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2 years ago

6

a box holds nine smaller boxes. each of the smaller boxes holds nine plastic containers. each plastic contai.er holds nine bags.

each bag holds nine elastic bands. how many ekastic bands are in the laregrst bix?

1 answer:

svp [43]2 years ago

4 0

So, basically ..you just multiply
9*9*9*9
6561

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Division of two quantities is expressed as the ____ of those two quantities.

andreev551 [17]

Division of two quantities is expressed as the quotient of those two quantities.

The word quotient is derived from the Latin language. It is from the Latin word "quotiens" which means "how many times." A quotient is the answer to a divisional problem. A divisional problem describes how many times a number will go into another. The first time that this word was known to have been used in mathematics was around 1400 - 1500 AD in England.

There are two different ways to find the quotient of two numbers. One of them is through Fractions. The quotient of a fraction is the number obtained when the fraction is simplified. The other way to find a quotient is by employing the long division method where the quotient value is positioned above the divisor and dividend.

6 0

2 years ago

Calculate the value of x to one decimal place. inches

kakasveta [241]

Use the law of cosines: $c^2=a^2+b^2-2ab\cos C$

We have:

$c=x\\a=3in\\b=7in\\C=54^o$

substitute:

$\cos54^o\approx0.5878\\\\c^2=3^2+7^2-2\cdot3\cdot7\cdot0.5878\\\\c^2=9+49-24.6876\\\\c^2=33.3129\to c=\sqrt{33.3129}\\\\c\approx5.8$

5 0

2 years ago

Find the mass and center of mass of the lamina that occupies the region D and has the given density function rho. D = {(x, y) |

Bas_tet [7]

Answer:

$M=168k$

$(\bar{x},\bar{y})=(5,\frac{85}{28})$

Step-by-step explanation:

Let's begin with the mass definition in terms of density.

$M=\int\int \rho dA$

Now, we know the limits of the integrals of x and y, and also know that ρ = ky², so we will have:

$M=\int^{9}_{1}\int^{4}_{1}ky^{2} dydx$

Let's solve this integral:

$M=k\int^{9}_{1}\frac{y^{3}}{3}|^{4}_{1}dx$

$M=k\int^{9}_{1}\frac{y^{3}}{3}|^{4}_{1}dx$

$M=k\int^{9}_{1}21dx$

$M=21k\int^{9}_{1}dx=21k*x|^{9}_{1}$

So the mass will be:

$M=21k*8=168k$

Now we need to find the x-coordinate of the center of mass.

$\bar{x}=\frac{1}{M}\int\int x*\rho dydx$

$\bar{x}=\frac{1}{M}\int^{9}_{1}\int^{4}_{1}x*ky^{2} dydx$

$\bar{x}=\frac{k}{168k}\int^{9}_{1}\int^{4}_{1}x*y^{2} dydx$

$\bar{x}=\frac{1}{168}\int^{9}_{1}x*\frac{y^{3}}{3}|^{4}_{1}dx$

$\bar{x}=\frac{1}{168}\int^{9}_{1}x*21 dx$

$\bar{x}=\frac{21}{168}\frac{x^{2}}{2}|^{9}_{1}$

$\bar{x}=\frac{21}{168}*40=5$

Now we need to find the y-coordinate of the center of mass.

$\bar{y}=\frac{1}{M}\int\int y*\rho dydx$

$\bar{y}=\frac{1}{M}\int^{9}_{1}\int^{4}_{1}y*ky^{2} dydx$

$\bar{y}=\frac{k}{168k}\int^{9}_{1}\int^{4}_{1}y^{3} dydx$

$\bar{y}=\frac{1}{168}\int^{9}_{1}\frac{y^{4}}{4}|^{4}_{1}dx$

$\bar{y}=\frac{1}{168}\int^{9}_{1}\frac{255}{4}dx$

$\bar{y}=\frac{255}{672}\int^{9}_{1}dx$

$\bar{y}=\frac{255}{672}8=\frac{2040}{672}$

$\bar{y}=\frac{85}{28}$

Therefore the center of mass is:

$(\bar{x},\bar{y})=(5,\frac{85}{28})$

I hope it helps you!

3 0

2 years ago

Jackson has ten dollar bills and one dollar bills in his pocket. The number of ten dollar bills is five more than six times the

Katyanochek1 [597]

Answer:

Let t be the number of ten dollar bills.

o×6+5=t

If you need it without t, then it's:

o×6+5

Step-by-step explanation:

Let t be the number of ten dollar bills.

o×6+5=t

We get this because the question said that there were more ten-dollar bills than 6 times the number of one-dollar bills.

5 0

1 year ago

I am a factor of 18.the other factor is 9.what number am i?

Liula [17]

The answer to this question is 2

3 0

1 year ago