s)=12r-15rs

Now, distribute in the first part of the equation.

2r(-6r-9s)=-12r-18rs

Combine this together.

12r-15sr-12r-18rs

Combine like terms. (12r and -12r will cancel out to 0.)

-33rs

Hope this helps! Have a wonderful day ^^

4 0

3 years ago

Read 2 more answers

Y = 3x + 10 Y = 2x - 8 find the substitution Show your work please ill give brainlyest

Nina [5.8K]

Answer:

x = -18, Y = -44

Step-by-step explanation:

Since Y = 2x - 8, we can substitute that into the first equation:

2x - 8 = 3x + 10

3x - 2x = -8 - 10

x = -18

Now we plug x = -18 back into the second equation:

Y = 2(-18) - 8

Y = -36 - 8

Y = -44

7 0

2 years ago

A sports company wants to package a ball with a 1.5-inch radius in sets of two. They have two options: a cylinder or a square pr

guajiro [1.7K]

Step 1

Find the Volume of the two balls

we know that

the Volume of a sphere is equal to

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

In this problem we have

$r=1.5\ in$

Substitute

$V=\frac{4}{3} \pi 1.5^{3}=4.5\pi\ in^{3}$

the volume of the two balls is equal

$V=2*4.5\pi=9 \pi\ in^{3}$

Step 2

Find the volume of a cylinder

we know that

The Volume of a cylinder is equal to

$V=\pi r^{2}h$

In this problem we have

$r=1.5\ in$

$h=4*r=4*1.5=6\ in$

Substitute

$V=\pi (1.5^{2})(6)=13.5 \pi\ in^{3}$

Step 3

Find the amount of wasted space in the cylinder

Subtract the volume of the two balls from the volume of the cylinder

$13.5\pi\ in^{3}-9\pi\ in^{3}=4.5\pi\ in^{3}=14.14\ in^{3}$

Step 4

Find the volume of the square prism

The volume of the prism is equal to

$V=Bh$

where

B is the area of the base of the prism

h is the height of the prism

In this problem we have

$B=(2*1.5)^{2}=9\ in^{2}$ -----> the base is equal to the diameter

$h=4*r=4*1.5=6\ in$

Substitute

$V=9*6=54\ in^{3}$

Step 5

Find the amount of wasted space in the prism

Subtract the volume of the two balls from the volume of the prism

$54\ in^{3}-9\pi\ in^{3}=25.73\ in^{3}$

Step 6

Compare the amount of wasted space

the amount of wasted space in the cylinder is $14.14\ in^{3}$

the amount of wasted space in the prism is $25.73\ in^{3}$

Find the difference

$25.73\ in^{3}-14.14\ in^{3}=11.59\ in^{3}=11.6\ in^{3}$

The package that has the least amount of wasted space is the cylinder

therefore

the answer is the option

the cylinder because it has approximately 11.6 in.3 less wasted space than the prism

5 0

3 years ago

Read 2 more answers