Adam has been asked to post a parcel for a family member. He has been told that the parcel weighs 12 lbs 8 oz. 1lb = 16 oz 1lb = 454 g How much does the parcel weigh to the nearest whole kilogram?

The answer is false due to the object is changes shape not size

Answer:

The density of the object is [tex]0.1067[/tex] grams per cubic centimeter. (Rounded off to 4th decimal)

Step-by-step explanation:

Density = [Mass/Volume]

Here we have been given the dimensions of the object.

Volume = Height * Width * Length

Volume = [tex]3.0cm*15.0cm*5.0cm[/tex]

             =[tex]225cm^{3}[/tex]

Now we can substitute the Mass and the Volume to the density equation.

Density = [Mass/Volume]

             =[tex]\frac{24grams}{225cm^{3} }[/tex]

             =[tex]0.1067[/tex] grams per cubic centimeter. (Rounded off to 4th decimal)

when you are buying toothbrushes and s is boxes of toothbrushes, and there are 100 tooth brushes per box, 974 > 878

To begin, Arnold should sort on whether or not it is a screw, nut, bolt.

This might only be # 1 and 3 but you should be able to put together a number line with the information I provided. Some of the #s may look different but that is because I used the Least Common Denominator. Hope this helps!

Screws: Least to greatest (top = least)

4 Screws 4/64” x 3” long

12 Screws 5/64” x 3” long

3 Screws 6/64” x 4” long

6 Screws 11/64” x 4” long

16 Screws 24/64” x 2” long

1 Screw 28/64” x 3” long

1 Screw 32/64” x 3” long  

1 Screw 48/64” x 2” long

Hex Bolts: Least to greatest by diameter (top = least)

4 Hex bolts 1.5” x 4” long

4 Hex bolts 1.875” x 4” long

1 Hex bolt 2.25” x 2” long

6 Hex bolts 3.375” x 3” long

4 Hex bolts 3.75” x 3” long

g(θ) = 20θ − 5 tan θ

To find out critical points we take first derivative and set it =0

g(θ) = 20θ − 5 tan θ

g'(θ) = 20 − 5 sec^2(θ)

Now we set derivative =0

20 − 5 sec^2(θ)=0

Subtract 20 from both sides

− 5 sec^2(θ)=0 -20

Divide both sides by 5

sec^2(θ)= 4

Take square root on both sides

sec(θ)= -2  and sec(θ)= +2

sec can be written as 1/cos

so sec(θ)= -2  can be written as cos(θ)= -1/2

Using unit circle the value of θ is [tex]\frac{2\pi}{3} and\frac{4\pi}{3}[/tex]

sec(θ)= 2  can be written as cos(θ)=1/2

Using unit circle the value of θ is [tex]\frac{\pi}{3}and\frac{5\pi}{3}[/tex]

For general solution we add 2npi

So critical points are

[tex]\frac{2\pi}{3} + 2n\pi,\frac{4\pi}{3} + 2n\pi , \frac{\pi}{3} + 2n\pi, \frac{5\pi}{3} + 2n\pi[/tex]

The critical points of a function are where its derivative is zero or undefined. The critical points for the function g(θ) = 20θ − 5 tan θ can be found by taking its derivative, setting it to zero, and solving for θ. Additional critical points may occur when the derivative does not exist.

In mathematics, the critical points of a function are those where its derivative is zero or undefined. We can find the critical points for the function g(θ) = 20θ − 5 tan θ by taking its derivative and setting it equal to zero.

To find these points, we differentiate the function g(θ) to obtain g'(θ) = 20 - 5/(cos^2 θ). Setting this equal to zero and solving for θ will provide the critical points. In this case, the function is undefined whenever cos θ=0 , providing additional potential critical points. Thus, with trigonometric functions like the tangent function in this question, the critical points occur when the derivative is zero or the derivative does not exist.

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The area of the stretched triangle is four times the area of the original triangle

A triangle is a plane figure or polygon with three sides and three angles.

A Triangle has three vertices and the sum of the interior angles add up to 180°

Let the Triangle be ΔABC , such that

∠A + ∠B + ∠C = 180°

The area of the triangle = ( 1/2 ) x Length x Base

For a right angle triangle

From the Pythagoras Theorem , The hypotenuse² = base² + height²

if a² + b² = c² , it is a right triangle

if a² + b² < c² , it is an obtuse triangle

if a² + b² > c² , it is an acute triangle

Given data ,

Let's assume that the original triangle has sides of length a, b, and c, and that its area is A. Without loss of generality, we can assume that side a is the shortest side.

After the rubber is stretched, each side of the enlarged triangle will be twice the length of its original length. Thus, the lengths of the sides of the stretched triangle will be 2a, 2b, and 2c.

Since the rubber is stretched equally from all sides, we know that each angle of the original triangle will be doubled. Therefore, the stretched triangle will be similar to the original triangle.

The area of a triangle is given by the formula:

A = (1/2) x base x height

where the base and height are the lengths of any two sides of the triangle that are perpendicular to each other.

