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Ulleksa [173]
3 years ago
8

How Can you represent writing a check for $10 as an integer?A b c or d

Mathematics
1 answer:
Lynna [10]3 years ago
6 0
The answer to represent a check for 10$ is A. -10
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A polygon is reflected across a line. The angles in the original figure are compared to the corresponding angles in the image to
irina [24]
None. <span>They will always be congruent.</span>
5 0
3 years ago
Read 2 more answers
A road repair crew spread 7/8 of a ton of gravel evenly over 7 feet of road. How many tons of gravel did they spread on each foo
Alex
7/8 duh it there u just have to read it nicely
5 0
3 years ago
Find the coordinates of the endpoint of the image?
ra1l [238]

Given:

The graph of a line segment.

The line segment AB translated by the following rule:

(x,y)\to (x+4,y-3)

To find:

The coordinates of the end points of the line segment A'B'.

Solution:

From the given figure, it is clear that the end points of the line segment AB are A(-2,-3) and B(4,-1).

We have,

(x,y)\to (x+4,y-3)

Using this rule, we get

A(-2,-3)\to A'(-2+4,-3-3)

A(-2,-3)\to A'(2,-6)

Similarly,

B(4,-1)\to B'(4+4,-1-3)

B(4,-1)\to B'(8,-4)

Therefore, the endpoint of the line segment A'B' are A'(2,-6) and B'(8,-4).

5 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.

4 0
3 years ago
Help asap with Arcs! 30 Points
alexira [117]
The answer is 15

these chords are congruent because the radius of the circle are perpendicular to the them
so we can find EF by making an equation:
3x + 5 = x + 35 \\ 3x - x = 35 - 5 \\ 2x = 30 \\ x = \frac{30}{2} = 15

so EF is = 3(15)+5=50
good luck
6 0
3 years ago
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