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2 years ago

Question 2 (2 points)

Mathematics

1 answer:

Sati [7]2 years ago

5 0

Answer:

Step-by-step explanation:

this question has already been answered on gauth math

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What’s the perimeter of this rectangle 2,5+D and 4-1,3D

koban [17]

Answer:

The result can be shown in multiple forms.

Exact Form:

2,5+D,4-1,3D

Decimal form;

2,5+D,3,3D

Step-by-step explanation:

To find the perimeter of a rectangle, add the lengths of the rectangle's four sides. If you have only the width and the height, then you can easily find all four sides (two sides are each equal to the height and the other two sides are equal to the width). Multiply both the height and width by two and add the results.

6 0

3 years ago

How much of the circle is shaded? Write your answer as a fraction in simplest form. 1/3 and 1/7

Ludmilka [50]

Answer:

11/21

Step-by-step explanation:

1/3=7/21

1/7=3/21

(7+3)/21=10/21

21/21-10/21=11/21

Thus 11/21 of the circle is shaded.

6 0

3 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.