Since the stretched triangle is similar to the original triangle, we can use the ratios of corresponding sides to find the lengths of the corresponding altitudes. Specifically, we have:

(altitude of stretched triangle corresponding to side 2a) = 2 x (altitude of original triangle corresponding to side a)

(altitude of stretched triangle corresponding to side 2b) = 2 x (altitude of original triangle corresponding to side b)

(altitude of stretched triangle corresponding to side 2c) = 2 x (altitude of original triangle corresponding to side c)

Therefore, the area of the stretched triangle is given by:

A' = (1/2) x (2a) x (2b) x (2/2) + (1/2) x (2b) x (2c) x (2/2) + (1/2) x (2c) x (2a) x (2/2)

Simplifying this expression gives:

A' = 4A

Hence , the area of the stretched triangle is four times the area of the original triangle.

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The equation that matches the information provided is:

In this equation, we have two variables: s represents the number of small tables, and t represents the number of large tables.

The equation states that the number of small tables (s) is equal to two times the number of large tables (t). In other words, for every large table, there are two small tables.

For example, if there are 3 large tables (t = 3), then according to the equation, there would be 2 times 3 small tables, which is 6 small tables (s = 6).

This equation captures the relationship between the number of small tables and large tables in the conference room, where the number of small tables is always double the number of large tables.

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The answer is 8.608 because you would just divide 860800 by 100k to get this answer

Answer:

-5.76x=-27.33

-6x=-27

Step-by-step explanation:

you would bring the 3.98 over to the other side, so it would be -3.98+6.58=2.6-8.39=-5.76

One of these doesn't seem to be required information for a feasibility region. Are you sure that these are the correct equations?

They will need 11 yards of rope

Answer:

Step-by-step explanation:

The second 1 is eight hundred and ninedey seven then it’s 899+90+70

Divide total miles driven by the total amount of gas used to find the number of miles per gallon.

Then multiply that by the number of gallons you want to find out:

268 miles / 12 gallons = 22.333 miles per gallon.

22.333 miles per gallon x 9 gallons = 201 total miles.

-2(6-x) is the remainder.

Answer:

Step-by-step explanation:

Let's first find out how many workstations there would be.

It says in the problem that there are two classrooms and that they both have the same number of workstations.

There are seven workstations in each of the two classrooms.

so 50 pages and 2 stamps in one page

The dimensions of the garden, based on the given condition and the total perimeter of 132 meters, are determined to be 46 meters in length and 20 meters in width.

In order to find the dimensions of the garden, you have to solve two equations. The first equation comes from the definition of the perimeter of a rectangle, which is 2*length + 2*width = perimeter, so 2*length + 2*width = 132 meters. The second equation comes from the given information that the length is 6 meters longer than twice the width so length = 2*width + 6.

Now you can use substitution to solve these equations. Plug (2*width + 6) in for length in the perimeter equation and get: 2*(2*width + 6) + 2*width = 132. Simplifying this gives 4*width +12 + 2*width = 132, which further simplifies to 6*width + 12 = 132. Solving for width gives width = 20 meters.

Substituting width = 20 meters into the equation for length gives length = 2*20 + 6 = 46 meters. So the dimension of the garden is 46 meters by 20 meters.

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.475 because .4 and .4 are equal but then you get too .47 and .44 which would make .475 larger.

The answer to your question is .44

Hopes this helps.

; P

(5,1)

x=5

y=1

m=1/2

Use this formula

y-y1=m(x-x1)

y-1=1/2(x-5)

y-1=x/2-5/2

Isolate for y

y=x/2-3/2

Hope this helps.

Answer:x-2y=3

Step-by-step explanation:

Given

Lines passes  through (5,1) and slope of 0.5

i.e. [tex] tan\theta =0.5[/tex]

where [tex]\theta [/tex]is the angle made by line with horizontal

equation of line can be given by

[tex]\frac{y-1}{x-5}=tan\theta [/tex]

2y-2=x-5

x-2y=3 is the required line

so lets say base = x

Then 1*x=2.5

Therefore the base equals 2.5

Hope this helps!

i think you add 232 and 184 then divide by 13.

Hope this helps!

It would go under Utilities Expense (right column, second from the bottom) as a Debit

√70 = 8.37

It would be between 8 and 9.

1) 4/5

2) -62/13

3) 1/9

4) 527/42

I only answered these four questions.

Hope this helps. By the way they are all in the exact form.

To solve the multiplication problem and simplify the fractions, follow these steps:

Multiply 32/5 × 513/1:

(32/5) × (513/1) = (32 × 513) / (5 × 1) = 16416 / 5

Now, let's find how many novels Jeffrey owns. He owns 230 books, and of them are novels:

Number of novels = (Number of novels) / Total books

Number of novels = (3/9) × 230

Number of novels = (1/3) × 230

Number of novels = 230 / 3

Number of novels = 76⅔ (mixed number)

To simplify the mixed number 76⅔, remember that ⅔ is equivalent to 2/3.

76⅔ = 76 + ⅔ = 76 + 2/3 = (76 × 3 + 2) / 3 = 230 / 3

So, Jeffrey owns 76⅔ novels.

For the standardized test practice, we have two fractions:

Fraction of students wearing school colors = 1/4

Fraction of students wearing school colors in the 6th grade = 3/20

To find the fraction of students wearing school colors in the 6th grade, we can multiply these fractions:

Fraction = (1/4) × (3/20) = 3/80

So, the fraction of students at the school dance who wore the school colors and were in the 6th grade is 3/80.

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