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3 years ago

Draw a picture to show 9/7

2 answers:

5 0

You could draw a box & divide the box into 7 sections and color the whole box in.Then draw an other box & divide that into 7 as well. Lastly color 2of the 7 sections

maksim [4K]3 years ago

3 0

9
_
7 or nine sevenths

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ABC Bookstore uses a storage unit that measures 13 ft x 16 ft x 12 ft. What is

Ludmilka [50]

Answer:

2496 cubic feet

Step-by-step explanation:

Dimension of storage unit 13 ft x 16 ft x 12 ft.

Since there are three sides it can be safely assumed that storage unit is rectangular in shape.

Volume of rectangular body is given by length * width * height

Using the above formula in the given dimension we have

volume of storage unit = length of storage unit * width of storage unit * height of storage unit

volume of storage unit = 13 ft * 16 ft * 12 ft = 2,496 cubic feet

5 0

3 years ago

Customers of a phone company can choose between two service plans for long distance calls. The first plan has a $28 monthly fee

aniked [119]

Answer:

POINTS

Step-by-step explanation:

7 0

10 months ago

Cuanto es (2x+2y=20) (-2x-6y=-52) con método de igualación y método de suma y resta

Nataly_w [17]

x =2 y =8

2x + 2y=20

2x=20-2y

x=10-y

Ahora len la otra ecuacion remplazamos x por (10-y)

-2(10-y)-6y=-52

-20+2y-6y=-52

2y-6y= -32 (-52+20= -32)

-4y=-32

-y=-8

y=8 x=10-y x=2

2*2+2*8=20

4+16=20

-2*2-6*8= -52

-4-48= -52

6 0

3 years ago

Check the picture please

slavikrds [6]

Answer:

ez no picture!!!!

Step-by-step explanation:

7 0

2 years ago

13. The least common multiple of two non-zero integers a and b is the unique positive integer m such that (i) m is a common mult

Vlad [161]

Answer:

[a,b] divides n

Step-by-step explanation:

Let us denote the least common multiple of a and b [a,b]=m.

We want to prove that m divides n, where n is a multiple of a and b.

We suppose m does not divide n, then by the Division Theorem, there exists q and r integers such that:

(1) ... n=mq+r, where 0<r<m

As n is a multiple of a and b, there exists s and t integers such that:

sa=n and tb=n

Same thing happens to m as it is the least common multiple, there exists u and v such that:

ua=m and vb=m

So (1) has the following form:

n=mq+r ⇒ sa=uaq+r ⇒sa-uaq=r⇒(s-uq)a=r and

n=mq+r ⇒ tb=vbq+r ⇒ tb-vbq=r⇒ (t-vq)b=r

So r is a multiple of a and b, but r<m which is a contradiction as, m is the least common multiple of a and b. So this concludes the proof.

So this means that $\frac{ab}{m}$ is and integer.

As m= vb, then $\frac{m}{b}$ is an integer, lets say $\frac{m}{b}$ =v; and as m=ua, then $\frac{m}{a}$ =u.

So $\frac{ab}{m}$ v= $\frac{ab}{m}$ $\frac{m}{b}$ =a, so $\frac{ab}{m}$ divides a; on the other hand, $\frac{ab}{m}$ u= $\frac{ab}{m}$ $\frac{m}{a}$ =b, so $\frac{ab}{m}$ divides b. From this we can conclude that $\frac{ab}{m}$ is a common divisor of a and b.

4 0

3 years ago